<?xml version='1.0' encoding='UTF-8'?><?xml-stylesheet href="http://www.blogger.com/styles/atom.css" type="text/css"?><feed xmlns='http://www.w3.org/2005/Atom' xmlns:openSearch='http://a9.com/-/spec/opensearchrss/1.0/' xmlns:blogger='http://schemas.google.com/blogger/2008' xmlns:georss='http://www.georss.org/georss' xmlns:gd="http://schemas.google.com/g/2005" xmlns:thr='http://purl.org/syndication/thread/1.0'><id>tag:blogger.com,1999:blog-5244458674117363159</id><updated>2024-10-03T09:31:07.548-07:00</updated><category term="Math GRE"/><category term="Physics GRE"/><title type='text'>Daily GRE</title><subtitle type='html'>Two GRE questions a day. One Math. One Physics.</subtitle><link rel='http://schemas.google.com/g/2005#feed' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/posts/default'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default?redirect=false'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/'/><link rel='hub' href='http://pubsubhubbub.appspot.com/'/><link rel='next' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default?start-index=26&amp;max-results=25&amp;redirect=false'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><generator version='7.00' uri='http://www.blogger.com'>Blogger</generator><openSearch:totalResults>70</openSearch:totalResults><openSearch:startIndex>1</openSearch:startIndex><openSearch:itemsPerPage>25</openSearch:itemsPerPage><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-2984017844797252225</id><published>2011-09-05T21:08:00.000-07:00</published><updated>2011-09-05T21:11:32.578-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Math GRE"/><category scheme="http://www.blogger.com/atom/ns#" term="Physics GRE"/><title type='text'>Zai geen, sayonara, goodbye (for now at least).</title><content type='html'>&lt;br /&gt;
With the start of school looming so close, I fear that I won&#39;t have enough time to update this blog daily. Thus, I have decided to take a break on posting questions &#39;til winter vacation comes. I&#39;ll still be in the background gathering questions, but they won&#39;t be released to the public until December.&lt;br /&gt;
&lt;br /&gt;
Thanks for all the comments, you guys really taught me a lot. I hope I was able to help all of you too (even if it was just the slightest bit).&lt;br /&gt;
&lt;br /&gt;
- Paul&lt;br /&gt;
&lt;br /&gt;
&lt;span class=&quot;Apple-style-span&quot; style=&quot;color: white;&quot;&gt;PS: The title of this post is from a song my friend wrote. His alias is LittleColumbus. This text is in white so that I can properly attribute him while not turning this post into a bad advertisement.&lt;/span&gt;&lt;br /&gt;
&lt;span class=&quot;Apple-style-span&quot; style=&quot;color: white;&quot;&gt;PPS: It would be awesome if you guys spread the word about this blog. I&#39;d love to get more people commenting so we can have more discussions.&lt;/span&gt;&lt;br /&gt;
&lt;span class=&quot;Apple-style-span&quot; style=&quot;color: white;&quot;&gt;&lt;br /&gt;&lt;/span&gt;</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/2984017844797252225/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/09/zai-geen-sayonara-goodbye-for-now-at.html#comment-form' title='5 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/2984017844797252225'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/2984017844797252225'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/09/zai-geen-sayonara-goodbye-for-now-at.html' title='Zai geen, sayonara, goodbye (for now at least).'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><thr:total>5</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-3826647488649221106</id><published>2011-09-05T20:53:00.001-07:00</published><updated>2011-09-05T20:53:55.743-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Math GRE"/><title type='text'>Math GRE - #34</title><content type='html'>$f$ is the function defined by: \[
f(x)=\begin{cases}
xe^{-x^{2}-x^{-2}} &amp;amp; \text{if } x\neq0\\
0 &amp;amp; \text{otherwise.}\end{cases}\]

At how many values of $x$ does $f$ have a horizontal tangent line?

&lt;br /&gt;
&lt;br /&gt;
&lt;ol&gt;
&lt;li&gt;None&lt;/li&gt;
&lt;li&gt;One&lt;/li&gt;
&lt;li&gt;Two&lt;/li&gt;
&lt;li&gt;Three&lt;/li&gt;
&lt;li&gt;Four&lt;/li&gt;
&lt;/ol&gt;
&lt;br /&gt;
&lt;div style=&quot;margin: 5px 20px 20px;&quot;&gt;
&lt;div class=&quot;smallfont&quot; style=&quot;margin-bottom: 2px;&quot;&gt;
Solution : &lt;input onclick=&quot;if (this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display != &#39;&#39;) { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;&#39;; this.innerText = &#39;&#39;; this.value = &#39;Hide&#39;; } else { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;none&#39;; this.innerText = &#39;&#39;; this.value = &#39;Show&#39;; }&quot; style=&quot;font-size: 10px; margin: 0px; padding: 0px; width: 60px;&quot; type=&quot;button&quot; value=&quot;Show&quot; /&gt; &lt;/div&gt;
&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;
&lt;div style=&quot;display: none;&quot;&gt;
&lt;b&gt;Choice 4 is the answer.&lt;/b&gt;&lt;br /&gt;
&lt;br /&gt;
$f$ has a horizontal tangent line only when the derivative of $x$ is 0. The derivative of $f$ is:&lt;br /&gt;
\[f^\prime(x) = e^{-x^2-x^{-2}}+(-2x+2x^{-3})xe^{-x^2-x^{-2}}=(1-2x^2+2x^{-2})e^{-x^2-x^{-2}}=0.\] This is only zero in two cases: in the limit $\displaystyle\lim_{x\rightarrow 0}&amp;nbsp;f^\prime(x)$ and when $1-2x^2+2x^{-2}=0$.&lt;br /&gt;
&lt;br /&gt;
The quadratic-like equation above can be solved by solving $x^2-2x^4+2=0.$ This equation has two non-zero solutions. These solutions, along with the with the solution obtained in the limit above, gives us three horizontal tangent lines in total.&lt;br /&gt;
&lt;br /&gt;&lt;/div&gt;
&lt;/div&gt;
&lt;/div&gt;
</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/3826647488649221106/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/09/math-gre-34.html#comment-form' title='2 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/3826647488649221106'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/3826647488649221106'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/09/math-gre-34.html' title='Math GRE - #34'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><thr:total>2</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-3991167029997850066</id><published>2011-09-05T20:53:00.000-07:00</published><updated>2011-09-05T20:53:51.727-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Physics GRE"/><title type='text'>Physics GRE - # 34</title><content type='html'>Eigenfunctions for a rigid dumbbell rotating about its centre have a $\phi$ dependence of the form $\psi(\phi)=Ae^{im\phi}$, where $m$ is a quantum number and $A$ is a constant. Which of the following values of $A$ will properly normalize the eigenfunction?&lt;br /&gt;
&lt;br /&gt;
&lt;ol&gt;
&lt;li&gt;$\sqrt{2\pi}$&lt;/li&gt;
&lt;br /&gt;
&lt;li&gt;$2\pi$&lt;/li&gt;
&lt;br /&gt;
&lt;li&gt;$(2\pi)^2$&lt;/li&gt;
&lt;br /&gt;
&lt;li&gt;$\dfrac{1}{\sqrt{2\pi}}$&lt;/li&gt;
&lt;br /&gt;
&lt;li&gt;$\dfrac{1}{2\pi}$&lt;/li&gt;
&lt;/ol&gt;
&lt;br /&gt;
&lt;div style=&quot;margin: 5px 20px 20px;&quot;&gt;
&lt;div class=&quot;smallfont&quot; style=&quot;margin-bottom: 2px;&quot;&gt;
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&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;
&lt;div style=&quot;display: none;&quot;&gt;
&lt;b&gt;Choice 4 is the answer.&lt;/b&gt;&lt;br /&gt;
&lt;br /&gt;
For $\psi(\phi)$ to be normalized on $a$ to $b$, recall the normalization condition: \[\int_a^b{\left|\psi(\phi)\right|^2}d\phi=1.\] Since $\phi$ ranges from $0$ to $2\pi$, we have:&amp;nbsp;\[\int_0^{2\pi}{\left|\psi(\phi)\right|^2}d\phi=\int_0^{2\pi}{\left|Ae^{im\phi}\right|^2}d\phi=\int_0^{2\pi}{|A|^2}d\phi=1\] where the second equality holds as $|e^{im\phi}|=1$ for any $m$ and any $\phi$.&lt;br /&gt;
&lt;br /&gt;
Thus, we know that: \[\int_0^{2\pi}{|A|^2}d\phi=2\pi|A|^2=1\implies |A|=\dfrac{1}{\sqrt{2\pi}}.\]&lt;br /&gt;&lt;/div&gt;
&lt;/div&gt;
&lt;/div&gt;
</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/3991167029997850066/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/09/physics-gre-34.html#comment-form' title='2 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/3991167029997850066'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/3991167029997850066'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/09/physics-gre-34.html' title='Physics GRE - # 34'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><thr:total>2</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-2744933952941354006</id><published>2011-09-04T21:43:00.001-07:00</published><updated>2011-09-04T21:43:58.700-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Math GRE"/><title type='text'>Math GRE - #33</title><content type='html'>How many integers from 1 to 1000 are divisible by 30 by not by 16?&lt;br /&gt;
&lt;br /&gt;
&lt;ol&gt;
&lt;li&gt;29&lt;/li&gt;
&lt;li&gt;31&lt;/li&gt;
&lt;li&gt;32&lt;/li&gt;
&lt;li&gt;33&lt;/li&gt;
&lt;li&gt;38&lt;/li&gt;
&lt;/ol&gt;
&lt;br /&gt;
&lt;div style=&quot;margin: 5px 20px 20px;&quot;&gt;
&lt;div class=&quot;smallfont&quot; style=&quot;margin-bottom: 2px;&quot;&gt;
Solution : &lt;input onclick=&quot;if (this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display != &#39;&#39;) { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;&#39;; this.innerText = &#39;&#39;; this.value = &#39;Hide&#39;; } else { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;none&#39;; this.innerText = &#39;&#39;; this.value = &#39;Show&#39;; }&quot; style=&quot;font-size: 10px; margin: 0px; padding: 0px; width: 60px;&quot; type=&quot;button&quot; value=&quot;Show&quot; /&gt; &lt;/div&gt;
&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;
&lt;div style=&quot;display: none;&quot;&gt;
&lt;b&gt;Choice 1 is the answer.&lt;/b&gt;&lt;br /&gt;
&lt;br /&gt;
There are 33 multiples of 30 under 1000 (1000/30 = 33.33..). So we only need to focus our attention on choices 1 to 4.&lt;br /&gt;
&lt;br /&gt;
Note that the factorization of 30 is: $2\cdot 3\cdot 5.$&lt;br /&gt;
Now lets consider multiples of 30 in the form: $30n$. For $30n$ to be divisible by 16, $n$ must be divisible by 8 (since we will have a factor of 2 from 30 and another factor of 8 from $n$ to make a factor of 16).&lt;br /&gt;
Since $n$ can only go from 1 to 33 (as there are only 33 multiples of 30 under 1000), we only need to find numbers under 33 that are divisible by 8.&lt;br /&gt;
&lt;br /&gt;
There are four such numbers: 8, 16, 24, 32. These numbers correspond to the multiples: 240, 480, 720, 960.&lt;br /&gt;
&lt;br /&gt;
All the multiples remaining are not divisible by 16. Thus there are 33 - 4 = 29 multiples of 30 under 1000 that are divisible by 30 but not 16.&lt;/div&gt;
&lt;/div&gt;
&lt;/div&gt;
</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/2744933952941354006/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/09/math-gre-33.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/2744933952941354006'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/2744933952941354006'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/09/math-gre-33.html' title='Math GRE - #33'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-3622182345809571987</id><published>2011-09-04T21:43:00.000-07:00</published><updated>2011-09-04T21:43:54.957-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Physics GRE"/><title type='text'>Physics GRE - #33</title><content type='html'>A radioactive nucleus decays, with the activity shown in the plot below. The half-life of the nucleus is:&lt;br /&gt;
&lt;br /&gt;
&lt;div class=&quot;separator&quot; style=&quot;clear: both; text-align: center;&quot;&gt;
&lt;a href=&quot;https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhVJuOG2SyU1Rx7vuk3rYftQ1Y547P3UcXSsfjhAZJwHAZrKNMkWDLJUNY5UGMNs9RoUoRF538CswQE3zyBu04Lf2vSTfp_GN8y6KVDKBbMgFu4ffgykVfQJNhb4DVpZl4hE-_fIA0V-gFk/s1600/DecayPlot.png&quot; imageanchor=&quot;1&quot; style=&quot;margin-left: 1em; margin-right: 1em;&quot;&gt;&lt;img border=&quot;0&quot; height=&quot;395&quot; src=&quot;https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhVJuOG2SyU1Rx7vuk3rYftQ1Y547P3UcXSsfjhAZJwHAZrKNMkWDLJUNY5UGMNs9RoUoRF538CswQE3zyBu04Lf2vSTfp_GN8y6KVDKBbMgFu4ffgykVfQJNhb4DVpZl4hE-_fIA0V-gFk/s400/DecayPlot.png&quot; width=&quot;400&quot; /&gt;&lt;/a&gt;&lt;/div&gt;
&lt;br /&gt;
&lt;ol&gt;
&lt;li&gt;2 min&lt;/li&gt;
&lt;li&gt;7 min&lt;/li&gt;
&lt;li&gt;11 min&lt;/li&gt;
&lt;li&gt;18 min&lt;/li&gt;
&lt;li&gt;23 min&lt;/li&gt;
&lt;/ol&gt;
&lt;br /&gt;
&lt;div style=&quot;margin: 5px 20px 20px;&quot;&gt;
&lt;div class=&quot;smallfont&quot; style=&quot;margin-bottom: 2px;&quot;&gt;
Solution : &lt;input onclick=&quot;if (this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display != &#39;&#39;) { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;&#39;; this.innerText = &#39;&#39;; this.value = &#39;Hide&#39;; } else { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;none&#39;; this.innerText = &#39;&#39;; this.value = &#39;Show&#39;; }&quot; style=&quot;font-size: 10px; margin: 0px; padding: 0px; width: 60px;&quot; type=&quot;button&quot; value=&quot;Show&quot; /&gt; &lt;/div&gt;
&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;
&lt;div style=&quot;display: none;&quot;&gt;
&lt;b&gt;Choice 2 is the answer.&lt;/b&gt;&lt;br /&gt;
&lt;br /&gt;
The initial activity is $6\cdot 10^3$ (yes, 6 since the line marked $10^3$ is really $1\cdot 10^3$ and you start counting up from there). At $3\cdot 10^3,$ the time is around 5 to 10 minutes. Thus the only possible answer is choice 2.&lt;br /&gt;&lt;/div&gt;
&lt;/div&gt;
&lt;/div&gt;
</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/3622182345809571987/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/09/physics-gre-33.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/3622182345809571987'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/3622182345809571987'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/09/physics-gre-33.html' title='Physics GRE - #33'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><media:thumbnail xmlns:media="http://search.yahoo.com/mrss/" url="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhVJuOG2SyU1Rx7vuk3rYftQ1Y547P3UcXSsfjhAZJwHAZrKNMkWDLJUNY5UGMNs9RoUoRF538CswQE3zyBu04Lf2vSTfp_GN8y6KVDKBbMgFu4ffgykVfQJNhb4DVpZl4hE-_fIA0V-gFk/s72-c/DecayPlot.png" height="72" width="72"/><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-856432370211991440</id><published>2011-09-03T19:57:00.001-07:00</published><updated>2011-09-03T19:57:48.640-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Math GRE"/><title type='text'>Math GRE - #32</title><content type='html'>Let $\mathbb{R}$ be the set of real numbers and let $f$ and $g$ be functions from $\mathbb{R}$ into $\mathbb{R}$. Which of the following is the negation of the statement:&lt;br /&gt;
&lt;br /&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;
&quot;For each $s$ in&amp;nbsp;$\mathbb{R}$, there exists an $r$ in&amp;nbsp;$\mathbb{R}$ such that if $f(r)&amp;gt;0$, then $g(s)&amp;gt;0$.&quot;&lt;/div&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;
&lt;br /&gt;&lt;/div&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;
&lt;/div&gt;
&lt;ol&gt;
&lt;li style=&quot;text-align: left;&quot;&gt;For each $s$ in&amp;nbsp;$\mathbb{R}$, there does not exist an $r$ in&amp;nbsp;$\mathbb{R}$ such that if $f(r)&amp;gt;0$, then $g(s)&amp;gt;0$.&lt;/li&gt;
&lt;li style=&quot;text-align: left;&quot;&gt;For each $s$ in&amp;nbsp;$\mathbb{R}$, there exists an $r$ in&amp;nbsp;$\mathbb{R}$ such that $f(r)&amp;gt;0$ and $g(s)\leq0$.&lt;/li&gt;
&lt;li style=&quot;text-align: left;&quot;&gt;There exists an $s$ in&amp;nbsp;$\mathbb{R}$ such that for each $r$ in&amp;nbsp;$\mathbb{R}$ such that $f(r)&amp;gt;0$ and $g(s)\leq0$.&lt;/li&gt;
&lt;li style=&quot;text-align: left;&quot;&gt;There exists an $s$ in&amp;nbsp;$\mathbb{R}$ and there exists an $r$ in&amp;nbsp;$\mathbb{R}$ such that $f(r)\leq0$ and $g(s)\leq0$.&lt;/li&gt;
&lt;li style=&quot;text-align: left;&quot;&gt;For each $r$ in&amp;nbsp;$\mathbb{R}$, there exists an $s$ in&amp;nbsp;$\mathbb{R}$ such that $f(r)\leq0$ and $g(s)\leq0$.&lt;/li&gt;
&lt;/ol&gt;
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&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;
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&lt;br /&gt;&lt;/div&gt;
&lt;div style=&quot;text-align: left;&quot;&gt;
&lt;b&gt;Choice 3 is the answer.&lt;/b&gt;&lt;/div&gt;
&lt;div style=&quot;text-align: left;&quot;&gt;
&lt;b&gt;&lt;br /&gt;&lt;/b&gt;&lt;/div&gt;
&lt;div style=&quot;text-align: left;&quot;&gt;
If you&#39;ve got an innate talent for logic, you could probably tell that choice 3 is the answer after a bit of thinking.&amp;nbsp;&lt;/div&gt;
&lt;div style=&quot;text-align: left;&quot;&gt;
&lt;br /&gt;&lt;/div&gt;
&lt;div style=&quot;text-align: left;&quot;&gt;
Another way is to use quantifiers to translate the statement into formal logic.&lt;/div&gt;
&lt;div style=&quot;text-align: left;&quot;&gt;
The statement:&lt;/div&gt;
&lt;div style=&quot;text-align: left;&quot;&gt;
&lt;br /&gt;&lt;/div&gt;
&lt;div style=&quot;text-align: left;&quot;&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;
&quot;For each $s$ in&amp;nbsp;$\mathbb{R}$, there exists an $r$ in&amp;nbsp;$\mathbb{R}$ such that if $f(r)&amp;gt;0$, then $g(s)&amp;gt;0$.&quot;&lt;/div&gt;
&lt;div&gt;
&lt;br /&gt;&lt;/div&gt;
&lt;div&gt;
can be translated as: \[\forall s\in\mathbb{R},\;\exists r\in\mathbb{R},\; f(r)&amp;gt;0\rightarrow g(s)&amp;gt;0.\]&lt;/div&gt;
&lt;div&gt;
&lt;br /&gt;&lt;/div&gt;
&lt;div&gt;
We can negate this statement (imagine the negation as an arrow, flipping each quantifier as it flies past it) to be: \[\exists s\in\mathbb{R},\;\forall r\in\mathbb{R},\; f(r)&amp;gt;0\wedge g(s)\leq0.\]&lt;/div&gt;
&lt;div&gt;
&lt;br /&gt;&lt;/div&gt;
&lt;div&gt;
This is choice 3 precisely.&lt;/div&gt;
&lt;/div&gt;
&lt;/div&gt;
&lt;/div&gt;
&lt;/div&gt;
</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/856432370211991440/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/09/math-gre-32.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/856432370211991440'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/856432370211991440'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/09/math-gre-32.html' title='Math GRE - #32'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-8741318480314334133</id><published>2011-09-03T19:57:00.000-07:00</published><updated>2011-09-03T19:57:42.605-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Physics GRE"/><title type='text'>Physics GRE - #32</title><content type='html'>As shown below, a coaxial cable having radii $a$, $b$, and $c$ carries equal and opposite currents of magnitude $i$ on the inner and outer conductors. What is the magnitude of the magnetic induction at a point $P$ outside the cable at a distance $r$ from the axis?

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&lt;a href=&quot;https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgVDVQZZP9i54FA1QaLrhg6c5mvaziaQRUuMBdgIGIJJhDpuaUm_rgV24qNjXcQSqIdhCRY5AzF-LKkH0AW0ZVx-CMk59O5YWHVv0G3DWl-R-Mlc6itTZVSU1ztrwzGjzZRwmwq5n-hrVY7/s1600/AmperesLawCable.png&quot; imageanchor=&quot;1&quot; style=&quot;margin-left: 1em; margin-right: 1em;&quot;&gt;&lt;img border=&quot;0&quot; height=&quot;277&quot; src=&quot;https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgVDVQZZP9i54FA1QaLrhg6c5mvaziaQRUuMBdgIGIJJhDpuaUm_rgV24qNjXcQSqIdhCRY5AzF-LKkH0AW0ZVx-CMk59O5YWHVv0G3DWl-R-Mlc6itTZVSU1ztrwzGjzZRwmwq5n-hrVY7/s320/AmperesLawCable.png&quot; width=&quot;320&quot; /&gt;&lt;/a&gt;&lt;/div&gt;
&lt;div class=&quot;separator&quot; style=&quot;clear: both; text-align: left;&quot;&gt;
&lt;br /&gt;&lt;/div&gt;
&lt;div class=&quot;separator&quot; style=&quot;clear: both; text-align: left;&quot;&gt;
&lt;/div&gt;
&lt;ol&gt;
&lt;li&gt;$0$&lt;/li&gt;
&lt;li&gt;$\dfrac{\mu_0 ir}{2\pi a^2}$&lt;/li&gt;
&lt;li&gt;$\dfrac{\mu_0 i}{2\pi r}$&lt;/li&gt;
&lt;li&gt;$\dfrac{\mu_0 i}{2\pi r}\dfrac{c^2-r^2}{c^2-b^2}$&lt;/li&gt;
&lt;li&gt;$\dfrac{\mu_0 i}{2\pi r}\dfrac{r^2-b^2}{c^2-b^2}$&lt;/li&gt;
&lt;/ol&gt;
&lt;br /&gt;
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&lt;b&gt;Choice 1 is the answer.&lt;/b&gt;&lt;/div&gt;
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&lt;br /&gt;&lt;/div&gt;
The currents cancel out the magnetic fields created by each other, resulting in a net field of zero. More formally, this is an application of &lt;a href=&quot;http://en.wikipedia.org/wiki/Amp%C3%A8re&#39;s_circuital_law#Integral_form&quot;&gt;Ampere&#39;s Law&lt;/a&gt;.&lt;br /&gt;&lt;/div&gt;
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</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/8741318480314334133/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/09/physics-gre-32.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/8741318480314334133'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/8741318480314334133'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/09/physics-gre-32.html' title='Physics GRE - #32'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><media:thumbnail xmlns:media="http://search.yahoo.com/mrss/" url="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgVDVQZZP9i54FA1QaLrhg6c5mvaziaQRUuMBdgIGIJJhDpuaUm_rgV24qNjXcQSqIdhCRY5AzF-LKkH0AW0ZVx-CMk59O5YWHVv0G3DWl-R-Mlc6itTZVSU1ztrwzGjzZRwmwq5n-hrVY7/s72-c/AmperesLawCable.png" height="72" width="72"/><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-7068493888674851380</id><published>2011-09-01T22:34:00.000-07:00</published><updated>2011-09-01T22:34:19.193-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Math GRE"/><title type='text'>Math GRE - #31</title><content type='html'>Let $\text{F}$ be a constant force that is given by the vector $\left(\begin{array}{c}-1\\0\\1\end{array}\right)$. What is the work done by $\text{F}$ on a particle that moves along the path given by  $\left(\begin{array}{c}t\\t^2\\t^3\end{array}\right)$ between time $t=0$ and $t=1$?&lt;br /&gt;
&lt;br /&gt;
&lt;ol&gt;
&lt;li&gt;$-\dfrac{1}{4}$&lt;/li&gt;
&lt;br /&gt;
&lt;li&gt;$-\dfrac{1}{4\sqrt{2}}$&lt;/li&gt;
&lt;br /&gt;
&lt;li&gt;$0$&lt;/li&gt;
&lt;br /&gt;
&lt;li&gt;$\sqrt{2}$&lt;/li&gt;
&lt;br /&gt;
&lt;li&gt;$3\sqrt{2}$&lt;/li&gt;
&lt;/ol&gt;
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&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;
&lt;div style=&quot;display: none;&quot;&gt;
&lt;b&gt;Choice 3 is the answer.&lt;/b&gt;&lt;br /&gt;
&lt;br /&gt;
Recall that &lt;a href=&quot;http://en.wikipedia.org/wiki/Work_(physics)#Force_and_displacement&quot;&gt;work&lt;/a&gt; can be calculated according to the line integral: \[W=\int_C{\text{F}\cdot d\text{r}}\] along a curve $C$ with force $\text{F}$ and position vector $\text{r}$. In this case, we have \[\text{r}=\left(\begin{array}{c}t\\t^2\\t^3\end{array}\right)\implies d\text{r}=\left(\begin{array}{c}1\\2t\\3t^2\end{array}\right)dt.\] The curve goes from 0 to 1, so our integral is: \[W=\int_0^1{\left(\begin{array}{c}-1\\0\\1\end{array}\right)\cdot&amp;nbsp;\left(\begin{array}{c}1\\2t\\3t^2\end{array}\right)}dt=\int_0^1{(-1+0+3t^2)}dt=0\] which gives zero work as we wanted.&lt;/div&gt;
&lt;/div&gt;
&lt;/div&gt;
</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/7068493888674851380/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/09/math-gre-31.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/7068493888674851380'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/7068493888674851380'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/09/math-gre-31.html' title='Math GRE - #31'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-8654515744337974705</id><published>2011-09-01T22:33:00.000-07:00</published><updated>2011-09-01T22:34:55.441-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Physics GRE"/><title type='text'>Physics GRE - #31</title><content type='html'>Which of the following is nearly the mass of the Earth? The radius of the Earth is about $6.4\cdot10^6\,\text{m}$.
&lt;br /&gt;
&lt;br /&gt;
&lt;ol&gt;
&lt;li&gt;$6\cdot10^{24}\,\text{kg}$&lt;/li&gt;
&lt;li&gt;$6\cdot10^{27}\,\text{kg}$&lt;/li&gt;
&lt;li&gt;$6\cdot10^{30}\,\text{kg}$&lt;/li&gt;
&lt;li&gt;$6\cdot10^{33}\,\text{kg}$&lt;/li&gt;
&lt;li&gt;$6\cdot10^{36}\,\text{kg}$&lt;/li&gt;
&lt;/ol&gt;
&lt;br /&gt;
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&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;
&lt;div style=&quot;display: none;&quot;&gt;
&lt;b&gt;Choice 1 is the answer.&lt;/b&gt;&lt;br /&gt;
&lt;br /&gt;
If you don&#39;t remember the mass of the earth off the bat, recall that at ground level, \[mg=\frac{GmM_E}{R_E^2}\implies M=\frac{gR_E^2}{G}\] where $g$, $G$, and $R_E$ are given constants (a table of constants is given during the exam). We can plug in the numbers and compute \[M_E\approx6\cdot10^{24}\,\text{kg}.\]&lt;/div&gt;
&lt;/div&gt;
&lt;/div&gt;
</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/8654515744337974705/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/09/physics-gre-31.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/8654515744337974705'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/8654515744337974705'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/09/physics-gre-31.html' title='Physics GRE - #31'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-4567190559240589243</id><published>2011-08-31T23:03:00.000-07:00</published><updated>2011-08-31T23:05:39.588-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Math GRE"/><title type='text'>Math GRE - #30</title><content type='html'>What is \[\int_{-3}^3{|x+1|\,dx}?\]&lt;br /&gt;
&lt;ol&gt;&lt;li&gt;0&lt;/li&gt;
&lt;li&gt;5&lt;/li&gt;
&lt;li&gt;10&lt;/li&gt;
&lt;li&gt;15&lt;/li&gt;
&lt;li&gt;20&lt;/li&gt;
&lt;/ol&gt;&lt;br /&gt;
&lt;div style=&quot;margin: 5px 20px 20px;&quot;&gt;&lt;div class=&quot;smallfont&quot; style=&quot;margin-bottom: 2px;&quot;&gt;Solution : &lt;input onclick=&quot;if (this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display != &#39;&#39;) { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;&#39;; this.innerText = &#39;&#39;; this.value = &#39;Hide&#39;; } else { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;none&#39;; this.innerText = &#39;&#39;; this.value = &#39;Show&#39;; }&quot; style=&quot;font-size: 10px; margin: 0px; padding: 0px; width: 60px;&quot; type=&quot;button&quot; value=&quot;Show&quot; /&gt; &lt;/div&gt;&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;&lt;div style=&quot;display: none;&quot;&gt;&lt;b&gt;Choice 3 is the answer.&lt;/b&gt;&lt;br /&gt;
&lt;br /&gt;
This is a very basic question. The usual method of dealing with absolute values in integrals is to split it into a sum of two integrals and remove the absolute values. Since $|x+1|$ is negative on $[-3, 1)$ and positive on $[-1, 3]\,\,\,$ , we can split the integral into: \[\int_{-3}^3{|x+1|\,dx}=-\int_{-3}^{-1}{(x+1)\,dx}+\int_{-1}^3{(x+1)\,dx}=10.\] Another way of doing this problem is to imagine the graph of $|x+1|$ and realize that the integral is the sum of two 45-45-90 triangles (one with base 2 and height 2, &amp;nbsp;the other with base 4 and height 4). Simply add the area of triangles up and we&#39;re done.&lt;br /&gt;
&lt;br /&gt;
&lt;b&gt;Bonus question:&lt;/b&gt;&amp;nbsp;Given that $c$ is a constant, what is \[\int_{-\infty}^{\infty}{e^{-|x+c|}\,dx}?\]&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/4567190559240589243/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/08/math-gre-30.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/4567190559240589243'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/4567190559240589243'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/08/math-gre-30.html' title='Math GRE - #30'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-6782993956931480407</id><published>2011-08-31T23:02:00.000-07:00</published><updated>2011-08-31T23:02:47.658-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Physics GRE"/><title type='text'>Physics GRE - #30</title><content type='html'>The energy levels of the hydrogen atom are given in terms of the principal quantum number $n$ and a positive constant $A$ by the expression:&lt;br /&gt;
&lt;br /&gt;
&lt;ol&gt;&lt;li&gt;$A\left(n+\frac{1}{2}\right)$&lt;/li&gt;

&lt;li&gt;$A(1-n^2)$&lt;/li&gt;

&lt;li&gt;$A\left(\frac{1}{n^2}-\frac{1}{4}\right)$&lt;/li&gt;

&lt;li&gt;$An^2$&lt;/li&gt;

&lt;li&gt;$-\frac{A}{n^2}$&lt;/li&gt;
&lt;/ol&gt;&lt;br /&gt;
&lt;div style=&quot;margin: 5px 20px 20px;&quot;&gt;&lt;div class=&quot;smallfont&quot; style=&quot;margin-bottom: 2px;&quot;&gt;Solution : &lt;input onclick=&quot;if (this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display != &#39;&#39;) { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;&#39;; this.innerText = &#39;&#39;; this.value = &#39;Hide&#39;; } else { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;none&#39;; this.innerText = &#39;&#39;; this.value = &#39;Show&#39;; }&quot; style=&quot;font-size: 10px; margin: 0px; padding: 0px; width: 60px;&quot; type=&quot;button&quot; value=&quot;Show&quot; /&gt; &lt;/div&gt;&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;&lt;div style=&quot;display: none;&quot;&gt;&lt;b&gt;Choice 5 is the answer.&lt;/b&gt;&lt;br /&gt;
&lt;br /&gt;
Recall that the energy levels of the hydrogen atom is given by \[E_n = -\frac{13.6\,\text{eV}}{n^2}.\] It&#39;s just one of those things that you&#39;ve got to remember, as it takes time you don&#39;t have to derive it from first principals (I only ever remember that $E_n\propto \frac{1}{n^2}\,$ since the constant of 13.6 eV is usually unimportant).&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/6782993956931480407/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/08/physics-gre-30.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/6782993956931480407'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/6782993956931480407'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/08/physics-gre-30.html' title='Physics GRE - #30'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-6143652772787684490</id><published>2011-08-30T21:52:00.001-07:00</published><updated>2011-08-30T21:54:58.323-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Math GRE"/><title type='text'>Math GRE - #29</title><content type='html'>If $z=e^{2\pi i/5},$ then \[1+z+z^2+z^3+5z^4+4z^5+4z^6+4z^7+4z^8+5z^9=\] &lt;br /&gt;
&lt;ol&gt;&lt;li&gt;$0$&lt;/li&gt;
&lt;li&gt;$4e^{3\pi i/5}$&lt;/li&gt;
&lt;li&gt;$5e^{4\pi i/5}$&lt;/li&gt;
&lt;li&gt;$-4e^{2\pi i/5}$&lt;/li&gt;
&lt;li&gt;$-5e^{3\pi i/5}$&lt;/li&gt;
&lt;/ol&gt;&lt;br /&gt;
&lt;div style=&quot;margin: 5px 20px 20px;&quot;&gt;&lt;div class=&quot;smallfont&quot; style=&quot;margin-bottom: 2px;&quot;&gt;Solution : &lt;input onclick=&quot;if (this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display != &#39;&#39;) { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;&#39;; this.innerText = &#39;&#39;; this.value = &#39;Hide&#39;; } else { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;none&#39;; this.innerText = &#39;&#39;; this.value = &#39;Show&#39;; }&quot; style=&quot;font-size: 10px; margin: 0px; padding: 0px; width: 60px;&quot; type=&quot;button&quot; value=&quot;Show&quot; /&gt; &lt;/div&gt;&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;&lt;div style=&quot;display: none;&quot;&gt;&lt;b&gt;Choice 5 is the answer.&lt;/b&gt;&lt;br /&gt;
&lt;br /&gt;
Recall that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Root_of_unity#Summation&quot;&gt;sum of all the $n-th$ roots of unity&lt;/a&gt; for $n&amp;gt;1$ ($n=5$ in this case) is 0. For this particular question, this means that \[1+z+z^2+z^3+z^4 = 0.\] This motivates us to write the sum in the question as:&amp;nbsp;\[\begin{eqnarray*}&lt;br /&gt;
(1+z+z^2+z^3+z^4)+4z^4(1+z+z^2+z^3+z^4)+5z^9 &amp; = &amp; 0+4z^4\cdot0+5z^9 \\&lt;br /&gt;
&amp; = &amp; 5z^9.&lt;br /&gt;
\end{eqnarray*}\] Now we can easily evaluate the sum: \[\begin{eqnarray*}&lt;br /&gt;
5z^9=5\left(e^{2\pi i/5}\right)^9 &amp; = &amp; 5\left(e^{2\pi i/5}\right)^4\cdot\left(e^{2\pi i/5}\right)^5 \\&lt;br /&gt;
&amp; = &amp; 5\left(e^{2\pi i/5}\right)^4 \\&lt;br /&gt;
&amp; = &amp;5e^{8\pi i/5}=-5e^{3\pi i/5}.&lt;br /&gt;
\end{eqnarray*}\]&lt;br /&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/6143652772787684490/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/08/math-gre-29.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/6143652772787684490'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/6143652772787684490'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/08/math-gre-29.html' title='Math GRE - #29'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-1256693050606112022</id><published>2011-08-30T21:52:00.000-07:00</published><updated>2011-08-31T17:29:20.935-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Physics GRE"/><title type='text'>Physics GRE - #29</title><content type='html'>Solid argon is held together by which of the following bonding mechanisms?&lt;br /&gt;
&lt;br /&gt;
&lt;ol&gt;&lt;li&gt;Ionic bond only.&lt;/li&gt;
&lt;li&gt;Covalent bond only.&lt;/li&gt;
&lt;li&gt;Partly covalent and partly ionic.&lt;/li&gt;
&lt;li&gt;Metallic bond.&lt;/li&gt;
&lt;li&gt;Van der Waals bond&lt;/li&gt;
&lt;/ol&gt;&lt;br /&gt;
&lt;div style=&quot;margin: 5px 20px 20px;&quot;&gt;&lt;div class=&quot;smallfont&quot; style=&quot;margin-bottom: 2px;&quot;&gt;Solution : &lt;input onclick=&quot;if (this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display != &#39;&#39;) { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;&#39;; this.innerText = &#39;&#39;; this.value = &#39;Hide&#39;; } else { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;none&#39;; this.innerText = &#39;&#39;; this.value = &#39;Show&#39;; }&quot; style=&quot;font-size: 10px; margin: 0px; padding: 0px; width: 60px;&quot; type=&quot;button&quot; value=&quot;Show&quot; /&gt; &lt;/div&gt;&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;&lt;div style=&quot;display: none;&quot;&gt;&lt;b&gt;Choice 5 is the answer.&lt;/b&gt;&lt;br /&gt;
&lt;br /&gt;
Argon is not charged and stable as a monoatomic element (due to it&#39;s full orbitals). Thus, because it is stable alone, it does not ionically bond with itself since argon does not exist naturally as an ion. This eliminates choice 1.&lt;br /&gt;
&lt;br /&gt;
Because covalent bonding occurs with electron sharing and the stability from filling unfilled orbitals, Argon does not covalently bond with itself since it already has full orbitals. This eliminates choice 2.&lt;br /&gt;
&lt;br /&gt;
It cannot be partly ionic and partly covalent as we just stated above that it is to stable to do either. This eliminates choice 3.&lt;br /&gt;
&lt;br /&gt;
Argon is not a metal and so it does not participate in metallic bonding (metallic bonding occurs in metals due to massive electron sharing among many metal atoms, but again, due to argon&#39;s completely filled orbitals, it is too stable to do this). This eliminates choice 4.&lt;br /&gt;
&lt;br /&gt;
Van der Waals bonding occurs in all chemicals and compounds as they are mostly caused by instantaneously inducted dipoles due to movement of electrons in the atom. Argon is no exception to this rule. Hence choice 5 is the answer. Note that although Van der Waals bonding is always present, it is by no means the strongest bonding mechanism. If this question was applied to other elements, Van der Waals may not be the answer since it could be&amp;nbsp;negligible&amp;nbsp;compared to other bonding mechanisms.&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/1256693050606112022/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/08/physics-gre-29.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/1256693050606112022'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/1256693050606112022'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/08/physics-gre-29.html' title='Physics GRE - #29'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-2639533714179695559</id><published>2011-08-29T22:07:00.001-07:00</published><updated>2011-08-30T18:46:27.080-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Math GRE"/><title type='text'>Math GRE - #28</title><content type='html'>What is the volume of the solid formed by revolving about the x-axis the region in the first quadrant of the xy-plane bounded by the coordinate axes and the graph of the equation: \[y=\frac{1}{\sqrt{1+x^2}}?\]&lt;br /&gt;
&lt;ol&gt;&lt;li&gt;$\dfrac{\pi}{2}$&lt;/li&gt;
&lt;li&gt;$\pi$&lt;/li&gt;
&lt;li&gt;$\dfrac{\pi^2}{4}$&lt;/li&gt;
&lt;li&gt;$\dfrac{\pi^2}{2}$&lt;/li&gt;
&lt;li&gt;$\infty$&lt;/li&gt;
&lt;/ol&gt;&lt;br /&gt;
&lt;div style=&quot;margin: 5px 20px 20px;&quot;&gt;&lt;div class=&quot;smallfont&quot; style=&quot;margin-bottom: 2px;&quot;&gt;Solution : &lt;input onclick=&quot;if (this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display != &#39;&#39;) { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;&#39;; this.innerText = &#39;&#39;; this.value = &#39;Hide&#39;; } else { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;none&#39;; this.innerText = &#39;&#39;; this.value = &#39;Show&#39;; }&quot; style=&quot;font-size: 10px; margin: 0px; padding: 0px; width: 60px;&quot; type=&quot;button&quot; value=&quot;Show&quot; /&gt; &lt;/div&gt;&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;&lt;div style=&quot;display: none;&quot;&gt;&lt;b&gt;Choice 4 is the correct answer.&lt;/b&gt;&lt;br /&gt;
&lt;br /&gt;
Recall that the &lt;a href=&quot;http://en.wikipedia.org/wiki/Solid_of_revolution#Disc_method&quot;&gt;volume of revolution&lt;/a&gt; around the x-axis is given by the formula \[V=\int{\pi y(x)^2}\,dx.\] In this case, we can substitute for $y$ and plug in the limits of integration to obtain: \[V=\pi\int_0^\infty{\frac{1}{1+x^2}}\,dx.\]&lt;br /&gt;
Almost everyone with some calculus experience has had the great displeasure of memorizing: \[\int{\frac{1}{1+x^2}}\,dx = \arctan x + C.\] Now we can put it to good use! We can see that the answer is (if you&#39;ll pardon my poor notation): \[V=\pi\int_0^\infty{\frac{1}{1+x^2}}\,dx=\pi\left[\arctan\infty-\arctan0\right]=\pi\cdot\frac{\pi}{2}=\frac{\pi^2}{2}.\]&lt;br /&gt;
&lt;br /&gt;
&lt;b&gt;Interesting tidbit:&lt;/b&gt; &lt;a href=&quot;http://en.wikipedia.org/wiki/Gabriel&#39;s_Horn&quot;&gt;Here&#39;s&lt;/a&gt; a funny little thing you&#39;ll often see in first year calculus. It&#39;s called Gabriel&#39;s Horn. It&#39;s got finite volume but INFINITE surface area. Just something interesting I wanted to share.&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/2639533714179695559/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/08/math-gre-28.html#comment-form' title='4 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/2639533714179695559'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/2639533714179695559'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/08/math-gre-28.html' title='Math GRE - #28'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><thr:total>4</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-3791109668176780825</id><published>2011-08-29T22:07:00.000-07:00</published><updated>2011-08-29T22:08:48.289-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Physics GRE"/><title type='text'>Physics GRE - #28</title><content type='html'>lalalala test element $^A_ZX$ decays by natural radioactivity in two stages to&amp;nbsp;$^{A-4}_{Z-1}Y$. The two stages would most likely be which of the following?&lt;br /&gt;
&lt;br /&gt;
&lt;table&gt;&lt;tbody&gt;
&lt;tr&gt; &lt;td&gt;&lt;/td&gt; &lt;td&gt;&lt;b&gt;First stage&lt;/b&gt;&lt;/td&gt; &lt;td&gt;&lt;b&gt;Second stage&lt;/b&gt;&lt;/td&gt; &lt;/tr&gt;
&lt;tr&gt; &lt;td&gt;1.&lt;/td&gt; &lt;td&gt;$\beta^-$ emission with an antineutrino&lt;/td&gt; &lt;td&gt;$\alpha$ emission&lt;/td&gt; &lt;/tr&gt;
&lt;tr&gt; &lt;td&gt;2.&lt;/td&gt; &lt;td&gt;$\beta^-$ emission&lt;/td&gt; &lt;td&gt;$\alpha$ emission with a neutrino&lt;/td&gt; &lt;/tr&gt;
&lt;tr&gt; &lt;td&gt;3.&lt;/td&gt; &lt;td&gt;$\beta^-$ emission&lt;/td&gt; &lt;td&gt;$\gamma$ emission&lt;/td&gt; &lt;/tr&gt;
&lt;tr&gt; &lt;td&gt;4.&lt;/td&gt; &lt;td&gt;Emission of a deuteron&lt;/td&gt; &lt;td&gt;Emission of two neutrons&lt;/td&gt; &lt;/tr&gt;
&lt;tr&gt; &lt;td&gt;5.&lt;/td&gt; &lt;td&gt;$\alpha$ emission&lt;/td&gt; &lt;td&gt;$\gamma$ emission&lt;/td&gt; &lt;/tr&gt;
&lt;/tbody&gt;&lt;/table&gt;&lt;br /&gt;
&lt;div style=&quot;margin: 5px 20px 20px;&quot;&gt;&lt;div class=&quot;smallfont&quot; style=&quot;margin-bottom: 2px;&quot;&gt;Solution : &lt;input onclick=&quot;if (this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display != &#39;&#39;) { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;&#39;; this.innerText = &#39;&#39;; this.value = &#39;Hide&#39;; } else { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;none&#39;; this.innerText = &#39;&#39;; this.value = &#39;Show&#39;; }&quot; style=&quot;font-size: 10px; margin: 0px; padding: 0px; width: 60px;&quot; type=&quot;button&quot; value=&quot;Show&quot; /&gt; &lt;/div&gt;&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;&lt;div style=&quot;display: none;&quot;&gt;&lt;b&gt;Choice 1 is the answer.&lt;/b&gt;&lt;br /&gt;
&lt;br /&gt;
For choice 1, we have \[^A_ZX\rightarrow ^A_{Z+1}X+e^-+\bar{v}_e\] for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Beta_decay#.CE.B2.E2.88.92_decay&quot;&gt;$\beta^-$ emission&lt;/a&gt;, and then \[^A_{Z+1}X \rightarrow ^{A-4}_{Z-1}Y+^4_2\text{He}\] for the &lt;a href=&quot;http://en.wikipedia.org/wiki/Alpha_decay&quot;&gt;$\alpha$ emission&lt;/a&gt;. Everything works out, charge/lepton/baryon numbers are conserved as they should be.&lt;br /&gt;
&lt;br /&gt;
For choice 2, lepton number is not conserved as the $\beta^-$ emission does not occur with an antineutrino.&lt;br /&gt;
&lt;br /&gt;
For choice 3, the atomic number remains unchanged under both emissions.&lt;br /&gt;
&lt;br /&gt;
For choice 4, the emissions specified are extremely rare. In other words, this type of emission is impossible for the &#39;natural radioactivity&#39; specified in the question.&lt;br /&gt;
&lt;br /&gt;
For choice 5, the atomic numbers do not match up (there are fewer protons than needed with this decay path).&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/3791109668176780825/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/08/physics-gre-28.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/3791109668176780825'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/3791109668176780825'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/08/physics-gre-28.html' title='Physics GRE - #28'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-1134843924898956271</id><published>2011-08-28T23:33:00.000-07:00</published><updated>2011-08-28T23:33:02.574-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Physics GRE"/><title type='text'>Physics GRE - #27</title><content type='html'>Suppose that there is a very small shaft in the Earth such that a point mass can be placed at a radius of $R/2$ where $R$ is the radius of the Earth. If $F(r)$ is the gravitational force of the Earth on a point mass at a distance $r$, what is: \[\frac{F(R)}{F(2R)}?\]&lt;br /&gt;
&lt;br /&gt;
&lt;ol&gt;&lt;li&gt;32&lt;/li&gt;
&lt;li&gt;8&lt;/li&gt;
&lt;li&gt;4&lt;/li&gt;
&lt;li&gt;2&lt;/li&gt;
&lt;li&gt;1&lt;/li&gt;
&lt;/ol&gt;&lt;br /&gt;
&lt;div style=&quot;margin: 5px 20px 20px;&quot;&gt;&lt;div class=&quot;smallfont&quot; style=&quot;margin-bottom: 2px;&quot;&gt;Solution : &lt;input onclick=&quot;if (this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display != &#39;&#39;) { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;&#39;; this.innerText = &#39;&#39;; this.value = &#39;Hide&#39;; } else { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;none&#39;; this.innerText = &#39;&#39;; this.value = &#39;Show&#39;; }&quot; style=&quot;font-size: 10px; margin: 0px; padding: 0px; width: 60px;&quot; type=&quot;button&quot; value=&quot;Show&quot; /&gt; &lt;/div&gt;&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;&lt;div style=&quot;display: none;&quot;&gt;&lt;b&gt;Choice 3 is the answer.&lt;/b&gt;&lt;br /&gt;
&lt;br /&gt;
A very simple question today.&lt;br /&gt;
&lt;br /&gt;
Recall that the gravitational force between two objects is given by \[F(r)=\frac{GMm}{r^2}=\frac{k}{r^2}\] where $k$ is a constant.&lt;br /&gt;
Hence we have: \[\frac{F(R)}{F(2R)}=\left.\frac{k}{R^2}\right/\frac{k}{(2R)^2}=4.\]&lt;br /&gt;
&lt;br /&gt;
On the same exam, they also asked for \[\frac{F(R)}{F\left(\frac{R}{2}\right)}.\] Try it if you want some extra practice.&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/1134843924898956271/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/08/physics-gre-27.html#comment-form' title='6 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/1134843924898956271'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/1134843924898956271'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/08/physics-gre-27.html' title='Physics GRE - #27'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><thr:total>6</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-3496537254610331664</id><published>2011-08-28T23:29:00.000-07:00</published><updated>2011-08-28T23:32:56.961-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Math GRE"/><title type='text'>Math GRE - #27</title><content type='html'>Which of the following is the best approximation of $\sqrt{1.5}(266)^{3/2}$?&lt;br /&gt;
&lt;br /&gt;
&lt;ol&gt;&lt;li&gt;1000&lt;/li&gt;
&lt;li&gt;2700&lt;/li&gt;
&lt;li&gt;3200&lt;/li&gt;
&lt;li&gt;4100&lt;/li&gt;
&lt;li&gt;5300&lt;/li&gt;
&lt;/ol&gt;&lt;br /&gt;
&lt;div style=&quot;margin: 5px 20px 20px;&quot;&gt;&lt;div class=&quot;smallfont&quot; style=&quot;margin-bottom: 2px;&quot;&gt;Solution : &lt;input onclick=&quot;if (this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display != &#39;&#39;) { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;&#39;; this.innerText = &#39;&#39;; this.value = &#39;Hide&#39;; } else { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;none&#39;; this.innerText = &#39;&#39;; this.value = &#39;Show&#39;; }&quot; style=&quot;font-size: 10px; margin: 0px; padding: 0px; width: 60px;&quot; type=&quot;button&quot; value=&quot;Show&quot; /&gt; &lt;/div&gt;&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;&lt;div style=&quot;display: none;&quot;&gt;&lt;b&gt;Choice 5 is the answer.&lt;/b&gt;&lt;br /&gt;
&lt;br /&gt;
With some simplification, we note that: \[\begin{eqnarray*}&lt;br /&gt;
\sqrt{1.5}(266)^{3/2}=\sqrt{\frac{3}{2}}\cdot(266)^{3/2} &amp; = &amp; 266\sqrt{\frac{3\cdot 266}{2}} \\ &lt;br /&gt;
&amp; = &amp; 266\sqrt{399}\approx266\cdot20\approx 5300.&lt;br /&gt;
\end{eqnarray*}\]&lt;br /&gt;
Those of us familiar with Taylor series may try approximations with calculus, however this question teaches us that too much knowledge may be a little bit dangerous if misapplied. And that most of the time, there are simpler solutions than you&#39;d think.&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/3496537254610331664/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/08/math-gre-27.html#comment-form' title='2 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/3496537254610331664'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/3496537254610331664'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/08/math-gre-27.html' title='Math GRE - #27'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><thr:total>2</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-5982754944920156835</id><published>2011-08-27T22:24:00.001-07:00</published><updated>2011-08-27T22:25:02.604-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Physics GRE"/><title type='text'>Physics GRE - #26</title><content type='html'>An engine absorbs heat at a temperature of 727 degrees Celsius and exhausts heat at a temperature of 527 degrees Celcius. If the engine operates at maximum possible efficiency, for 2000 J of heat the amount of work the engine performs is most nearly:&lt;br /&gt;
&lt;br /&gt;
&lt;ol&gt;&lt;li&gt;400 J&lt;/li&gt;
&lt;li&gt;1450 J&lt;/li&gt;
&lt;li&gt;1600 J&lt;/li&gt;
&lt;li&gt;2000 J&lt;/li&gt;
&lt;li&gt;2760 J&lt;/li&gt;
&lt;/ol&gt;&lt;br /&gt;
&lt;div style=&quot;margin: 5px 20px 20px;&quot;&gt;&lt;div class=&quot;smallfont&quot; style=&quot;margin-bottom: 2px;&quot;&gt;Solution : &lt;input onclick=&quot;if (this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display != &#39;&#39;) { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;&#39;; this.innerText = &#39;&#39;; this.value = &#39;Hide&#39;; } else { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;none&#39;; this.innerText = &#39;&#39;; this.value = &#39;Show&#39;; }&quot; style=&quot;font-size: 10px; margin: 0px; padding: 0px; width: 60px;&quot; type=&quot;button&quot; value=&quot;Show&quot; /&gt; &lt;/div&gt;&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;&lt;div style=&quot;display: none;&quot;&gt;&lt;b&gt;Choice 1 is the answer.&lt;/b&gt;&lt;br /&gt;
&lt;br /&gt;
Recall that the&amp;nbsp;&lt;a href=&quot;http://en.wikipedia.org/wiki/Carnot_cycle#The_Carnot_cycle&quot;&gt;Carnot Efficiency&lt;/a&gt;&amp;nbsp;of a heat engine is: \[\eta=1-\frac{T_C}{T_H}\] where $T_C$ and $T_H$ are in Kelvins. For this problem, the efficiency is: \[\eta\approx1-\frac{527+273}{727+273}=1-\frac{800}{1000}=0.2.\] Thus the maximum amount of work we can get is \[W=\eta\cdot2000\,J=400\,J.\]&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/5982754944920156835/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/08/physics-gre-26.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/5982754944920156835'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/5982754944920156835'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/08/physics-gre-26.html' title='Physics GRE - #26'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-8010852420855698206</id><published>2011-08-27T22:24:00.000-07:00</published><updated>2011-08-27T22:24:24.135-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Math GRE"/><title type='text'>Math GRE - #26</title><content type='html'>A total of $x$ feet of fencing is to form three sides of a level rectangular yard. What is the maximum possible area of the yard, in terms of $x$?&lt;br /&gt;
&lt;br /&gt;
&lt;ol&gt;&lt;li&gt;$\dfrac{x^2}{9}$&lt;/li&gt;
&lt;li&gt;$\dfrac{x^2}{8}$&lt;/li&gt;
&lt;li&gt;$\dfrac{x^2}{4}$&lt;/li&gt;
&lt;li&gt;$x^2$&lt;/li&gt;
&lt;li&gt;$2x^2$&lt;/li&gt;
&lt;/ol&gt;&lt;br /&gt;
&lt;div style=&quot;margin: 5px 20px 20px;&quot;&gt;&lt;div class=&quot;smallfont&quot; style=&quot;margin-bottom: 2px;&quot;&gt;Solution : &lt;input onclick=&quot;if (this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display != &#39;&#39;) { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;&#39;; this.innerText = &#39;&#39;; this.value = &#39;Hide&#39;; } else { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;none&#39;; this.innerText = &#39;&#39;; this.value = &#39;Show&#39;; }&quot; style=&quot;font-size: 10px; margin: 0px; padding: 0px; width: 60px;&quot; type=&quot;button&quot; value=&quot;Show&quot; /&gt; &lt;/div&gt;&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;&lt;div style=&quot;display: none;&quot;&gt;&lt;b&gt;Choice 2 is the answer.&lt;/b&gt;&lt;br /&gt;
&lt;br /&gt;
Let&#39;s call one side of the fence $a$ and the other side $b$. Since we are only looking at 3 sides of a rectangular fence (e.g. we&#39;re fencing in our back yard and one side is the wall of the house), let&#39;s call the side length that we have to fence twice. So we have: \[x = 2a+b.\] The area of the fence is \[A = ab = a(x-2a)\] where the last equality comes from our equation for $x$. To maximize the area, we take the derivative of $A$ with respect to $a$ and set equal to 0 to obtain: \[0 = x-4a\implies a=\frac{x}{4}.\] We know this represents the maximum since the second derivative of $A$ is always negative. Using our equation for the perimeter, we now know that: \[x=2a + b\implies b=\frac{x}{2}.\] Hence the maximum possible area is: \[A=ab=\frac{x}{4}\cdot\frac{x}{2}=\frac{x^2}{8}.\]&lt;br /&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/8010852420855698206/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/08/math-gre-26.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/8010852420855698206'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/8010852420855698206'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/08/math-gre-26.html' title='Math GRE - #26'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-2287202443800130065</id><published>2011-08-26T18:20:00.001-07:00</published><updated>2011-08-26T18:21:52.678-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Physics GRE"/><title type='text'>Physics GRE - #25</title><content type='html'>A 2-kilogram box hangs by a massless rope from a ceiling. A force slowly pulls the box horizontally until the horizontal force is 10 N. The box is then in equilibrium as shown below.&lt;br /&gt;
&lt;br /&gt;
&lt;div class=&quot;separator&quot; style=&quot;clear: both; text-align: center;&quot;&gt;&lt;a href=&quot;https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi_wZr4zwI6lwzFEVPE_p5Ca_0A3MtPvunkzTkDMF7TD-tdbpKWUXg070DA9A6yLx_nczAyKalwGecL2tiX5_H33Evk5JcaBHaQkXSEh20SBpY-JovXDsY6AJmIrBsIhju-1I2U6QdJDcik/s1600/BoxInEqm.png&quot; imageanchor=&quot;1&quot; style=&quot;margin-left: 1em; margin-right: 1em;&quot;&gt;&lt;img border=&quot;0&quot; src=&quot;https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi_wZr4zwI6lwzFEVPE_p5Ca_0A3MtPvunkzTkDMF7TD-tdbpKWUXg070DA9A6yLx_nczAyKalwGecL2tiX5_H33Evk5JcaBHaQkXSEh20SBpY-JovXDsY6AJmIrBsIhju-1I2U6QdJDcik/s1600/BoxInEqm.png&quot; /&gt;&lt;/a&gt;&lt;/div&gt;&lt;br /&gt;
The angle that the rope makes with the vertical is closest to:&lt;br /&gt;
&lt;br /&gt;
&lt;ol&gt;&lt;li&gt;$\arctan0.5$&lt;/li&gt;
&lt;li&gt;$\arcsin0.5$&lt;/li&gt;
&lt;li&gt;$\arctan2.0$&lt;/li&gt;
&lt;li&gt;$\arcsin2.0$&lt;/li&gt;
&lt;li&gt;$45^\circ$&lt;/li&gt;
&lt;/ol&gt;&lt;br /&gt;
&lt;div style=&quot;margin: 5px 20px 20px;&quot;&gt;&lt;div class=&quot;smallfont&quot; style=&quot;margin-bottom: 2px;&quot;&gt;Solution : &lt;input onclick=&quot;if (this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display != &#39;&#39;) { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;&#39;; this.innerText = &#39;&#39;; this.value = &#39;Hide&#39;; } else { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;none&#39;; this.innerText = &#39;&#39;; this.value = &#39;Show&#39;; }&quot; style=&quot;font-size: 10px; margin: 0px; padding: 0px; width: 60px;&quot; type=&quot;button&quot; value=&quot;Show&quot; /&gt; &lt;/div&gt;&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;&lt;div style=&quot;display: none;&quot;&gt;&lt;b&gt;Choice 1 is the answer.&lt;/b&gt;&lt;br /&gt;
&lt;br /&gt;
Since the block is in equilibrium, we must have $\vec{F}+\vec{F_g}=\vec{T}$ as shown below.&lt;br /&gt;
&lt;br /&gt;
&lt;div class=&quot;separator&quot; style=&quot;clear: both; text-align: center;&quot;&gt;&lt;a href=&quot;https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhp8f7vpmrh3RPkwiSQaZUH-pepcUj6hCwG4ONukwvrWI0j6TItVC7hdB1ibmpOlgPThZiozz-3Q75Toiv8U5Y2AIYpobGPdYfDb9YSktntGuqHwuoOmGsd62bodR6YgWnt3iLVu5ABmXPD/s1600/Force+Diagram.png&quot; imageanchor=&quot;1&quot; style=&quot;margin-left: 1em; margin-right: 1em;&quot;&gt;&lt;img border=&quot;0&quot; height=&quot;241&quot; src=&quot;https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhp8f7vpmrh3RPkwiSQaZUH-pepcUj6hCwG4ONukwvrWI0j6TItVC7hdB1ibmpOlgPThZiozz-3Q75Toiv8U5Y2AIYpobGPdYfDb9YSktntGuqHwuoOmGsd62bodR6YgWnt3iLVu5ABmXPD/s320/Force+Diagram.png&quot; width=&quot;320&quot; /&gt;&lt;/a&gt;&lt;/div&gt;&lt;br /&gt;
Hence $\theta=\arctan\frac{|\vec{F}|}{|\vec{F_g}|}=\arctan\frac{10}{2g}\approx\arctan\frac{10}{20}=\arctan0.5$.&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/2287202443800130065/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/08/physics-gre-25.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/2287202443800130065'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/2287202443800130065'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/08/physics-gre-25.html' title='Physics GRE - #25'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><media:thumbnail xmlns:media="http://search.yahoo.com/mrss/" url="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi_wZr4zwI6lwzFEVPE_p5Ca_0A3MtPvunkzTkDMF7TD-tdbpKWUXg070DA9A6yLx_nczAyKalwGecL2tiX5_H33Evk5JcaBHaQkXSEh20SBpY-JovXDsY6AJmIrBsIhju-1I2U6QdJDcik/s72-c/BoxInEqm.png" height="72" width="72"/><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-8030630543322719802</id><published>2011-08-26T18:20:00.000-07:00</published><updated>2011-08-26T18:23:12.421-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Math GRE"/><title type='text'>Math GRE - #25</title><content type='html'>Let $k$ be the number of roots to the equation $f(x)=e^x+x-2$ in the interval $[0, 1]$, and let $n$ be the number of roots that are not in $[0,1]$. Which of the following is true?&lt;br /&gt;
&lt;br /&gt;
&lt;ul&gt;&lt;li&gt;$k=0$ and $n=1$&lt;/li&gt;
&lt;li&gt;$k=1$ and $n=0$&lt;/li&gt;
&lt;li&gt;$k=n=1$&lt;/li&gt;
&lt;li&gt;$k&amp;gt;1$&lt;/li&gt;
&lt;li&gt;$n&amp;gt;1$&lt;/li&gt;
&lt;/ul&gt;&lt;br /&gt;
&lt;div style=&quot;margin: 5px 20px 20px;&quot;&gt;&lt;div class=&quot;smallfont&quot; style=&quot;margin-bottom: 2px;&quot;&gt;Solution : &lt;input onclick=&quot;if (this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display != &#39;&#39;) { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;&#39;; this.innerText = &#39;&#39;; this.value = &#39;Hide&#39;; } else { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;none&#39;; this.innerText = &#39;&#39;; this.value = &#39;Show&#39;; }&quot; style=&quot;font-size: 10px; margin: 0px; padding: 0px; width: 60px;&quot; type=&quot;button&quot; value=&quot;Show&quot; /&gt; &lt;/div&gt;&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;&lt;div style=&quot;display: none;&quot;&gt;&lt;b&gt;Choice 2 is the answer.&lt;/b&gt;&lt;br /&gt;
&lt;br /&gt;
We can solve this problem by using the &lt;a href=&quot;http://en.wikipedia.org/wiki/Intermediate_value_theorem&quot;&gt;Intermediate Value Theorem&lt;/a&gt;.&lt;br /&gt;
By evaluating $f$ at 0 and 1, we see that: \[f(0)=e^0+0-2&amp;lt;0\] \[f(1)=e^1+1-2&amp;gt;0.\]&lt;br /&gt;
By the Intermediate Value Theorem, there exists a $c\in[0,1]$ such that $f(c)=0.$ Thus we know for sure that there is at least one root in [0,1].&lt;br /&gt;
&lt;br /&gt;
Taking the derivative of $f$, we note that: \[f^\prime(x)=e^x+1&amp;gt;0\] for all $x$. &lt;br /&gt;
Thus $f$ is increasing everywhere and so it must only have one root. Since this root is in [0,1], $k=1.$ As we cannot have roots anywhere outside of [0,1], we also have $n=0$. This gives choice 2 as the answer.&lt;br /&gt;
&lt;br /&gt;
&lt;b&gt;Bonus question:&lt;/b&gt; The Intermediate Value Theorem may seem obvious, but it can also be used to prove many fun facts. One such fact is that if you draw a circle anywhere in the universe, there will be two points directly opposite to each other on the circle (i.e. 180 deg. apart) that have the exact same temperature. Can you prove this?&lt;br /&gt;
&lt;br /&gt;
If you get stuck, here is some &lt;a href=&quot;http://hippomath.blogspot.com/2011/06/equal-temperatures-at-two-opposite.html&quot;&gt;intuition&lt;/a&gt; as to why this interesting fact is true.&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/8030630543322719802/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/08/math-gre-25.html#comment-form' title='2 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/8030630543322719802'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/8030630543322719802'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/08/math-gre-25.html' title='Math GRE - #25'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><thr:total>2</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-5310797094520613827</id><published>2011-08-25T22:53:00.000-07:00</published><updated>2011-08-26T04:02:25.186-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Math GRE"/><title type='text'>Math GRE - #24</title><content type='html'>What is the greatest possible area of a triangular region with one vertex at the centre of a unit circle and the other two vertices on the circle?&lt;br /&gt;
&lt;br /&gt;
&lt;ol&gt;&lt;li&gt;$\frac{1}{2}$&lt;/li&gt;
&lt;li&gt;$1$&lt;/li&gt;
&lt;li&gt;$\sqrt{2}$&lt;/li&gt;
&lt;li&gt;$\pi$&lt;/li&gt;
&lt;li&gt;$\frac{1+\sqrt{2}}{4}$&lt;/li&gt;
&lt;/ol&gt;&lt;br /&gt;
&lt;div style=&quot;margin: 5px 20px 20px;&quot;&gt;&lt;div class=&quot;smallfont&quot; style=&quot;margin-bottom: 2px;&quot;&gt;Solution : &lt;input onclick=&quot;if (this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display != &#39;&#39;) { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;&#39;; this.innerText = &#39;&#39;; this.value = &#39;Hide&#39;; } else { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;none&#39;; this.innerText = &#39;&#39;; this.value = &#39;Show&#39;; }&quot; style=&quot;font-size: 10px; margin: 0px; padding: 0px; width: 60px;&quot; type=&quot;button&quot; value=&quot;Show&quot; /&gt; &lt;/div&gt;&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;&lt;div style=&quot;display: none;&quot;&gt;Choice 1 is the answer.&lt;br /&gt;
&lt;br /&gt;
The &lt;a href=&quot;http://en.wikipedia.org/wiki/Triangle#Using_trigonometry&quot;&gt;area of any triangle&lt;/a&gt; with sides $a$, $b$, and angle $\theta$ between $a$ and $b$ is: \[A = \frac{1}{2}ab\sin\theta\]&lt;br /&gt;
Since the sides have the same length as the radius of the circle, $a=b=1.$ &lt;br /&gt;
Therefore, \[A=\frac{1}{2}\sin\theta\] which equals $\frac{1}{2}$ at the maximum of $\theta=\frac{\pi}{2}$.&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/5310797094520613827/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/08/math-gre-24.html#comment-form' title='2 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/5310797094520613827'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/5310797094520613827'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/08/math-gre-24.html' title='Math GRE - #24'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><thr:total>2</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-2686972628880611439</id><published>2011-08-25T22:52:00.000-07:00</published><updated>2011-08-25T22:53:12.244-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Physics GRE"/><title type='text'>Physics GRE - #24</title><content type='html'>Assume $A$, $T$, $\lambda$ are positive constants. The equation \[y=A\sin\left[2\pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)\right]\] represents a wave whose:&lt;br /&gt;
&lt;br /&gt;
&lt;ol&gt;&lt;li&gt;amplitude is $2A$&lt;/li&gt;
&lt;li&gt;velocity is in the negative x-direction&lt;/li&gt;
&lt;li&gt;period is $\frac{T}{\lambda}$&lt;/li&gt;
&lt;li&gt;speed is $\frac{x}{t}$&lt;/li&gt;
&lt;li&gt;speed is $\frac{\lambda}{T}$&lt;/li&gt;
&lt;/ol&gt;&lt;br /&gt;
&lt;div style=&quot;margin: 5px 20px 20px;&quot;&gt;&lt;div class=&quot;smallfont&quot; style=&quot;margin-bottom: 2px;&quot;&gt;Solution : &lt;input onclick=&quot;if (this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display != &#39;&#39;) { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;&#39;; this.innerText = &#39;&#39;; this.value = &#39;Hide&#39;; } else { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;none&#39;; this.innerText = &#39;&#39;; this.value = &#39;Show&#39;; }&quot; style=&quot;font-size: 10px; margin: 0px; padding: 0px; width: 60px;&quot; type=&quot;button&quot; value=&quot;Show&quot; /&gt; &lt;/div&gt;&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;&lt;div style=&quot;display: none;&quot;&gt;&lt;b&gt;Choice 5 is the answer.&lt;/b&gt;&lt;br /&gt;
&lt;br /&gt;
Choice 1 is incorrect because $\sin\theta\leq 1$ for any $\theta$ so $y\leq A$.&lt;br /&gt;
&lt;br /&gt;
To see that choice 2 is incorrect, suppose that we are following the minimum wave as time passes (by changing $x$ with respect to time to a specific $x(t)$) in such a way that the wave seems to stand still (i.e. we are travelling at the same velocity as the wave). For the wave to &#39;stand still&#39;, we need $x(t)$ to satisfy \[2\pi\left(\frac{t}{T}-\frac{x(t)}{\lambda}\right) = const. = C\] Satisfying this relationship ensures that the wave always looks the same to us as: \[y=A\sin\left[2\pi\left(\frac{t}{T}-\frac{x}{\lambda}\right)\right] = A\sin C = constant.\]&amp;nbsp;Since we are following one of the minimums of the wave, we can choose $C = 0$ to get the minimum at $\sin 0$. Thus we get:&amp;nbsp;\[2\pi\left(\frac{t}{T}-\frac{x(t)}{\lambda}\right) = 0 \implies x(t) = \frac{\lambda}{T}t.\] To get the sign of the velocity, we take the derivative of $x(t)$: \[\frac{dx}{dt} = \frac{\lambda}{T}&amp;gt;0\] since $\lambda,T &amp;gt; 0$. This also tells us choice 5 is correct since the speed is &amp;nbsp;$\frac{\lambda}{T}.$&lt;br /&gt;
&lt;br /&gt;
Choice 3 is wrong as the units do not work out.&lt;br /&gt;
&lt;br /&gt;
Choice 4 is wrong because we just calculated the speed to be $\frac{\lambda}{T}.$&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/2686972628880611439/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/08/physics-gre-24.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/2686972628880611439'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/2686972628880611439'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/08/physics-gre-24.html' title='Physics GRE - #24'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-3798092154419942923</id><published>2011-08-24T03:45:00.001-07:00</published><updated>2011-08-24T03:45:29.025-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Physics GRE"/><title type='text'>Physics GRE - #23</title><content type='html'>A particle is initially at rest at the top of a curved frictionless track.&lt;br /&gt;
&lt;br /&gt;
&lt;div class=&quot;separator&quot; style=&quot;clear: both; text-align: center;&quot;&gt;&lt;a href=&quot;https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEisODhIROHdgeBWfRzRDheeQfk-e5AXlodzQp0q4zHcRsov9G140IrhUmD0-_AcCMD3WGcVw-yTQJ_N4k52p6j9Nh3zxbkClf9DLi-PMrkU2dNUMelZr-J_BHT-QuE5cq2UaF6YrO9Nhifz/s1600/FrictionlessParabolicTrack.png&quot; imageanchor=&quot;1&quot; style=&quot;margin-left: 1em; margin-right: 1em;&quot;&gt;&lt;img border=&quot;0&quot; height=&quot;152&quot; src=&quot;https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEisODhIROHdgeBWfRzRDheeQfk-e5AXlodzQp0q4zHcRsov9G140IrhUmD0-_AcCMD3WGcVw-yTQJ_N4k52p6j9Nh3zxbkClf9DLi-PMrkU2dNUMelZr-J_BHT-QuE5cq2UaF6YrO9Nhifz/s200/FrictionlessParabolicTrack.png&quot; width=&quot;200&quot; /&gt;&lt;/a&gt;&lt;/div&gt;&lt;div class=&quot;separator&quot; style=&quot;clear: both; text-align: left;&quot;&gt;&lt;br /&gt;
&lt;/div&gt;&lt;div class=&quot;separator&quot; style=&quot;clear: both; text-align: left;&quot;&gt;The x and y coordinates of the track are related in dimensionless units by $y=\frac{x^2}{4}$, where the positive y-axis is in the vertical downward direction. As the particle slides down the track, what is its tangential acceleration?&lt;/div&gt;&lt;div class=&quot;separator&quot; style=&quot;clear: both; text-align: left;&quot;&gt;&lt;/div&gt;&lt;ol&gt;&lt;li&gt;$0$&lt;/li&gt;
&lt;li&gt;$g$&lt;/li&gt;
&lt;li&gt;$\frac{gx}{2}$&lt;/li&gt;
&lt;li&gt;$\frac{gx}{\sqrt{x^2+4}}$&lt;/li&gt;
&lt;li&gt;$\frac{gx^2}{\sqrt{x^2+16}}$&lt;/li&gt;
&lt;/ol&gt;&lt;br /&gt;
&lt;div style=&quot;margin: 5px 20px 20px;&quot;&gt;&lt;div class=&quot;smallfont&quot; style=&quot;margin-bottom: 2px;&quot;&gt;Solution : &lt;input onclick=&quot;if (this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display != &#39;&#39;) { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;&#39;; this.innerText = &#39;&#39;; this.value = &#39;Hide&#39;; } else { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;none&#39;; this.innerText = &#39;&#39;; this.value = &#39;Show&#39;; }&quot; style=&quot;font-size: 10px; margin: 0px; padding: 0px; width: 60px;&quot; type=&quot;button&quot; value=&quot;Show&quot; /&gt; &lt;/div&gt;&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;&lt;div style=&quot;display: none;&quot;&gt;&lt;b&gt;Choice 4 is the solution.&lt;/b&gt;&lt;br /&gt;
&lt;br /&gt;
First we note that choice 2 cannot be the answer as there is a normal force from the track.&lt;br /&gt;
&lt;br /&gt;
Now as $x\rightarrow\infty$, we should expect that the acceleration goes to $g$ as the tangent becomes nearly vertical. This makes choice 4 is the only possible answer.&lt;br /&gt;
&lt;br /&gt;
We can also solve this question elegantly without resorting to limits and boundary conditions (however it will take a good half page). Can you see how it can be done?&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/3798092154419942923/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/08/physics-gre-23.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/3798092154419942923'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/3798092154419942923'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/08/physics-gre-23.html' title='Physics GRE - #23'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><media:thumbnail xmlns:media="http://search.yahoo.com/mrss/" url="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEisODhIROHdgeBWfRzRDheeQfk-e5AXlodzQp0q4zHcRsov9G140IrhUmD0-_AcCMD3WGcVw-yTQJ_N4k52p6j9Nh3zxbkClf9DLi-PMrkU2dNUMelZr-J_BHT-QuE5cq2UaF6YrO9Nhifz/s72-c/FrictionlessParabolicTrack.png" height="72" width="72"/><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-5244458674117363159.post-2338304752184033724</id><published>2011-08-24T03:45:00.000-07:00</published><updated>2011-08-24T03:49:19.294-07:00</updated><category scheme="http://www.blogger.com/atom/ns#" term="Math GRE"/><title type='text'>Math GRE - #23</title><content type='html'>Suppose B is a basis for a real vector space V of dimension greater than 1. Which of the following statements could be true?&lt;br /&gt;
&lt;br /&gt;
&lt;ol&gt;&lt;li&gt;The zero vector of V is an element of B.&lt;/li&gt;
&lt;li&gt;B has a proper subset that spans V.&lt;/li&gt;
&lt;li&gt;B is a proper subset of a linearly independent subset of V.&lt;/li&gt;
&lt;li&gt;There is a basis for V that is disjoint from B.&lt;/li&gt;
&lt;li&gt;One of the vectors in B is a linear combination of the other vectors in B.&lt;/li&gt;
&lt;/ol&gt;&lt;br /&gt;
&lt;div style=&quot;margin: 5px 20px 20px;&quot;&gt;&lt;div class=&quot;smallfont&quot; style=&quot;margin-bottom: 2px;&quot;&gt;Solution : &lt;input onclick=&quot;if (this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display != &#39;&#39;) { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;&#39;; this.innerText = &#39;&#39;; this.value = &#39;Hide&#39;; } else { this.parentNode.parentNode.getElementsByTagName(&#39;div&#39;)[1].getElementsByTagName(&#39;div&#39;)[0].style.display = &#39;none&#39;; this.innerText = &#39;&#39;; this.value = &#39;Show&#39;; }&quot; style=&quot;font-size: 10px; margin: 0px; padding: 0px; width: 60px;&quot; type=&quot;button&quot; value=&quot;Show&quot; /&gt; &lt;/div&gt;&lt;br /&gt;
&lt;div class=&quot;alt2&quot; style=&quot;-moz-background-clip: -moz-initial; -moz-background-inline-policy: -moz-initial; -moz-background-origin: -moz-initial; background: #f5f5f5 none repeat scroll 0% 50%; border: 1px inset; color: #7f4500; line-height: 1.5em; margin: 0px; padding: 6px;&quot;&gt;&lt;div style=&quot;display: none;&quot;&gt;Let us first recall the definition of a basis:&lt;br /&gt;
&lt;div style=&quot;text-align: center;&quot;&gt;&quot;A basis for a vector space is a &lt;b&gt;linearly independent&lt;/b&gt; set of vectors that &lt;b&gt;spans the vector space&lt;/b&gt;&quot;.&lt;/div&gt;&lt;div style=&quot;text-align: left;&quot;&gt;&lt;br /&gt;
&lt;/div&gt;&lt;div style=&quot;text-align: left;&quot;&gt;B is linearly independent because it is a basis. This removes choice 5.&lt;br /&gt;
&lt;br /&gt;
Because a basis is linearly independent, the zero vector cannot be part of it. &lt;br /&gt;
This removes choice 1.&lt;br /&gt;
&lt;br /&gt;
For B to have a proper subset that spans V, there must exist vectors in B which are not linearly independent (since we can remove vectors and still span V). However, this is impossible since B is a basis (i.e. we cannot remove any vectors and still hope to span V). This removes choice 2.&lt;br /&gt;
&lt;br /&gt;
Similarly, we cannot add any more linearly independent vectors to B because it already spans all of V. Thus B cannot be a proper subset of a linearly independent subset of V. This removes choice 3.&lt;br /&gt;
&lt;br /&gt;
The only choice left is choice 4. Thus choice 4 is the answer (there are in fact an infinite number of bases that are disjoint from B).&lt;br /&gt;
&lt;br /&gt;
&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://dailygre.blogspot.com/feeds/2338304752184033724/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://dailygre.blogspot.com/2011/08/math-gre-23.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/2338304752184033724'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/5244458674117363159/posts/default/2338304752184033724'/><link rel='alternate' type='text/html' href='http://dailygre.blogspot.com/2011/08/math-gre-23.html' title='Math GRE - #23'/><author><name>Paul Liu</name><uri>http://www.blogger.com/profile/16809371907394009052</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='//blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjdEg4zzND8XBvuq12LG5Jh-8mNuTujrEF9DqCWYO3Pfg3vWi3X-gAe8sR2VDEvRIL14iHGpQuCvo1QE8PIzcPdrk5E-ugPOZwogZGGuYR4Dx-_Q4lTbSCpX8DwgWa8xYs/s220/profile+pic+-+Copy.jpg'/></author><thr:total>0</thr:total></entry></feed>