Homework Helper

Michelson-Morley 2

2012-10-18T20:07:00.002-07:00

Verify by calculation that the result of the Michelson-Morley experiment places an upper
limit on Earth’s speed relative to the ether of about 5 km/s.

Because M-M found that N must be less than or equal to 0.01, then his places a maximum velocity limit on this calculation:
Showing the limit mathematically by solving for v with the given information:
The numeric calculation for the above formula is 4.91 km/s

Because the Earth rotates at approximately a rate of 30 km/s, and because the velocity above is much smaller, our calculations imply that the "ether" does not exist. Meaning, there is no preferred reference frame.

Michelson-Morley Experiment 1

2012-10-18T19:54:00.004-07:00

A shift of one fringe in the Michelson-Morley experiment would result from a difference
of one wavelength or a change of one period of vibration in the round-trip travel of the light
when the interferometer is rotated by 90°. What speed would Michelson have computed for
Earth’s motion through the ether had the experiment seen a shift of one fringe?

Change in time for the two paths of light:
So we should find fringes for the extra distance that the light travels in time t:
Solving for v:
Given from the M-M Experiment:
Solving for v when N=1, or where the first fringe would have been detected:

Differential Equation, Drug mixture in blood stream

2011-12-23T13:57:00.000-08:00

Blood goes into an organ at a rate of 3 cm^3/sec (or cm3/sec) and leaves at the same rate. The organ has a liquid volume of 125 cm^3 (or cm3). A drug is introduced into the blood going into the organ at a concentration of 0.2 g/cm^3. Find a formula for the concentration of the drug at time t (t is time in sec after it is first introduced). When will the concentration of the drug in the organ reach 0.1 g/cm^3?

We will let x = grams of the drug, t = tine in seconds and the rate is g/sec

Follow the steps below to solve.

This is a differential equation mixture problem. We can set our initial equation based on (rate in) - (rate out) = dx/dt (or change of x over change in time)
We know our rate in as (.2 g/cm^3)(3 cm^3/sec). This does give us our correct rate units of g/sec):
Our rate out is (x/125 g/cm^3)(3 cm^3/sec). x/125 is the amount of drug in the organ at time t. This is always changing and therefore we assign it a variable:
Now simplified with only our values:
Collecting like terms to get the dx and the x together and then setting up the integral:
Integrate both sides. The left side is integrated as follows:

The completed integration equation is shown and the C (constant) has been collected on one side:
We can use the known initial condition x(0)=0, to solve for C. We know x(0)=0 because at time 0 there was no drug in the organ.
Rewriting the equation with our constant, C:
Because of the natural log, ln, we will raise each side to the power of e and we are also multiplying each side by (-3) to be able to solve:
Simplifying with using the laws of logarithms and exponents: Also note, we can drop the absolute value sign as the constraint of our problem requires it to be positive.
This only gives us x at time t, or the amount of the drug in grams. The question asks for the concentration of the drug at time t. To find concentration we divide this number by 125 (total capacity) to find the concentration:
Therefore, our concentration of the drug at time t is:
Find t when the concentration of the drug reached 0.1 g/cm^3 and simplifying:
Taking the natural log of both sides and solving for t:
Final answer for t is approx 28.88 seconds:

Find a function f such that f'(x) = x^3 and the line x + y = 0 is tangent to the graph of f.

2011-12-23T08:55:00.000-08:00

The problem states:

Find a function f such that f'(x) = x^3 and the line x + y = 0 is tangent to the graph of f.

Follow steps below to solve (C is the unknown constant):

Use x + y = 0 and solve for y we get y=-x:
Take the derivative of y and we see that y' = -1.
Since f'(x) = y', we can use y' = -1 to solve for x by saying -1 = x^3 and solving for x we get x = -1:
Using x = -1 in x + y = 0 we can find that y = 1:
Take the antiderivative of f'(x) and we get f(x) = (1/4)x^4 + C
Because we know x and y, we can use those to solve for C:
Using C and the antiderivative gives us our final answer for our function f:

Solve the Differential Equation: y'=(3x^2)/(y^3) where y(1)=2

2011-12-22T22:45:00.000-08:00

Solve the differential equation y'=3x^2/y^3 where y(1)=2:

Rewrite with dy/dx:
Multiply both sides by y^3
Multiply both sides by dx. We now have our variables separated and ready to integrate:
Integrating both sides:
Using y(1)=2 to solve for C, we find that C = 3.
Our new equation with C:
Multiplying both side by 4:
Taking the 4th root of both sides to solve for y gives us our final answer:

Integrate the following integral: x sin(x)

2011-12-22T18:44:00.000-08:00

Find the following indefinite integral:

This integral is easily solved by using the Integration by Parts formula. Follow the steps below to solve:

Reciting the Integration by Parts equation:
Set u=x and dv=sin(x) dx and solve for du and v:

Plugging in u, du, v and dv into the IBP equation above:
Simplify and then integrate the remaining integral we arrive at our final answer:

The final answer is -xcos(c)+sin(x)+C:

Find the indefinite integral x ln(2x) dx

2011-12-21T20:16:00.000-08:00

Integrate x ln(2x) dx:

We will use integration by parts to solve this equation.
Assign as follows: u = ln(2x) and dv = x dx. Then solve for du and v.

Now plug each part into the "Integration by Parts" equation above.
Simplified:
Integrating the right side will give us our final answer:

Solve the Differential Equation: y'=(3x^2)/Sin(y) when y(0)=0

2011-12-21T19:50:00.000-08:00

We have a variable-separable differential equation with an initial condition.

Rewrite y' to dy/dx to be able to separate the x and y variables.
Collect all like terms on each side by multiplying each side by sin(y) and dividing both sides by 3x^2.
Integrating both sides.
Solving for y.
Because we have an initial condition of y(0)=0, we can use this to solve for the constant C. Plug 0 in for x and y and solve for C. We find that C = 1
Rewriting our equation with 1 as the constant we get y=ArcCos[-x^3+1) as our final answer.

Find the derivative of y=10^(1-x^2)

2011-12-21T15:03:00.000-08:00

Find the derivative of the function y = 10^(1-x^2).

Original equation:
To setup the chain rule, we will set the inner function as (1-x^2) and the outer function as (10^u) where u=1-x^2.
Using the chain rule, we take the derivative of the outer function times the derivative of the inner function:
Substituting 1-x^2 back in for u we get our final answer:

Trig Integration Problem: Integral of (sin(x))^3/tan(x)

2011-12-15T09:08:00.000-08:00

To find the following indefinite integral:

May be written as: Integrate or integral of (Sin(x))^3/(tan(x)) or Integrate or integral of (sin(x))^3/tan(x)

Follow these steps:

We are trying to get this into a form where we can do u substitution. This is done by using trig identities and rewriting the integral.
First, rewrite this a (sin(x))^3/1 divided by sin(x)/cos(x):
Simplified from above:
Simplified further by dividing the sin(x) we now have this simplified integral:
Substitution can now be used. We see that if we let u=sin(x) this makem du=cos(x)dx. This takes care of the sin and the cos.
The integral rewritten with the u substitution:
Performing the integration with respect to u:
Now replacing our u with the value we specified above gives us our final answer of (1/3)(sin(x))^3 + C.

Differential Equation, Mixture Problem.

2011-12-09T23:37:00.000-08:00

A tank contains 1000 L of pure water. Brine that contains 0.05 kg of salt per liter of water enters the tank at a rate of 5 L/min. Brine that contains 0.04 kg of salt per liter of water enters the tank at a rate of 10 L/min. The solution is kept thoroughly mixed and drains from the tank at a rate of 15 L/min. How much salt is in the tank (a) after t minutes and (b) after one hour?

Setting our variables: S (or S(t)) = kg of salt at time t, t = time in minutes.

We need to setup our D.E. and solve for S(t) to find our salt concentration for any time, t.

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We set up our differential equation structure by saying dS/dt = (rate in) - (rate out). The rate is kg of salt per min.
Finding rate in is [(.05)(5)+(.04)(10)]. Notice liters cancel out to get kg/min.
Finding rate out is (S/1000)(15). S/1000 is the amount of total salt in the mixture. We use S because this changes with time. Again, notice liters cancel out to get our desired rate.
Combining our rate in with our rate out.
Combining the terms so we can multiple/divide with a common denominator.
Combining "like terms" on each side of the equation and setting up the integral.
Integrating each side.
Need to find the constant, C. We can use our initial condition of S(0) = 0 because at time = 0, we had 0 kg of salt.
Plugging in 0 for S and t.
Solve for C = -(1/3)ln(130)
Putting C into our equation
Multiplying each side by (-3) and then raising each side to the e power.
Remember that E^ln(x) = x, and that the addition of a power of e is the same as multiplying each addition term by e^(each term)
Again e^ln(x) = x. Also, we can remove the absolute value signs as our constraints of the problem require that the value to always be positive.
Solving for S, we get s(t)=-(130/3)e^(-3t/200)+(130/3). This is our equation for the amount of salt for time t.
Solving for S(60), we get 25.72 kg of salt in one hour.

Series Test. Find convergence or divergence.

2011-12-09T19:36:00.000-08:00

Find out if the sum of sqrt(n)/(n^3+1) as n goes from 1 to infinity is divergent or convergent.

We need to find a function, that we already know as convergent or divergent, that we can compare to. By looking at the leading n values, we can extract a function that is 1/n^(something) that is easily solved.

Rewrite the expression with only the leading/dominant values of n. We will get (n^(1/2))/(n^3).
Using the laws of exponents, this simplifies to 1/n^(5/2).
This series is convergent by the P-series test, which states that a series of 1/n^P is convergent when P > 1. Our P in this example is 5/2.
We can then use the Limit Comparison Test (LCT) to compare the original series with our P-series. The LCT states that if both series have positive terms and that dividing them gives us a finite number (C), and that finite number (C) is greater than zero then either both series diverge or converge.
We take the limit of our original series divided by our P-series comparison sum. We can see that when we simplify this down we get C=1. By the LCT these functions are either both convergent or divergent. Because our p-series comparison is convergent, our function is convergent.

Simple Oscillator Problem: Spring and a block

2011-12-08T17:20:00.000-08:00

Problem: A 0.200 Kg mass is attached to a spring of constant k = 5.6 N/m and set into oscillation with an amplitude of 0.25m. Solve the following circumstances:

A) Determine the frequency of the motion

Trying to solve for F (frequency). By using the free body diagram we sum the forces in the x direction (horizontal in the example).

By summing forces in the X axis (positive to the right) we get: Sum of Forces in x = Force of the spring. Steps shown below:

Sum of Forces in x = (mass)(acceleration)
Showing the only force in the X direction is the Force of the spring in the negative direction.
A force of a spring is the spring constant times the distance from rest.
Substituting the force of the spring in we get (mass)(accell.) = -(spring const.)(distance)

We are looking to get this into the form to where we can show what the second derivative of x (the position) is equal to a so we can extract omega from it. Acceleration (a) is the second derivative of position (x). Therefore we can solve for (a) from our forces above. Steps:

Start with our equation from our sum of forces above.
Solving for a we get: a = -(k/m)x
We need to find that second derivative of position equals negative omega times the position (second derivative of x = -(w^2)a). Acceleration is the second derivative of position. We replace a with (d^2/dt^2)x to show that. Notice that k/m is omega squared.
Because we got our equation in the correct form, we show that k/m = w^2. Solve for w (omega) we get sqrt(k/m) = w.
We know that w (omega) equals (2)(Pi)(Frequency). So we can easily solve for F to find out answer.
Plugging in the given values of the problem we can get a numerical answer which is approximately .842169 hertz.

B) Determine the Period

Frequency is defined as 1 divided by the period, T. So we use F=1/T for our equation to work with.
Solving for T we get T=1/F. So T equals 1/.842169. Therefore F equals 1.18741 seconds.

C) Determine Maximum Velocity

Using the velocity function of a simple oscillator we get this equation. From this, wx(not), is the maximum velocity (amplitude of the wave form)
Solving for Vmax we get (.842169)(.25) where .25 is the amplitude of the x function (given). Therefore we get .210542 m/s

D) Determine the Maximum Force

From our free-body diagram, we know our only force is the spring. Which the force of the spring = -kx.
So the max force would be k times the amplitude of x. So we have (5.6)(.25) which equals 1.4 Newtons.