Assignment 18  19

course Mth 174

I tried my hardest on these assignments thats why they are a little late. I have been working on keeping that C average up.

Cal 211-20-2007

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

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14:40:48 `questionNumber 180000 Query problem 11.3.4 (was 10.3.6) Euler y' = x^3-y^3, (0,0), `dx = .2, 5 steps

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RESPONSE --> Well first we will look at the function of y' = x^3 - y^3 and we see that once we go back to the original function we get y = (1/4)x^4 - (1/4)y^4

This is not valid. y^4 / 4 is not an antiderivative of y with respect to x. The derivative of y^4 / 4 with respect to x would be y^3 * dy/dx, not y^3.

once we look at the point that this passes through we can see the slope of the function and once we have the slope we can use it to get the y-valuewhen x =1 by using Euler's Mthod using the equation delta(y) = (slope)delta(x) which will give us the true y-value.

The problem asked you to use Euler's Method. I suggest you resubmit this question and include an Euler's Method solution.

Here you start with at the point (0, 0). 1. What is the value of y ' at this point? 2. y ' represents the slope of the graph of y vs. x. Assuming the slope remains constant over the `dx interval, what therefore is the change in y over this interval? 3. At what point do you then end up after moving `dx units in the x direction and `dy units in the y direction? Starting at this point, answer the same questions again. Continue to repeat the process until you have completed the specified number of steps. The first two iterations of this process give you the following answers to the three questions: First iteration: 1. y ' = x^3 - y^3. At (0, 0) this gives you y ' = 0^3 - 0^3 = 0. 2. If the slope remains 0 over the interval `dx = .2, the change in y is y ' * `dx = 0 * .2 = 0. 3. After moving .2 units in the x direction and 0 units in the y direction, starting from (0, 0) you end up at (.2, 0). Second iteration: 1. y ' = x^3 - y^3. At (.2, 0) this gives you y ' = .2^3 - 0^3 = 0 = .008. 2. If the slope remains .008 over the interval `dx = .2, the change in y is y ' * `dx = .008 * .2 = .0016. 3. After moving .2 units in the x direction and 0 units in the y direction, starting from (.2, 0) you end up at (.4, .008). Complete the remainins steps and let me know what you get.

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14:40:50 `questionNumber 180000 what is your estimate of y(1)?

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RESPONSE -->

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14:40:52 `questionNumber 180000 Describe how the given slope field is consistent with your step-by-step results.

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RESPONSE -->

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14:40:54 `questionNumber 180000 Is your approximation an overestimate or an underestimate, and what property of the slope field allows you to answer this question?

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RESPONSE -->

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14:41:26 `questionNumber 180000 Query problem 11.3.10 (was 10.3.10) Euler and left Riemann sums, y' = f(x), y(0) = 0

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RESPONSE --> to be honest i really do not understand the whole Eulers method so i need some help with this one.

See my notes on the preceding problem and see they help with this one.

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14:45:36 `questionNumber 180000 Query problem 11.4.40 (3d edition 11.4.39) (was 10.4.30) t dx.dt = (1 + 2 ln t ) tan x, 1st quadrant

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RESPONSE -->

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14:47:13 `questionNumber 180000 what is your solution to the problem?

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RESPONSE --> The solution to the problem would be abs(tanx) = Ae^(1/t(1+2ln(t)) then to find the B we would use the formula abs(tanx) = Be^(1/t(1+2ln(t)) which would be for any B.

Separation of variables is possible here, and gives you dx / tan(x) = (1 + 2 ln(t)) dt. Expressing this as (cos(x) / sin(x) ) * dx = (1 + 2 ln(t) ) dt. To solve you need to integrate both sides of this equation. What do you get for your general antiderivatives? Can you solve the resulting equation for x as a function of t?

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14:47:16 `questionNumber 180000 What is the general solution to the differential equation?

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RESPONSE -->

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14:47:24 `questionNumber 180000 Explain how you separated the variables for the problem.

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RESPONSE --> shown in the first screen

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14:47:31 `questionNumber 180000 What did you get when you integrated the separated equation?

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RESPONSE --> shown before

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14:48:04 `questionNumber 180000 Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE --> did not really understand the whole Euler's Method problems so if you have any suggestions i would greatly appreciate them.

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?????????O??{????·??·assignment #019 019. `query 19 Cal 2 11-20-2007

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14:51:10 `questionNumber 190000 Query problem 11.5.8 (3d edition 11.5.12) $1000 at rate r

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RESPONSE -->

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14:51:39 `questionNumber 190000 what differential equation is satisfied by the amount of money in the account at time t since the original investment?

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RESPONSE --> the differential equation here is dM/dt = 0.05M

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14:53:29 `questionNumber 190000 What is the solution to the equation?

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RESPONSE --> The solution of the equation is M = M0e^0.05t which using 1000 dollars we come up with the equation M = 1000e^0.05t and if we plug in the time until now 2007 we end up with the equation M = 1000e^0.05(7) is the answer

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14:54:29 `questionNumber 190000 Describe your sketches of the solution for interest rates of 5% and 10%.

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RESPONSE --> With the higher interest rates the money would not gain as much so they would start off low and end up going higher and higher but the 5% interest rate would be higher at first than the 10% one.

The solution to the equation is M = M0 e^(r t), where r is the rate. If you start with $1000, then since e^(r * 0) = e^0 = 0, M0 will be $1000 no matter what the rate is. The two functions would be M(t) = 1000 * e^(.05 t) and M(t) = 1000 * e^(.10 t). How do the graphs of these two functions compare?

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14:54:33 `questionNumber 190000 Does the doubled interest rate imply twice the increase in principle?

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RESPONSE -->

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14:59:34 `questionNumber 190000 Query problem 11.5.22 (3d edition 11.5.20) At 1 pm power goes out with house at 68 F. At 10 pm outside temperature is 10 F and inside it's 57 F. {}{}Give the differential equation you would solve to obtain temperature as a function of time.{}{}Solve the equation to find the temperature at 7 am.

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RESPONSE --> the differential equation here would be dT/dt = 1.0T the equation at 7 am would be T = T0e^1.0t which would turn into T = 57e^1.0(18) and that would be your answer and you would have to worry about your pipes freezing.

The equation is dT / dt = k * (T - T0), where t is the time and T the temperature, with T0 the outdoor temperature. This equation tells you that the rate of change of the temperature with respect to clock time is proportional to the difference between house and outdoor temperature. The equation is solved by separation of variables: dT / (T - T0) = k * dt; integration yields ln | T - T0 | = k * t + c, where c is an arbitrary constant. This yields e^( ln | T - T0 | ) = e^(k t + c) and by the standard rearrangement T - T0 = A e^(k t), A > 0 so that T = T0 + A e^(k t). If we let t = 0 at 1 p.m., then at 10 p.m. we have t = 9 hours. The outside temperature is T0 = So at t = 0 we have T = 68 deg and at t = 9 hr we have T = 57 deg. This yields two equations 68 deg = 10 deg + A e^(k * 0 hr) 57 deg = 10 deg + A e^(k * 9 hr). The first equation becomes 68 deg = 10 deg + A, so that A = 58 deg. The second equation is therefore 57 deg = 10 deg + 58 deg * e^( 9 * k hr) so that 58 deg * e^(9 * k hr) = 47 deg and e^(9 * k hr) = 47 / 58. Taking natural logs of both sides 9 * k hr = ln(47 / 58) so that k = ln(47 / 58) / (9 hr), which can be easily approximated using a calculator. The temperature function is therefore T = 57 deg + 58 deg * e^( k t), for this value of k. To find the temperature at a given time just substitute into the function.

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14:59:37 `questionNumber 190000 What assumption did you make about outside temperature, and how would your prediction of the 7 am temperature change if you refined your assumption?

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RESPONSE -->

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15:19:04 `questionNumber 190000 what is your differential equation for x = quantity of C at time t, and what is its solution for x(0) = 0?

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RESPONSE --> Do not understand this problem since it is not in the book.

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15:19:11 `questionNumber 190000 If your previous answer didn't include it, what is the solution in terms of a proportionality constant k?

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RESPONSE -->

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15:39:32 `questionNumber 190000 Query problem 11.6.25 (3d edition 11.6.20) F = m g R^2 / (R + h)^2.{}{}Find the differential equation for dv/dt and show that the Chain Rule dv/dt = dv/dh * dh/dt gives you v dv/dh = -gR^2/(R+h)^2.{}{}Solve the differential equation, and use your solution to find escape velocity.{}{}Give your solution.

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RESPONSE --> Newton's Law of Motion is stated by F = M*a mg - kv = m(dv/dt) dv/dt = - k/m (v - mg/k) so after that go to converting things and we end up with the equation of dv/dt = - gR^2/(R + h)^2 after some things are cancelled out. solving the differential equation we get v^2 = v0^2 + (2gR^2 / (R + h)) - 2gR the smallest value of the escape veloctiy is shown by the equation v0 = squrt(2gR)

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15:39:52 `questionNumber 190000 Query problem 11.6.20 THIS IS THE FORMER PROBLEM, VERY UNFORTUNATELY OMITTED IN THE NEW EDITION. rate of expansion of universe: (R')^2 = 2 G M0 / R + C; case C = 0

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RESPONSE -->

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15:39:54 `questionNumber 190000 what is your solution to the differential equation R' = `sqrt( 2 G M0 / R ), R(0) = 0?

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RESPONSE -->

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15:39:55 `questionNumber 190000 How the you determine the nature of the resulting long-term expansion of the universe, and what is your conclusion?

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RESPONSE -->

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15:40:05 `questionNumber 190000 Query problem 11.5.18 NOTE: THIS PROBLEM HAS BEEN OMITTED FROM THE NEW EDITION OF THE TEXT. VERY UNFORTUNATE. absorption of light in water

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RESPONSE -->

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15:40:07 `questionNumber 190000 what is your intensity function?

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RESPONSE -->

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15:40:08 `questionNumber 190000 If your previous answer didn't include it, what is the intensity function in terms of a proportionality constant k?

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RESPONSE -->

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15:40:09 `questionNumber 190000 if 50% is absorbed in 10 ft, how much is absorbed in 20 ft, and how much in 25 feet?

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RESPONSE -->

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15:40:25 `questionNumber 190000 Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

See my notes and please respond as appropriate.

end of document

Your work has not been reviewed. Please notify your instructor of the error, using the Submit Work form, and be sure to include the date 11-26-2007.

Good work.

`gr9

Assignment 18  19

course Mth 174

I tried my hardest on these assignments thats why they are a little late. I have been working on keeping that C average up.

Cal 211-20-2007

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

......!!!!!!!!...................................

14:40:48

`questionNumber 180000

Query problem 11.3.4 (was 10.3.6) Euler y' = x^3-y^3, (0,0), `dx = .2, 5 steps

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RESPONSE -->

Well first we will look at the function of

y' = x^3 - y^3 and we see that once we go back to the original function we get

y = (1/4)x^4 - (1/4)y^4

This is not valid.

y^4 / 4 is not an antiderivative of y with respect to x.

The derivative of y^4 / 4 with respect to x would be y^3 * dy/dx, not y^3.

once we look at the point that this passes through we can see the slope of the function and once we have the slope we can use it to get the y-valuewhen x =1 by using Euler's Mthod using the equation

delta(y) = (slope)delta(x) which will give us the true y-value.

The problem asked you to use Euler's Method. I suggest you resubmit this question and include an Euler's Method solution.

Here you start with at the point (0, 0).

1. What is the value of y ' at this point?

2. y ' represents the slope of the graph of y vs. x. Assuming the slope remains constant over the `dx interval, what therefore is the change in y over this interval?

3. At what point do you then end up after moving `dx units in the x direction and `dy units in the y direction?

Starting at this point, answer the same questions again.

Continue to repeat the process until you have completed the specified number of steps.

The first two iterations of this process give you the following answers to the three questions:

First iteration:

1. y ' = x^3 - y^3. At (0, 0) this gives you y ' = 0^3 - 0^3 = 0.

2. If the slope remains 0 over the interval `dx = .2, the change in y is y ' * `dx = 0 * .2 = 0.

3. After moving .2 units in the x direction and 0 units in the y direction, starting from (0, 0) you end up at (.2, 0).

Second iteration:

1. y ' = x^3 - y^3. At (.2, 0) this gives you y ' = .2^3 - 0^3 = 0 = .008.

2. If the slope remains .008 over the interval `dx = .2, the change in y is y ' * `dx = .008 * .2 = .0016.

3. After moving .2 units in the x direction and 0 units in the y direction, starting from (.2, 0) you end up at (.4, .008).

Complete the remainins steps and let me know what you get.

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14:40:50

`questionNumber 180000

what is your estimate of y(1)?

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RESPONSE -->

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14:40:52

`questionNumber 180000

Describe how the given slope field is consistent with your step-by-step results.

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RESPONSE -->

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14:40:54

`questionNumber 180000

Is your approximation an overestimate or an underestimate, and what property of the slope field allows you to answer this question?

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RESPONSE -->

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14:41:26

`questionNumber 180000

Query problem 11.3.10 (was 10.3.10) Euler and left Riemann sums, y' = f(x), y(0) = 0

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RESPONSE -->

to be honest i really do not understand the whole Eulers method so i need some help with this one.

See my notes on the preceding problem and see they help with this one.

.................................................

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14:45:36

`questionNumber 180000

Query problem 11.4.40 (3d edition 11.4.39) (was 10.4.30) t dx.dt = (1 + 2 ln t ) tan x, 1st quadrant

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RESPONSE -->

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14:47:13

`questionNumber 180000

what is your solution to the problem?

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RESPONSE -->

The solution to the problem would be

abs(tanx) = Ae^(1/t(1+2ln(t)) then to find the B we would use the formula abs(tanx) = Be^(1/t(1+2ln(t)) which would be for any B.

Separation of variables is possible here, and gives you

dx / tan(x) = (1 + 2 ln(t)) dt. Expressing this as

(cos(x) / sin(x) ) * dx = (1 + 2 ln(t) ) dt. To solve you need to integrate both sides of this equation. What do you get for your general antiderivatives? Can you solve the resulting equation for x as a function of t?

.................................................

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14:47:16

`questionNumber 180000

What is the general solution to the differential equation?

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RESPONSE -->

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14:47:24

`questionNumber 180000

Explain how you separated the variables for the problem.

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RESPONSE -->

shown in the first screen

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14:47:31

`questionNumber 180000

What did you get when you integrated the separated equation?

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RESPONSE -->

shown before

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14:48:04

`questionNumber 180000

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

did not really understand the whole Euler's Method problems so if you have any suggestions i would greatly appreciate them.

.................................................

?????????O??{????·??·assignment #019

019. `query 19

Cal 2

11-20-2007

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14:51:10

`questionNumber 190000

Query problem 11.5.8 (3d edition 11.5.12) $1000 at rate r

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RESPONSE -->

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14:51:39

`questionNumber 190000

what differential equation is satisfied by the amount of money in the account at time t since the original investment?

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RESPONSE -->

the differential equation here is dM/dt = 0.05M

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14:53:29

`questionNumber 190000

What is the solution to the equation?

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RESPONSE -->

The solution of the equation is M = M0e^0.05t

which using 1000 dollars we come up with the equation

M = 1000e^0.05t and if we plug in the time until now 2007 we end up with the equation

M = 1000e^0.05(7) is the answer

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14:54:29

`questionNumber 190000

Describe your sketches of the solution for interest rates of 5% and 10%.

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RESPONSE -->

With the higher interest rates the money would not gain as much so they would start off low and end up going higher and higher but the 5% interest rate would be higher at first than the 10% one.

The solution to the equation is M = M0 e^(r t), where r is the rate. If you start with $1000, then since e^(r * 0) = e^0 = 0, M0 will be $1000 no matter what the rate is.

The two functions would be

M(t) = 1000 * e^(.05 t) and

M(t) = 1000 * e^(.10 t).

How do the graphs of these two functions compare?

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14:54:33

`questionNumber 190000

Does the doubled interest rate imply twice the increase in principle?

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RESPONSE -->

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14:59:34

`questionNumber 190000

Query problem 11.5.22 (3d edition 11.5.20) At 1 pm power goes out with house at 68 F. At 10 pm outside temperature is 10 F and inside it's 57 F. {}{}Give the differential equation you would solve to obtain temperature as a function of time.{}{}Solve the equation to find the temperature at 7 am.

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RESPONSE -->

the differential equation here would be

dT/dt = 1.0T

the equation at 7 am would be

T = T0e^1.0t

which would turn into T = 57e^1.0(18) and that would be your answer and you would have to worry about your pipes freezing.

The equation is dT / dt = k * (T - T0), where t is the time and T the temperature, with T0 the outdoor temperature.

This equation tells you that the rate of change of the temperature with respect to clock time is proportional to the difference between house and outdoor temperature.

The equation is solved by separation of variables:

dT / (T - T0) = k * dt; integration yields

ln | T - T0 | = k * t + c, where c is an arbitrary constant. This yields

e^( ln | T - T0 | ) = e^(k t + c) and by the standard rearrangement

T - T0 = A e^(k t), A > 0 so that

T = T0 + A e^(k t).

If we let t = 0 at 1 p.m., then at 10 p.m. we have t = 9 hours. The outside temperature is T0 =

So at t = 0 we have T = 68 deg and at t = 9 hr we have T = 57 deg. This yields two equations

68 deg = 10 deg + A e^(k * 0 hr)

57 deg = 10 deg + A e^(k * 9 hr).

The first equation becomes

68 deg = 10 deg + A, so that A = 58 deg.

The second equation is therefore

57 deg = 10 deg + 58 deg * e^( 9 * k hr) so that

58 deg * e^(9 * k hr) = 47 deg and

e^(9 * k hr) = 47 / 58. Taking natural logs of both sides

9 * k hr = ln(47 / 58) so that

k = ln(47 / 58) / (9 hr),

which can be easily approximated using a calculator.

The temperature function is therefore

T = 57 deg + 58 deg * e^( k t), for this value of k.

To find the temperature at a given time just substitute into the function.

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14:59:37

`questionNumber 190000

What assumption did you make about outside temperature, and how would your prediction of the 7 am temperature change if you refined your assumption?

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RESPONSE -->

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15:19:04

`questionNumber 190000

what is your differential equation for x = quantity of C at time t, and what is its solution for x(0) = 0?

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RESPONSE -->

Do not understand this problem since it is not in the book.

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15:19:11

`questionNumber 190000

If your previous answer didn't include it, what is the solution in terms of a proportionality constant k?

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RESPONSE -->

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15:39:32

`questionNumber 190000

Query problem 11.6.25 (3d edition 11.6.20) F = m g R^2 / (R + h)^2.{}{}Find the differential equation for dv/dt and show that the Chain Rule dv/dt = dv/dh * dh/dt gives you v dv/dh = -gR^2/(R+h)^2.{}{}Solve the differential equation, and use your solution to find escape velocity.{}{}Give your solution.

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RESPONSE -->

Newton's Law of Motion is stated by

F = M*a

mg - kv = m(dv/dt)

dv/dt = - k/m (v - mg/k)

so after that go to converting things and we end up with the equation of dv/dt = - gR^2/(R + h)^2

after some things are cancelled out.

solving the differential equation we get

v^2 = v0^2 + (2gR^2 / (R + h)) - 2gR

the smallest value of the escape veloctiy is shown by the equation v0 = squrt(2gR)

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15:39:52

`questionNumber 190000

Query problem 11.6.20 THIS IS THE FORMER PROBLEM, VERY UNFORTUNATELY OMITTED IN THE NEW EDITION. rate of expansion of universe: (R')^2 = 2 G M0 / R + C; case C = 0

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RESPONSE -->

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15:39:54

`questionNumber 190000

what is your solution to the differential equation R' = `sqrt( 2 G M0 / R ), R(0) = 0?

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RESPONSE -->

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15:39:55

`questionNumber 190000

How the you determine the nature of the resulting long-term expansion of the universe, and what is your conclusion?

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RESPONSE -->

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15:40:05

`questionNumber 190000

Query problem 11.5.18 NOTE: THIS PROBLEM HAS BEEN OMITTED FROM THE NEW EDITION OF THE TEXT. VERY UNFORTUNATE. absorption of light in water

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RESPONSE -->

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15:40:07

`questionNumber 190000

what is your intensity function?

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RESPONSE -->

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15:40:08

`questionNumber 190000

If your previous answer didn't include it, what is the intensity function in terms of a proportionality constant k?

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RESPONSE -->

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15:40:09

`questionNumber 190000

if 50% is absorbed in 10 ft, how much is absorbed in 20 ft, and how much in 25 feet?

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RESPONSE -->

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15:40:25

`questionNumber 190000

Query Add comments on any surprises or insights you experienced as a result of this assignment.

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RESPONSE -->

See my notes and please respond as appropriate.

end of document

Your work has not been reviewed.

Please notify your instructor of the error, using the Submit Work form, and be sure to include the date 11-26-2007.

Good work.

&#

Let me know if you have questions. &#