Asst 171819

course Mth 174

e׻ױKassignment #017

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017. `query 17

Cal 2

08-01-2007

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13:14:56

Query problem 11.1.10 Find a value of omega such that y '' + 9 y = 0 is satisfied by y = cos(omega * t).{}{}Give your solution.

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RESPONSE -->

y = cos(omega*t)

y' = -omega*sin(omega*t)

y"" = -omega^2 * cos(omega*t)

(d^2t)/(dt^2) + 9y = 0

-(omega)^2 * cos(omega*t) + 9cos(omega*t) = 0

-(omega)^2 + 9 = 0

-(omega)^2 = -9

omega = +/- 3

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13:17:44

how did you show that the given function satisfies the given equation?

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RESPONSE -->

P = 1/(1+e^(-t))

P' = (1+e^(-t))^(-2)

P' = e^-t * (1+e^(-t))^(-2)

(dP)/(dt) = (1-P)

1/(1+e^(-t))*(1-(1/(1+e^(-t)))) = 1/(1+e^(-t)) - [1/(1+e^(-t))]^2

P>0

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13:17:57

What is the derivative dP/dt?

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RESPONSE -->

(dP)/(dt) = (1-P)

1/(1+e^(-t))*(1-(1/(1+e^(-t)))) = 1/(1+e^(-t)) - [1/(1+e^(-t))]^2

STUDENT SOLUTION WITH INSTRUCTOR COMMENT

First I took the derivative of 1/(1+e^(-t)) and got e^(-t)/(1+e^(-t))^2. So then I set up the equation the following way:

e^(-t) / (1 + e^(-t))^2 = (1/(1+e^(-t)) * ( 1 - 1/(1+e^(-t)))

e^(-t) / (1 + e^(-t))^2 = (1/(1+e^(-t)) * e^-t/(1+e^(-t))

e^(-t) / (1 + e^(-t))^2 = e^(-t) / (1+e^(-t))^2

This shows that it does satisfy the equation.

INSTRUCTOR COMMENT

The right-hand side is

(1/(1+e^(-t)) * ( 1 - 1/(1+e^(-t))).

Expressing 1 - 1/(1+e^(-t)) as

(1 + e^(-t))/(1 + e^-t) - 1 / (1 + e^(-t)) = ((1 + e^(-t) - 1) / (1 + e^(-t)) = e^(-t) / (1 + e^(-t)

we have

(1/(1+e^(-t)) * e^(-t) / ( 1 + e^(-t) ) = e^(-t) / ( 1 + e^(-t) ) ^ 2,

which agrees with the left-hand side.

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13:18:28

Query problem 11.1.16 (3d edition 11.1.15) (was 10.1.1) match equation with solution (cos x, cos(-x), x^2, e^x+e^-x, `sqrt(2x) )

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RESPONSE -->

y = cosx -->

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13:24:20

which solution(s) correspond to the equation y'' = y and how can you tell?

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RESPONSE -->

(I) y = cosx y' = -sinx y"" = -cosx --> [d] y"" = -y

(II) y = cos(-x) y' = sin(-x) y"" = -cos(-x) --> [d] y"" = -y

(III) y = x^2 y' = 2x y"" = 2 --> [e] x^2 * y"" - 2y = 0

(IV) y - e^x + e^(-x) --> [c] y' = 1/y & [a] y''= y

(V) y = sqrt(2x) y' = x(2x)^(-1/2) --> [c] y' = 1/y

III because I found the derivatives

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13:24:50

which solution(s) correspond to the equation y' = -y and how can you tell

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RESPONSE -->

(I) y = cosx y' = -sinx y"" = -cosx --> [d] y"" = -y

(II) y = cos(-x) y' = sin(-x) y"" = -cos(-x) --> [d] y"" = -y

(III) y = x^2 y' = 2x y"" = 2 --> [e] x^2 * y"" - 2y = 0

(IV) y - e^x + e^(-x) --> [c] y' = 1/y & [a] y''= y

(V) y = sqrt(2x) y' = x(2x)^(-1/2) --> [c] y' = 1/y

none correspond to y'= -y

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13:25:48

which solution(s) correspond to the equation y' = 1/y and how can you tell

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RESPONSE -->

(I) y = cosx y' = -sinx y"" = -cosx --> [d] y"" = -y

(II) y = cos(-x) y' = sin(-x) y"" = -cos(-x) --> [d] y"" = -y

(III) y = x^2 y' = 2x y"" = 2 --> [e] x^2 * y"" - 2y = 0

(IV) y - e^x + e^(-x) --> [c] y' = 1/y & [a] y''= y

(V) y = sqrt(2x) y' = 2*(2x)^(-1/2) --> [c] y' = 1/y

y = sqrt (2x)

i found the first derivative

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13:26:18

which solution(s) correspond to the equation y''=-y and how can you tell

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RESPONSE -->

(I) y = cosx y' = -sinx y"" = -cosx --> [d] y"" = -y

(II) y = cos(-x) y' = sin(-x) y"" = -cos(-x) --> [d] y"" = -y

(III) y = x^2 y' = 2x y"" = 2 --> [e] x^2 * y"" - 2y = 0

(IV) y - e^x + e^(-x) --> [c] y' = 1/y & [a] y''= y

(V) y = sqrt(2x) y' = x(2x)^(-1/2) --> [c] y' = 1/y

(I) y = cosx y' = -sinx y"" = -cosx --> [d] y"" = -y

(II) y = cos(-x) y' = sin(-x) y"" = -cos(-x) --> [d] y"" = -y

I know because I found their derivatives

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13:27:14

which solution(s) correspond to the equation x^2 y'' - 2y = 0 and how can you tell

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RESPONSE -->

y = x^2

i know because y' = 2x y""= 2

x^2(2) - 2(x^2) = 0

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13:29:19

Query problem 11.2.5 (3d edition 11.2.4) The slope field is shown. Plot solutions through (0, 0) and (1, 4).{}{}Describe your graphs. Include in your description any asymptotes or inflection points, and also intervals where the graph is concave up or concave down.

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RESPONSE -->

there is a horisontal asymptote at 10 the graph is horiazontal against x= 0 and then dependin on the slope curves a little then goes towards y = 10 the graph is concave up y>10 and concave down when y<10.

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13:30:57

describe the slope field corresponding to y' = x e^-x

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RESPONSE -->

y' = x e^-x

they are all tending towards x = 3 there is a littls dib as it approaches y-axis then curves up from t-acis til x= 3

For x = 0 we will have y ' = 0, so as we approach the x axis slopes will approach 0 and the slope field lines will approach the x axis.

For all x, e^(-x) is positive, so the sign of y ' is the same at that of x. Therefore y ' will be positive to the right of the y axis, negative to the left.

For positive x the value of e^-x rapidly approaches 0 as x increases, overwhelming the steady increases in the value of x, so that y ' must approach 0 as x increases.

For negative x the values of e^-x rapidly increase, so that the slopes, while negative, will become nearly vertical as we move to the left.

The only graph with these characteristics is iii.

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13:31:34

describe the slope field corresponding to y' = sin x

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RESPONSE -->

y' = sin x

it is the graph of -cosx and does not pass through (0,0)

| sin x | <= 1, so the slope never exceeds that of a 45-degree line. (sin x) cycles through alternative intervals of negative and positive values, so the slope lines will move from positive to 0 to negative then back to 0 then positive again.

The only graph with these characteristic is are i and ii.

The sine function is odd, with values to the left of the y axis equal and opposite to values taken at equal distances to the right of the y axis. So the slopes on one side of the y axis will be negative when the slopes at equal distances on the other side will be positive. This is consistent with graph i.

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13:31:58

describe the slope field corresponding to y' = cos x

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RESPONSE -->

y' = cos x

is the graph of sinx

passes through (0,0)

This shares the charactistics of the preceding sine graph, but the cosine function is even, with values to one side of the y axis equal to those taken at equal distances on the other side. This is consistent with graph ii.

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13:32:50

describe the slope field corresponding to y' = x^2 e^-x

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RESPONSE -->

y' = x^2 e^-x

is the graph of -x^2e^(-x)

it's concave down and has a horisontal asymptote around x=3

This function shares most of the characteristics of the previous function y ' = x e^-x, except that x^2 is always positive so that the slopes will all be positive, on both sides of the y axis. This is consistent only with graph vi.

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13:33:45

describe the slope field corresponding to y' = e^-(x^2)

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RESPONSE -->

y' = e^-(x^2)

is the graph of y = 02xe^(-x^2)

the gpach is sriahg until it goes up appoximately 3 ten levels back out. it passes though (0,0)

For x = 0, we would have y ' = e^0 = 1, so that the slopes through the y axis would all be 1.

e^-(x^2) is always positive so there would be no negative slopes.

As we move away from the y axis in either direction, e^(-x^2) rapidly approaches 0.

So the graph must be that depicted in v.

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13:34:29

describe the slope field corresponding to y' = e^-x

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RESPONSE -->

y' = e^-x

is the graph of y = -e^(-x)

it is con cave down and has a veritcal asymptote around x = -2

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}Ŕ̈́

assignment #018

018. `query 18

Cal 2

08-01-2007

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13:38:04

what is your estimate of y(1)?

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RESPONSE -->

y(1) = 0.5417

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13:40:35

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

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RESPONSE -->

x y

0 0

.2 0.0016

.4 0.0656

.6 0.2813

.8 0.7710

1 0.5417

** The table below summarizes the calculation. x values starting at 0 and changing by `dx = .2 are as indicated in the x column. y value starts at 0.

The value of y ' is found by evaluating x^3 - y^3. `dy = y ' * `dx and the new y is the previous value of y plus the change `dy.

Starting from (0,0):

m = 0, delta y = 0.2*0 = 0, new point (0.2,0)

m = 0.2^3 = 0.008, delta y = 0.2*.008 = 0.0016, new point (0.4, 0.0016)

m = 0.4^3 - 0.0016^3 = 0.064, delta y = 0.2*0.064 = 0.0128, new point (0.6, 0.0144)

m = 0.6^3 - 0.0144^3 = 0.216, delta y = 0.2*0.216 = 0.0432, new point (0.8,0.0576)

m = 0.8^3 - 0.0576^3 = 0.512, delta y = 0.2*0.512 = 0.1024, new point (1.0, 0.16)

y(1) = 0.16

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13:42:20

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 -->

over estimate

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** If you follow the slope field starting at x = 0 and move to the right until

you reach x = 1 your graph will at first remain almost flat, consistent with

the fact that the values in the Euler approximation don't reach .1 until x has

exceeded .6, then rise more and more quickly. The result y(1) = .448 is very

plausible. **

** The slope field is rising and concave up, so that the slope at the left-hand endpoint will be less than the slope at any other point of a given interval. It follows that an approximation based on slope at the left-hand endpoint of each interval will be an underestimate. **

** Euler's method multiplies the value of y ' at the left-hand endpoint of each subinterval to calculate the approximate change dy in y, using the approximation dy = y ' * `dx. This result is calculated for each interval and the changes dy are added to approximate the total change in y over the interval.

The left-hand Riemann sum for y ' is obtained by calculating y ' at the left-hand endpoint of each subinterval then multiplying that value by `dx to get the area y ' `dx of the corresponding rectangle. These areas are added to get the total area over the large interval.

So both ways we are totaling the same y ' `dx results, obtaining identical final answers. **

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13:44:18

what is your solution to the problem?

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RESPONSE -->

B= 75e^(2-2t) + 25

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13:47:21

What is the general solution to the differential equation?

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RESPONSE -->

B= 75e^(2-2t)+ 25

1 = 75e^(2-2t)+ 25

8/25 = e^(2-2t)

ln(8/25) = (2-2t) lne

-1.139 = 2-2t

-3.139 = -2t

t = 1.570

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13:48:38

Explain how you separated the variables for the problem.

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RESPONSE -->

B= 75e^(2-2t)+ 25

dB/dt +2B = 50

dB + 2B = 50 dt

Good. The following is included for your reference:

** We can separate variables.

We get dB/dt = 50 - 2 B, so that dB / (50 - 2B) = dt.

Integrating both sides we obtain

-.5 ln(50 - 2B) = t + c, where c is an arbitrary integration constant.

This rearranges to

ln | 50 - 2B | = -2 (t + c) so that

| 50 - 2B | = e^(-2(t+c)). Since B(1) = 100 we have, for t = 1, 50 - 2B =

-150 so that |50 - 2 B | = 2 B - 50. Thus

2B - 50 = e^(-2(t+c)) and

B = 25 + .5 * e^(-2(t+c)).

If B(1) = 100 we have

100 = 25 + .5 * e^(-2 ( 1 + c) ) so that

e^(-2 ( 1 + c) ) = 150 and

-2(1+c) = ln(150). Solving for c we find that

c = -1/2 * ln(150) - 1.

Thus

B(t) = 25 + .5 e^(-2(t - 1/2 ln(150) - 1))

= 25 + .5 e^(-2t + ln(150) + 2))

= 25 + .5 e^(-2t) * e^(ln(150) * e^2

= 25 + 75 e^2 e^(-2t)

= 25 + 75 e^(-2 (t 1) ).

Note that this checks out:

B(1) = 25 + 75 e^2 e^(-2 * 1)

= 25 + 75 e^0 = 25 + 75 = 100.

Note also that starting with the expression

| 50 - 2B | = e^(-2(t+c)) we can write

| 50 - 2B | = e^(-2c) * e^(-2 t). Since e^(-2c) can be any positive number we replace this factor by C, with the provision that C > 0. This gives us

| 50 - 2B | = C e^(-2 t) so that

50 2 B = +- C e^(-2 t), giving us solutions

B = 25 + C e^(-2t) and

B = 25 C e^(-2t).

The first solution gives us B values in excess of 25; the second gives B values less than 25.

Since B(1) = 100, the first form of the solution applies and we have

100 = 25 + C e^(-2), which is easily solved to give

C = 75 e^2.

The solution corresponding to the given initial condition is therefore

B = 25 + 75 e^-2 e^(-2t), which is simplified to give us

B = 25 + 75 e^(-2(t 1) ). **

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13:48:44

What did you get when you integrated the separated equation?

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RESPONSE -->

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13:53:25

what is your solution to the problem?

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RESPONSE -->

t dx.dt = (1 + 2 ln t ) tan x, 1st quadrant

I was very confused on this problem I went round and round and couldn't get the antiderivative of t/(1+2lnt)

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13:53:27

What is the general solution to the differential equation?

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RESPONSE -->

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13:53:29

Explain how you separated the variables for the problem.

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RESPONSE -->

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13:53:31

What did you get when you integrated the separated equation?

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RESPONSE -->

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Jی걛EnEWÅ

assignment #019

019. `query 19

Cal 2

08-01-2007

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13:54:54

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 -->

M= 1000^r(t-2000)

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13:56:38

What is the solution to the equation?

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RESPONSE -->

dM/dt = 1000e^(r(t-2000))

1/1000 dM = e^r(t-2000) dt

M/1000 = e^(r(t-2000)

** Compare the details with the following:

We separate variables to obtain

dM / M = r * dt so that

ln | M | = r * t + c and

M = e^(r * t + c) = e^c * e^(r t), where c is an arbitrary real number.

For any real c we have e^c > 0, and for any real number > 0 we can find c such

that e^c is equal to that real number (c is just the natural log of the

desired positive number). So we can replace e^c with A, where it is

understood that A > 0.

We obtain general solution

M = A e^(r t) with A > 0.

Specifically we have M ( 0 ) = 1000 so that

1000 = A e^(r * 0), which tells us that 1000 = A. So our function is

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

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14:00:17

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

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RESPONSE -->

dM/dt = 1000e^(r(t-2000)

1000e^(0.05(2030-2000))= $4481.70

1000e^(0.1(2030-2000))= $200085.54

the 5% graph grows much more slowly than than 10% graph.

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14:00:55

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

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RESPONSE -->

Very close, yes

** We see that at t = 1 the doubled interest rate r = .10 results in an increase of $105.17 in principle, which is more than twice the $51.27 increase we get for r = .05. **

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14:05:11

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 -->

dH/dt = -k (H-10)

H-10 = Be^(-kt)

68-58e^(-k7)

1.1724 = e ^(-7k)

ln(1.1724) = -7klne

.1591 = -7k

-0.02 = k

Close but you didn't find the temperature at 7 am and your numbers might not be completely accurate. Compare with the following:

** Assuming that dT / dt = k * (T - 10) we find that

dT / (T - 10) = k dt. Integrating both sides we get

ln | T - 10 | = k t + c so that

| T 10 | = e^(k t + c) = e^c * e^kt. Since e^c can be any positive number we use A = e^c and write our solution as

| T 10 | = A e^(k t).

If T > 10 (temperature above 10 deg F) then | T 10 | = T 10 so that our solution is

T 10 = A e^(k t) and

T = A e^(k t) + 10.

Counting clock time t from 1 pm we have

T(0) = 68 and

T(9) = 57

giving us equations

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

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

We subtract 10 from both sides and note that e*(k * 0) = e^0 = 1 to get

58 = A and

47 = A e^(9 k)

The first equation tells us that A = 58. The second equation becomes

47 = 58 e^(9 k) so that

e^(9 k) = 47 / 58 and

9 k = ln(47 / 58) so that

k = 1/9 * ln(47/58) = -.0233, approx..

Our equation is therefore

T(t) = 58 * e^(-.0233 t) + 10.

At 7 am the clock time will be t = 16 so our temperature will be

T(16) = 58 * e^(-.0233 * 16) = 50.0, approx.. (actual result 49.95 rounds to 3 significant figures to give us 50.0).

At 7 a.m. the temperature will be about 50.0 deg. **

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14:05:23

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 -->

that it was zero

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14:12:05

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 -->

This problem really confused me from the begining. I' no sure how to respond to what it's asking.

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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14:15:06

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 -->

I got hte problems at the beginning and the ones you gave have thrown me can you help explain this one also. Thanks.

rate of expansion of universe: (R')^2 = 2 G M0 / R + C; case C = 0

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

F = m*a and a = dv/dt, so

F = m*(dv/dt).

F = mgR^2/(R+h)^2 is given. Substituting this expression for F in the preceding, and considering that the acceleration is back toward the Earth so that this force is in the direction opposite increasing r and hence negative, we get the differential equation

m*(dv/dt) = -mgR^2/(R+h)^2

Dividing both sides by m:

dv/dt = -gR^2/(R+h)^2.

Since h is the position variable its derivative with respect to t is the velocity--i.e., dh/dt = v. Thus since dv/dt = dv/dh * dh/dt we have

dv/dt = dv/dh * v.

Substituting for dv/dt in differential equation from above:

v*dv/dh = -gR^2/(R+h)^2. We separate variables to get

v dv = -gR^2/(R+h)^2 dh the integrate both sides to get

(v^2)/2 = -gR^2*[-1/(R+h)] + C; the two negatives on the right-hand side are multiplied so we have

(v^2)/2 = gR^2/(R+h) + C.

Since v = v0 (v0 being the escape velocity) at h = 0

(v0^2)/2 = gR^2/(R+0) + C so

(v0^2)/2 = gR + C and

C = (v0^2)/2 - gR.

Thus,

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

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

As h>>infinity, 2*gR^2/(R+h) ->0, leaving

v^2 = v0^2 - 2*gR.

If the velocity v approaches anything greater than 0 as h -> infinity, the initial velocity must have been greater than minimum escape velocity. So for minimum escape velocity v, the limiting velocity as h -> infinity, is 0.

Therefore

v0^2 - 2*gR = 0 so

v0^2 = 2 g R and

v0 = +- sqrt( 2 g R).

Since v0 is positive we have

v0 = sqrt(2 g R).

Minimum escape velocity is therefore sqrt (2gR).

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Good work overall. See my notes and let me know if you have additional questions.