Assignment 19

course Mth 174

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019. `query 19

Cal 2

08-05-2008

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09:32:47

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

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09:32:49

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

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09:32:51

What is the solution to the equation?

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09:33:06

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

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09:33:10

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

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09:33:32

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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09:33:34

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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09:33:55

Query problem NOT IN 4th EDITION???!!! 11.6.16 (was 10.6.10) C formed at a rate proportional to presence of A and of B, init quantities a, b the same

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09:33:58

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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09:34:00

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

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09:34:09

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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09:34:21

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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09:34:23

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

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09:34:25

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

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09:34:40

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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09:34:42

what is your intensity function?

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09:34:44

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

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09:34:48

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

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09:34:49

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

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assignment #019

019. `query 19

Cal 2

08-05-2008

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12:02:51

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

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12:36:15

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

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I first tried to use the P(0) * e^-kt, but I ended up getting a bad answer and it seemed wrong. So my equation was M = P(0) * (1 + r)^t. dP/dt = P(1 + r)^t. (1/p) dP = (1 + r)^t dt. (My first attempt went bad because I took the integrals and on the right side I had (1 / t+1) * (1 + r) ^ (t+1), and this caused problems, but I will show my attempt.)

ln[abs(1/P)] = (1 / t+1) * (1 + r) ^ (t+1) + C -> abs(1/P) = e ^ [(1 / t+1) * (1 + r) ^ (t+1)] * e^C -> 1/P = B * e^[(1 / t+1) * (1 + r) ^ (t+1)] where A = e^C and B = +-A. After multiplying then dividing by P I have P = 1 / [B * e^[(1 / t+1) * (1 + r) ^ (t+1)]]

The equation is dM/dt = r * M.

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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12:40:55

What is the solution to the equation?

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12:57:30

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

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I didn't answer the last question. So i will answer it now. I solved for be to a decimal answer which I do not like, so I just used this B = 1 / (P * e^[(1 / t+1) * (1 + r)^(t + 1)]) When t = 0 and r = .05 and P = 1000, B = 1 / (e^(1.05)(1000)). After thrity years I found the value of the money to be $12968.13.

The graph of 1000 e^(.05 t) is an exponential graph passing through (0,
1000) and (1, 1000 * e^.05), or approximately (1, 1051.27).

The graph of 1000 e^(.10 t) is an exponential graph passing through (0, 1000)
and (1, 1000 * e^.10), or approximately (1, 1105.17).

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13:00:45

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

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I think it implies it, but I do not agree with it. I also do not know how to refute it though.

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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13:17:37

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My equation was dH/dt = -k(H - 10).

dH / H - 10 = -k dt

-> ln[abs(H - 10)] = -kt + C -> abs(H - 10) = e^-kt * e^C

-> H - 10 = Be^-kt where B = +-A and A = e^C

-> To solve for B: 68 - 10 = B * e^-k(0) = 58 = B

-> To solve for k, plug in B, 57 for H, and 9 for t: 57 - 10 = 58 * e^-9k

-> 47/58 = e^-9k

-> ln(47 / 58) = -9k

-> ln(47 / 58) / -9 = k is approx. .023366

At seven a.m., which is 18 hours later: H - 10 = 58 * e^-k(18)

-> H - 10 = 58 * e^[-(ln(47 / 58) / -9) * 18]

-> H = 58 * e^[-.023366 * 18] - 10 = 28.08631421

Assuming that dT / dt = k * (T – 10) we find that T(t) = 10 + A e^(k t).

Counting clock time t from 1 pm we have

T(0) = 68 and
T(9) = 57

giving us equations

68 = 10 + A e^(k * 0) and
57 = 10 + A e^(k * 9).

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

57 = 10 + 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) = -.0234, approx..

Our equation is therefore

T(t) = 10 + 58 * e^(-.0234 t).

At 7 am the clock time will be t = 18 so our temperature will be

T(18) = 10 + 58 * e^(-.0234 * 18) = 48, approx..

At 7 a.m. the temperature will be about 48 deg.

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13:18:16

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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I made the assumption that it did not get colder outside. I think that as the night moved, say towards1 am, it became colder than 10 degrees.

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13:18:22

Query problem NOT IN 4th EDITION???!!! 11.6.16 (was 10.6.10) C formed at a rate proportional to presence of A and of B, init quantities a, b the same

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13:18:25

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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13:18:27

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

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13:19:25

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I couldn't figure this out because I couldn't find how to translate mgR^2 / (R + h)^2 into velocity and time.

Good student solution:

F = ma and a = dv/dt, so
F = m(dv/dt)
F = mgR^2/(R+h)^2
Substituting for F:

m(dv/dt) = -mgR^2/(R+h)^2 ( - because gravity is a force of attraction, so that the acceleration will be in the direction opposite the direction of increasing h)

Dividing both sides by m:
dv/dt = -gR^2/(R+h)^2
Using dv/dt = dv/dh dh/dt, where dh/dt is just the velocity v:

dv/dt = dv/dh * v

Substituting for dv/dt in differential equation from above:
vdv/dh = -gR^2/(R+h)^2 so
v dv = -gR^2/(R+h)^2 dh.

Integral of v dv = Integral of -gR^2/(R+h)^2 dh
Integral of v dv = -gR^2 Integral of dh/(R+h)^2
(v^2)/2 = -gR^2[-1/(R+h)] + C
(v^2)/2 = gR^2/(R+h) + C.

Since v = v_0 at H = 0

(v_0^2)/2 = gR^2/(R+0) + C
(v_0^2)/2 = gR + C
C = (v_0^2)/2 – gR.

Thus,

(v^2)/2 = gR^2/(R+h) + (v_0^2)/2 - gR
v^2 = 2gR^2/(R+h) + v_0^2 - 2gR

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

v^2 = v_0^2 - 2*gR.

v must be >= 0 for escape, therefore v_0^2 must be >= 2gR (since a negative v_0 is not possible).

Minimum escape velocity occurs when v_0^2 = 2gR,
Thus minimum escape velocity = sqrt (2gR).

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13:19:39

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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13:19:41

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

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13:19:43

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

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13:19:48

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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13:19:50

what is your intensity function?

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13:19:52

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

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13:19:54

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

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13:20:29

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

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I didn't get alot out of 11.6 other than a good amount of practice. I feel like I know what I am doing, but I could be wrong.

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Assignment 19

The graph of 1000 e^(.05 t) is an exponential graph passing through (0, 1000) and (1, 1000 * e^.05), or approximately (1, 1051.27). The graph of 1000 e^(.10 t) is an exponential graph passing through (0, 1000) and (1, 1000 * e^.10), or approximately (1, 1105.17).

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.

&#Good work. See my notes and let me know if you have questions. &#

The graph of 1000 e^(.05 t) is an exponential graph passing through (0,
1000) and (1, 1000 * e^.05), or approximately (1, 1051.27).

The graph of 1000 e^(.10 t) is an exponential graph passing through (0, 1000)
and (1, 1000 * e^.10), or approximately (1, 1105.17).