Assignment 19

course Mth 174

ǩx܄ᷭzassignment #019

019. `query 19

Cal 2

06-26-2009

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10:26:25

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

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RESPONSE -->

dM/dt = .05M

M = M(sub0) e^(.05t)

M = 1000 e^(.05)(2000)

M = 1000 e^ 100

M =2.689 * 10^46

Your solution is OK, but t = 2000 is not a value associated with this problem.

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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10:26:37

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 -->

dM/dt = .05M

M = M(sub0) e^(.05t)

M = 1000 e^(.05)(2000)

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10:26:44

What is the solution to the equation?

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RESPONSE -->

M =2.689 * 10^46

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10:31:02

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

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RESPONSE -->

My graph has time as the x-axis and amount in dollars as the y-axis. Both lines begin at (0, 1000) and the one representing 10% interest is decreasing at a slower rate than the one at 5%

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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10:32:06

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

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RESPONSE -->

The doubled interest rate means that the rate of compounding money will decrease more slowley than the lower rate.

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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10:57:40

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 -->

Instead of T I used H so that t would be time.

dH/dt = -k(H- 10)

dH/ (H - 10) = -k dt

after integrating both sides I got

ln(H - 10) = -kt dt

H- 10 = e^-kt e^c

H - 10 = Be^-(kt)

Then I found B

68 - 10 = Be^-k(0) = B

B = 58

H - 10 = 58e^-(kt)

Since it is 57degrees F at 10:00 p.m.:

t = 10 - 1 = 9 hours

57 - 10 = 58e^(-k(9))

ln(47/58) = -9k

-.21 = -9k

k = .02

So the new equation is:

H - 10 = 58e^(-.02 * t)

At 7 am t= 1pm - 10 pm the next day = 18 hours

H - 10 = 58e^(-.02 * 18)

H - 10 = 40

H = 50 degrees F at 7 am the next morning

Good.

You rounded off your value of k to only one significant figure. Check the solution below, which uses three significant figures (though 2 would probably suffice).

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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10:58:44

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 -->

I made the assumption that the outside temperature was constant 10 degrees F without fluctuation. If I assumed that the temp. got colder in the night I would assume that the temperature in the house at 7am would be colder than what I found.

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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 was very confusing for me. I wasn't sure where to start. I tried taking the equation and integrating both sides but I could never find the answer.

Good student solution:

F = m*a 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:

v*dv/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 = vzero at H = 0

(vzero^2)/2 = gR^2/(R+0) + C

(vzero^2)/2 = gR + C

C = (vzero^2)/2 gR.

Thus,

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

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

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

So

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

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

Minimum escape velocity occurs when vzero^2 = 2gR,

Thus minimum escape velocity = sqrt (2gR).

&#Your work looks good. See my notes. Let me know if you have any questions. &#