Assignment 19 Query

course Mth 174

?????y???????assignment #019019. `query 19

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Cal 2

12-15-2007

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09:17:33

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

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RESPONSE -->

a) I found the differential equation to be:

dM / dt = rM

b) Solving the differential equation:

dM / dt = rM

dM / M = r dt

ln |M| = rt + C

M = e^(rt + C)

M = e^(rt) e^C (e^C = A)

M = Ae^(rt)

Since M=1000 at t=0, A = 1000:

M = 1000e^(rt)

c) With a rate of 5%, the graph slopes upwards reaching a maximum of 4481.69 at t=30. The graph is concave up with no inflection points, no asymptotes, etc.

With a rate of 10%, the graph has a greater rate and reaches a maximum of 20,085.54 at t=30 with the same properties of the other graph.

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09:17:45

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 = rM

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09:18:05

What is the solution to the equation?

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RESPONSE -->

M = 1000e^(rt)

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09:18:11

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

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RESPONSE -->

See above.

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09:19:00

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

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RESPONSE -->

Since it is exponential, doubling the interest rate slighly more than quadruples the increase in principle over 30 years.

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09:30:46

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 -->

a) I used the differential equation:

dT / dt = a(T - 10)

b) To solve it, I needed to adjust the sign of a because the change is negative (a higher temperature going to a lower temperature). Next I needed to seperate variables:

dT / (T-10) = -a dt

ln |T-10| = -at + C

T-10 = Ae^(-at)

T = 10 + Ae^(-at)

Now I can solve with T=68 at t=0:

68 = 10 + A

A = 58

Next I can solve for a by using T=57 at t=9:

57 = 10 + 58e^( -a(9) )

ln | 47 / 58 | = -a(9)

a = 0.0234

Now I have the complete equation:

T = 10 + 58e^(-0.0234t)

Solving for t=18:

T = 48

c) The assumption here is that the external temperature stayed the same from 1pm to 10pm. In fact, the external temperature probably dropped during this time period, widening the difference between the internal and external temperatures and speeding up the decrease in internal temperature. I would then revise the 7am estimate down.

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09:30:52

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 -->

See above.

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09:50:13

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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RESPONSE -->

Hmm. I think it goes like this:

dC / dt = k(A + B)

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10:09:49

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 -->

I missed something, because to solve this equation I need to have C somewhere. I am just not sure where.

The problem states that C is formed by the combination of a molecule of A and a molecule of B, and that the rate of combination is proportional to the product of the numbers of molecules of A and B present.

If x molecules of C have been formed, then x molecules of A and of B will have been used up.

If the initial numbers of A and B are respectively a and b, then if x molecules of C have been formed there will be a – x molecules of A and b – x molecules of B.

This product of the numbers of molecules of A and of B present will be (a – x) * (b – x), so the equation will be

dx/dt = k (a - x)(b - x).

If a and b are the same then a = b and we can write

dx / dt = k ( a - x) ( b - x) = k ( a - x) ( a - x).

The equation is therefore

dx / dt = k ( a-x)^2. Separating variables we have

dx / (a - x)^2 = k dt. Integrating we have

1/(a-x) = k t + c so that

a - x = 1 / (k t + c) and

x = a - 1 / (kt + c).

If x(0) = 0 then we have

0 = a - 1 / (k 0 * c) so that

c = 1 / a.

Thus

x = a - 1 / (k t + 1/a) = a - a / (a k t + 1) = a ( 1 - 1/(akt + 1) ). **

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10:10:42

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

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10:56:17

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 -->

a) Based on previous information in the chapter:

F = m (dv/dt) and here it says

F = m g R^2 / (R + h)^2

Setting these equal to one another:

m (dv/dt) = m g R^2 / (R + h)^2

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

b) Working backwards:

dv/dh = gR^2/( v (R+h)^2 )

So dh/dt must equal:

dh/dt = -v

c) Solving the last differential equation gives me:

v dv = gR^2/(R+h)^2 dh

1/2 v^2 = -gR^2/(R+h)

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

But this isn't right, and I don't know what I missed.

Good attempt. The remaining details are included in this following very good student solution:

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

As h -> infinity, 2gR^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).

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10:58:47

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 -->

Interesting.

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11:27:36

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

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RESPONSE -->

I am assuming R' = dR/dx so I have a variable of integration.

dR/dx = `sqrt( 2 G M0 / R )

dR/dx = 'sqrt( 2 G M0) / 'sqrt R

dR / 'sqrt R = 'sqrt( 2 G M0) dx

2 'sqrt R = 'sqrt( 2 G M0) x

4 R = ( 2 G M0) x^x

R = ( 2 G M0 x^2 ) / 4

This then satisfies for x=0, R=0.

R ' = `sqrt( 2 G M0 / R ) gives you

dR/dt = 'sqrt( 2 G M0) / 'sqrt R so that
dR * sqrt(R) = sqrt(2 G M0) dt and
2/3 R^(3/2) = sqrt(2 G M0) t + c.

Since R(0) = 0, c = 0 and we have

R = ( 3/2 sqrt( 2 G M0) t)^(2/3).

This can be simplified, but the key is that R is proportional to t^(2/3), which increases without bound as t -> infinity.

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11:30:36

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 -->

To determine the long-term nature, I think all I need to do is look at the value of R as x -> infinity (to take the limit, in other words).

With the equation I found, there is no limit to the growth of R.

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11:30:53

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 -->

Ah, another interesting one. This is problem 11.6.8 in 4th edition.

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12:52:55

what is your intensity function?

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RESPONSE -->

Well, we can assume that the farther light goes in, the more is absorbed and so the rate of absorbtion decreases the farther in light goes. So we have P being the absorbtion value and x being the depth. Then:

dP / dx = -k(100-P)

The rate is negative because the absorbtion value is dropping. The function is then:

100 - P = Ae^(-kx)

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12:54:05

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 -->

To find k, the function needs to be rearranged:

k = - ln | (100 - P)/A | / x

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19:31:29

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 -->

First I need to find a value of A. Since at x=0, P=0:

A = 100

To find k, just substitute known values of P, A, and x using P=50 and x=10:

k = - ln | (100 - 50) / 100 | / 10 = 0.0693

The equation becomes:

100 - P = 100e^(-0.0693x)

P = 100 - 100e^(-0.0693x)

So for x=20, P=75%.

For x=25, P=82.3%.

The rate of absorption, in units of intensity per unit of distance, is proportional to the intensity of the light.

If I is the intensity then the equation is

dI/dx = k I.

This is the same form as the equation governing exponential population growth and the solution is easily found by separating variables. We get

I = C e^(-k x).

If the initial intensity is I0 then the equation becomes

I = I0 e^(-k x).

If 50% is absorbed in the first 10 ft then the intensity at that position will be .50 I0 and we have

.5 I0 = I0 e^(-k x) so that

e^(-k * 10 ft) = .5 and
k * 10 ft = ln(2) so that
k = ln(2) / (10 ft).

Then at x = 20 ft we have

I = I0 * e^(- ln(2) / 10 ft * 20 ft) = I0 * e^-(2 ln(2) ) = .25 I0, i.e., 25% will be left

At x = 25 ft we have

I = I0 * e^(- ln(2) / 10 ft * 25 ft) = I0 * e^-(2.5 ln(2) ) = .18 I0 (approx), i.e., about 18% will be left.

Note that the expressions e^(-2 ln(2)) and e^(-2.5 ln(2)) reduce to .5^2 and .5^(2.5), since e^(-ln(2)) = 1/2.

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19:31:40

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

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RESPONSE -->

Neat applications.

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Assignment 19 Query

Good attempt. The remaining details are included in this following very good student solution:

I agree; unfortunately some of the neatest have been omitted from the latest edition. Thanks for locating that absorption problem.

I agree; unfortunately some of the neatest have been omitted from the latest edition.

Thanks for locating that absorption problem.