Assignment 19

course MTH 174

£Ãìùä‘mßäÃÝ’¢ìÆ Ì„Šý¬x¼assignment #019

019. `query 19

Cal 2

11-19-2008

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21:22:25

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

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RESPONSE -->

a) dM/dt = M * 0.05

b) 1/M dM = 0.05 dt

integral 1/M dM = integral 0.05 dt

ln |M| = 0.05t + C

M = e^(0.05t + C)

M = e^0.05t * e^C

At t= 0, M = 1000

1000 = e^0.05(0) * e^C

1000 = e^C

So M = 1000 * e^0.05t

c) The sketch starts out with both graphs starting at y=1000. The graph of 10% grows more quickly, while the graph of 5% grows more slowly. At t=30, the 10% graph is approx. $20085 and the 5% graph is approx $4881

Good. A more general solution in terms of r, which is completely analogous to your solution but is worth looking over:

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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21:23:34

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 -->

a) dM/dt = M * 0.05

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21:23:49

What is the solution to the equation?

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RESPONSE -->

b) 1/M dM = 0.05 dt

integral 1/M dM = integral 0.05 dt

ln |M| = 0.05t + C

M = e^(0.05t + C)

M = e^0.05t * e^C

At t= 0, M = 1000

1000 = e^0.05(0) * e^C

1000 = e^C

So M = 1000 * e^0.05t

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21:23:57

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

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RESPONSE -->

c) The sketch starts out with both graphs starting at y=1000. The graph of 10% grows more quickly, while the graph of 5% grows more slowly. At t=30, the 10% graph is approx. $20085 and the 5% graph is approx $4881

Good. Again for comparison:

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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21:25:14

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

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RESPONSE -->

No, it is actually much more. It is closer to 4.5 times more

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21:48:51

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)

dH/dt = k(10 - H)

1/(10 - H) * dH = k * dt

integral of 1/(10 - H) * dH = integral of k * dt

ln |10 -H| = kt + C

10 - H = e^(kt + C)

10 - H = e^C * e^kt

At t=0, H = 68

10 - 68 = e^C * e^k(0)

10 - 68 = e^C = -58

Continuing on:

10 - H = -58 * e^kt

- H = -58 * e^kt - 10

H = 10 + 58 * e^kt

At 10pm, t = 9 and H = 57

57 = 10 + 58 * e^k(9)

47 = 58 * e^k(9)

ln (47/58) = 9k

ln (47/58)/9 = k

k = approx. -0.0234

At 7am, t = 18:

H = 10 + 58 * e^(-0.0234*18)

H = 48 F

c)The above work assumes that the temperature outside is a constant 10 F, which is unlikely because it was likely warmer outside at 1pm than it was at 10pm. It will probably not drop to 48 inside before the morning because the estimate would have exagerated the drop from 68 to 57 because we assumed it was 10 outside the whole time. However, at 10pm, the temperature might drop a bit more before the morning, though probably not enough to make the pipes freeze inside the house. I think the model would work through the night because temperature outside is unlikely to change a lot during those 8 hours.

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21:49:38

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 -->

c)The above work assumes that the temperature outside is a constant 10 F, which is unlikely because it was likely warmer outside at 1pm than it was at 10pm. It will probably not drop to 48 inside before the morning because the estimate would have exagerated the drop from 68 to 57 because we assumed it was 10 outside the whole time. However, at 10pm, the temperature might drop a bit more before the morning, though probably not enough to make the pipes freeze inside the house.

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19:45:18

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 -->

As C gets bigger, the quantities of a and b get smaller.

From problem 9, the equation looks like this:

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

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

Since both quantities of a and b are the same, we can use just a as the variable leaving

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

dx /(a - x)^2 = k dt

Using w = a - x:

dw/dx = -1

-dw = dx

-dw/w^2 = k dt

integral of -dw/w^2 = integral of k dt

1/w = kt + C

1/(a - x) = kt + C

At x(0)=0:

1/(a - 0) = k(0) + C

1/a = C

Substituting in for C:

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

1 = (kt + 1/a) * (a - x)

1 = akt -xkt + 1 -x/a

1 = -xkt -x/a + akt + 1

1 = -x(kt + 1/a) + akt + a

1 - akt - a = -x(kt + 1/a)

(1 - akt - a) / (kt + 1/a) = -x

-(1 - akt - a) / (kt + 1/a) = x

(a + akt - 1) / (kt + 1/a) = x

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19:46:18

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 -->

1/(a - x) = kt + C

At x(0)=0:

1/(a - 0) = k(0) + C

1/a = C

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19:46:24

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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20:21:02

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)

F = mgR^2 / (R + h)^2

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

However, the object is accelerating in a direction opposite to g, so F = -m dv/dt

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

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

b)

Applying the chain rule: dv/dt = dv/dh * dh/dt

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

dh/dt is the rate of change in height, or velocity, so:

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

c)

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

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

integral of v * dv = integral of -gR^2 / (R + h)^2 * dh

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

Using w = R + h:

dw/dx = 1

dw = dx

v^2 / 2 = -gR^2 integral dw w^-2

v^2 / 2 = -gR^2 * -w^-1 + C

v^2 / 2 = -gR^2 * -1/w + C

v^2 / 2 = gR^2 / w + C

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

When h=0

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

v0^2 = 2gR^2 / R + C

v0^2 = 2gR + C

C = v0^2 - 2gR

Substituting back in for C:

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

As h -> infinity, 2gR^2 / (R + h) gets closer to 0

v^2 = v0^2 - 2gR

v^2 + 2gR = v0^2

If v was 0, then `sqrt(2gR) = v0

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

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

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

It will not take infinitely long to escape the gravitational pull, so v0 = sqrt(2gR) ensures that v will not be 0.

a)

F = mgR^2 / (R + h)^2

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

However, the object is accelerating in a direction opposite to g, so F = -m dv/dt

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

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

b)

Applying the chain rule: dv/dt = dv/dh * dh/dt

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

dh/dt is the rate of change in height, or velocity, so:

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

c)

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

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

integral of v * dv = integral of -gR^2 / (R + h)^2 * dh

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

Using w = R + h:

dw/dx = 1

dw = dx

v^2 / 2 = -gR^2 integral dw w^-2

v^2 / 2 = -gR^2 * -w^-1 + C

v^2 / 2 = -gR^2 * -1/w + C

v^2 / 2 = gR^2 / w + C

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

When h=0

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

v0^2 = 2gR^2 / R + C

v0^2 = 2gR + C

C = v0^2 - 2gR

Substituting back in for C:

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

As h -> infinity, 2gR^2 / (R + h) gets closer to 0

v^2 = v0^2 - 2gR

v^2 + 2gR = v0^2

If v was 0, then `sqrt(2gR) = v0

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

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

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

It will not take infinitely long to escape the gravitational pull, so v0 = `sqrt(2gR) ensures that v will not be 0.

The escape velocity, v0 is `sqrt(2gR)

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20:49:50

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 -->

(R')^2 = 2 G M0 / R + C where C = 0

dR/dt = (2 G M0 / R)^1/2

dR/dt = (2 G M0)^1/2 / R^1/2

R^1/2 dR = (2 G M0)^1/2 dt

integral of R^1/2 dR = integral of `sqrt(2 G M0) dt

2/3 * R^3/2 = `sqrt(2 G M0) * t + C

Assuming R=0 at t=0:

2/3 * (0)^3/2 = `sqrt(2 G M0) * 0 + C

C = 0

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

4/9 * R^3 = 2 G M0 * t^2

R^3 = 9/2 G M0 * t^2

R = (9/2 G M0 * t^2)^1/3

R expands in a near linear fashion according to this equation.

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20:49:58

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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20:50:04

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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21:27:36

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 -->

The rate at which light is absorbed in water is proportional to the intensity

a)

dQ/dx = kQ

integral of 1/Q dQ = kdx

ln |Q| = kx + C

Q = e^(kx + C)

Q = e^C * e^(kx)

1/2 Q0 = Q0e^(10k)

1/2 = e^(10k)

ln (1/2) = 10k

k = (ln (1/2))/10

k = -0.069

At 20 ft, e^(-0.069 * 20) = approx 25.2%

At 25 ft, e^(-0.069 * 25) = approx 17.8%

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21:28:57

what is your intensity function?

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RESPONSE -->

Q = Q0 * e^(-0.069x)

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21:29:06

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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21:29:20

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 -->

At 20 ft, e^(-0.069 * 20) = approx 25.2%

At 25 ft, e^(-0.069 * 25) = approx 17.8%

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21:29:42

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

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RESPONSE -->

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Excellent work. See my notes and let me know if you have questions.