Assignment 19

course Mth 174

???|???y?r??L?r?C??assignment #019019. `query 19

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Cal 2

04-15-2007

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14:37:37

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

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14:38:31

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

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dM/dt = k*M

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14:45:48

What is the solution to the equation?

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dM/dt = r*M

dM/M = r*dt

Integral of dM/M = Integral of r*dt

ln M = rt + C

M = e^(rt) + C

M = Ce^(rt)

Since M = Mzero when t=0,

C = Mzero

M = Mzero*e^(rt)

Mzero =$1000

M = 1000e^(rt)

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14:55:47

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

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RESPONSE -->

Both curves begin at t= 0, M = $1000, both are concave up. For r = 5%, curve increases slowly, with M at 2010 = $1649, M at 2020 = $2718, M at 2030 = $4482. For r =- 10%, curves increases more rapidly, with M at 2010 = $2718, M at 2020 = $7389, M at 2030 = $20,086 (all M approx.).

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15:02:39

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

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No. Because these are exponential functions, doubling the interest rate will result in a doubled increase in principle at some point, after which the principle will be more than double (for the doubled interest rate compared to the other) as the curves grow further and further apart.

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15:19:50

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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T= temperature in house (degrees F)

t = time (in hours from 1 p.m.)

dT/dt = -k(T-10)

dT/T-10 = -k dt

ln (T-10) = -kt + C

T-10 = Be^(-kt)

T = Be^(-kt) + 10

At t = 0, T = 68

68 = Be^(0) + 10

B = 58

T = 58e^(-kt) + 10

At t = 9 (10 p.m.), T = 57

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

47/58 = e^(-k*9)

ln (47.58) = -9k

k = 0.0234 (approx.)

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

At 7 a.m., t = 18

T(18) = 58e^[-0.0234(18)] + 10

t = 48 degrees F (approx.)

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15:23:10

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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The assumption is that the outside termperature is constant, when in fact this is likely not the case. Under normal conditions (i.e., no fronts or storms coming through), the outside temperature would probably rise slightly from 1 p.m. for 2 hours or so, then drop quite a bit during the night. Therefore, if the change in outside temperature is accounted for, the interior temperature would likely be lower at 7 a.m. than predicted by the equation used.

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15:25:03

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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15:54:20

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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I don't know that there is enough information given, so I made some assumptions:

(1) The quantity of A is changing at some fixed rate--similar to an interest rate, so that the change in quantity of A = A*r(sub A).

(2) Same for B, change in quantity of B = B*r(sub B).

Since the initial quantity of A = B, the total quantity of A + B at any time is [2A + A*r(sub A) + B*r(sub B)] (if the presence of A or B is decreasing the value of the corresponding r would be negative, so using + signs is not a problem.)

Thus,

dx/dt = k*[2A + A*r(sub A) + B*r(sub B)]

Solving:

dx = k*[2A + A*r(sub A) + B*r(sub B)] dt

x = k*[2A + A*r(sub A) + B*r(sub B)]*t + D

Since x(0) = 0,

x(0) = 0 k*[2A + A*r(sub A) + B*r(sub B)]*0 + D

D = 0

x = k*[2A + A*r(sub A) + B*r(sub B)]*t

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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15:55:14

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

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RESPONSE -->

k = x/[2A + A*r(sub A) + B*r(sub B)]*t

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16:20:31

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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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 accelation is measured in opposite direction from force of gravity)

Dividing both sides by m:

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

Using dv/dt = dv/dh * dh/dt and because dh/dt = v

Substituting v for dh/dt:

dv/dt = dv/dh * v

Substituting for dv/dt in differential equation from above:

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

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

Leaving,

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).

Excellent solution.

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16:23:32

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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16:46:35

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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R' = sqrt (2GM0/R)

dR/dt = sqrt(2GM0)/sqrt(R)

sqrt(R) dR = sqrt(2GM0) dt

Integrate both sides:

(2/3)R^(3/2) = sqrt(2GM0)*t + C

Since R(0) = 0, C = 0

R^(3/2) = (3/2)sqrt(2GM0)*t

Raise both sides to (2/3) power:

R = [(3/2)sqrt(2GM0)*t]^(2/3)

Since the right side is a constant * t raised to the (2/3),

the equation can be rewritten as: R = (kt)^(2/3).

As t>>infinity, R will grow without bound but at a constantly decreasing rate, thus, assuming the equation is valid, the universe will continue to expand without bound but at a decreasing rate.

R = kt^(2/3)

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16:48:56

what is your intensity function?

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Not enough information to do problem. I think something got chopped off the question.

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16:49:09

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

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Cannot do problem.

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16:49:25

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

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Cannot do problem.

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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16:50:31

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

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Apparently the last problem given is missing some information.

Very interesting stuff. It's amazing what can be done with differentiatial equations (and I'm sure this stuff is just the tip of the iceberg).

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If you've got multivariate calculus in your curriculum then chances are excellent that you've got a differential equations course in your future. It is extremely interesting stuff.