Assignment 19

#$&*

course MTH 174

4:26 pm 7/25/10.

Calculus IIAsst # 19

11-30-2002

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  ****   Query problem 11.5.12 (was 10.5.8)  $1000 at rate r

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Your solution: The differential equation here is dM/dt = rM. The solution to this is given by: 1/M dM = r dt --> ln |M| = rt + C --> M = e^(rt)* e^C= Ae^(rt). For our needs here the e^c = A for all A>0. A is also equal to the initial balance in the account since e^(r*0)= 1. So $1000= A (1) --> A= 1000. Thus the solution is M= 1000 e^(rt). Where r= rate and t= years with t=0= year 2000.

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  ****   what differential equation is satisfied by the amount of money in the

account at time t since the original investment?

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Your solution:This is also given above, and is: dM/dt= M*r.

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 dM/dt = 0.05M

** The equation is dM/dt = r * M.

The question is not posed for the specific value r = .05. **

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  ****   What is the solution to the equation

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Your solution: The solution, also given above, is: M=1000 e^(rt).

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M = 1000e^(0.05*t)

t is in years

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

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Your solution:These are obviously exponential graphs. Both with y-intercept of $1000. However the 10% graph climbs much much more quickly that the 5%. The 5% passes through the point (30, 4,482) which means that in the year 2030 the balance is $4,482. And for r=10% we get (30, 20,086) which means that in the year 2030 we have $20,085.

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  They're just exponential graphs

 

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

From the 30 year values we can see that doubling the interest rate corresponds to a much much greater return than just doubling. The 10% rate actually gives a return of approximately 4.48 times greater than the 5% rate.

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  No

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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Query problem 11.5.25 was 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.

  ****   What is your prediction of the temperature at 7 am and how did you

get it? Would you revise your estimate up or down in considering the

assumption you made in writing the differential equation?

 Your solution: Using the equation from the book we start with: dT/dt = a*∆T . For our scenario the temperature difference is: ∆T= T- 10º F. Now we are losing heat so we need to have a negative sign here so we use -k and get: dT/dt= -k(T-10). Using the separation of variables method we solve this equation and end up with: T = 10 + A e^(-kt). Solving for B at t=o and T= 68º F tells us that B= 58. Then to find k we use t= 9 and T= 57º F we get k ≈ 0.0234. Now to find T at t= 18 (7 am is 18 hours after 1 pm) is simply a matter of plugging in the value of t to get: T ≈ 48º F. This is not cold enough to worry about freezing pipes. Our approximate value is approximate because we used the outside temperature of 10º F throughout our calculations, which is the assumption that the outside temperature was constant; however, we know very well that the outside temperature would also vary with time and would most likely be greater than 10º F at 1 pm, so for a more accurate approximation we would have to find a function in terms of time to model the temperature variance outside.

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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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  ****   What is your differential equation?

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dT/dt = -k(T - 10)

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  ****   How did you solve your differential equation?

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I just worked it out by a general solution given in the book

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  ****  11.6.15 was 11.6.11 (3d edition 11.6.6).  20 cal/day maintains weight; rate of wt change is prop to difference with prop const 1/3500

  ****   What is the differential equation you would solve to get W(t) for intake I cal/day?

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Your solution: What I started with was the fact that the net calories, which directly affects any change in weight, was I - 20W. This gives the calories that will create a change in weight so it is directly proportional to change in weight and since we're given the proportionality constant the equation is then: dW/dt = 1/3500 (I- 20W). Using the method of variable separation we easily find the solution to this equation, and it is: W= I/20 - Ae^(-t/175). With our initial values we find A: 160 = 3000/20 - A(e^0) --> A= -10 --> W= 150 + 10e^(-t/175). It easy to see that this approaches 150 from above as a limit as t-->∞. This is what we see confirmed in the graph of the function. The function very slowly goes toward 150. At the end of 90 days the persons weight would be approximately 150.6 lbs. So we have a graph with y-intercept of 160 and passing through the point (90,150.6).

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  dW/dt = w/3500

** The rate of change is is proportional to the difference between the number of calories consumed and the number required to maintain weight.

The number of daily calories required to maintain weight is weight * 20, or using the notation W(t) for weight, the number of calories required to maintain weight is W(t) * 20.

If the daily intake is I then the difference between the number of calories consumed and the number required to maintain weight is I - W(t) * 20. 

Thus to say that rate of change is is proportional to the difference between the number of calories consumed and the number required to maintain weight is to say that 

dW/dt = k ( I - 20 * W).

We are told that k = 1/3500 so that

dW/dt = 1/3500 ( I - 20 W). 

The equation is solved by separation of variables:

dW / (I - 20 W) = 1/3500 * dt.  Integrating both sides we get

-1/20 ln | I - 20 W | = t / 3500 + c.  We solve for W.  First multiplying both sides by -20 we get

ln | I - 20 W | = -20 t /3500 + c  (-20 * c is still just an arbitrary

constant so we still call the result c).

| I - 20 W | = e^(-1/175 * t + c) or

| I - 20 W | = A e^(-1/175 * t), with A > 0.

If I > 20 W then we have

 I - 20 W  = A e^(-1/175 * t), with A > 0 so that

 W = I / 20 - A e^(-1/175 * t), with A > 0.

 Thus if the person consumes more than 20 W calories per day weight will approach the limit I / 20 from below. 

If I < 20 W then we have

 -(I - 20 W)  = -A e^(-1/175 * t), with A > 0 so that

 W = I / 20 + A e^(-1/175 * t), with A > 0.

 Thus if the person consumes fewer than 20 W calories per day weight will approach the limit I / 20 from above. 

If W = 160 when I = 3000 then I < 20 W and we have

 W = I / 20 + A e^(-1/175 * t), with A > 0.  At t = 0 we have

 160 = 3000 / 20 + A e^(-1/175 * 0), or so that

 160 = 150 + A so we have

 A = 10. 

Now the weight function is 

 W(t) = 150 + 10 e^(-1/175 * t). 

This graph starts at W = 150 when t = 0.  The 'half-life' for e^(-1/175 t) occurs when -1/175 t = ln(1/2), at t = -175 * ln(1/2) = 121.3 approx..  At this point 10 e^(-1/175 * t) = 10 e^(-1/175 * 121.3) = 5 and W(t) = 150 + 5 = 155.

The graph will therefore exponentially approach W = 160, passing thru points (0,150) and (121.3,155). **

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  ****   How did you solve the equation?

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  Multiplied out the dW and dt and treated them as normal variables, then

integrated

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  ****   Describe the graph of a 160 lb. person (init wt) eating 3000 cal/day.

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An exponential function

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NOTE: I don't have this problem so I can't work it but I will look over the solution to still get an idea about what is going on.

  ****   (no longer present as of 4th edition) Query   problem 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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 OK

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  ****   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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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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  ****   If your previous answer didn't include it, what is the solution in

terms of a proportionality constant k?

 

Query   problem 11.6.29 was 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.

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Your solution: It is easy to show that dv/dt = -gR^2/(R+h)^2 because we know that F= ma= m dv/dt. Here we have m(gR^2/(R+h)^2. Dividing out the mass on both sides leaves gR^2/(R+h)^2, but our acceleration must be negative since gravity is constantly degrading the velocity which means dv/dt= -a= -gR^2/(R+h)^2. The next part is simple if you just recognize that dh/dt = velocity = v. This just says that the velocity is the change of height with respect to time. So we have v dv/dh = dv/dt = -gR^2/(R+h)^2. To solve this equation we multiply through with dt. This gives us the familiar variable separation scenario. Integrating both sides makes a general solution, which is: 1/2 v^2 = gR^2(1/(R+h) + C. To find a more useful form we need to solve for C. We can do this like we have done many other times and that is to use initial values. Here we know that v=v0 at h=0. Filling in produces: 1/2 v0^2 = gR+C--> C= 1/2 v0^2 - gR --> 1/2v^2 = gR^2/(R+h)^2 + 1/2 v0^2 -gR --> v^2 = 2gR^2/(R+h)^2 + v0^2 - 2gR. This is a very usable general solution. The minimum escape velocity is found by looking at happens as h grows very large. So as h->∞ gR^2/(R+h)^2 -> 0. This makes the solution equation take the form v^2 = v0^2 - 2gR. Here we must note that the velocity must be >0 in order to escape which implies that v0^2 ≥ 2gR --> v0≥ √(2gR). This has a minimum value when v0=√(2gR) and thus this is our minimum escape velocity.

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RESPONSE --> 

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).

NOTE: I don't have this problem so I can't work it but I will look over the solution to still get an idea about what is going on.

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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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.

Note that this problem is in the fourth edition and it is number 11.6.8.

Problem 11.6.13 from 3d edition

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

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Your solution: Here we know dI/dy = kI. We also know that k needs to be negative because the light intensity is being degraded. So we have dI/dy = -kI. Separating, integrating and solving for I makes this the familiar: I = Ae^(-kt). Since we are given I as a percentage, we can use A = 100 : I= 100*e^(-kt) to get an answer in percent. At I=50%= 100(e^-k*10) --> -k≈ 0.069. Then I(20) = 25= 25 % and I(25)= 17.8 ≈ 18%.

RESPONSE --> 

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

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RESPONSE --> It's just really interesting how so many different things can be modeled in such simple terms. I know that we eventually will complicate these models more since there are always more factors involved than what we have considered so far. It also speaks to the power of the exponential function e.

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The process is always the same: figure out the equation for the rate of change, cross your fingers and see if it can be solved.

The equations don't have to get much more complicated before we start running into unsolvable cases. The majority of differential equations can't be solved. But those that can are very powerful and useful.