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course Mth 173

‚ïš¯–ËÓy‘“Á®äì”ö ¸lè¹«üÓÏassignment #008

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Calculus I

02-11-2007

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15:33:54

What functions f(z) and g(t) express the function 2^(3t-5) as a composite f(g(t))?

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f(x) = 2 ^ x

g(x) = 3t - 5

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15:34:01

** g(t) = 3t - 5, f(z) = 2^z.

The z is a 'dummy' variable; when we find f(g(t)), the g(t) is substituted for z and we get f(g(t)) = 2^(g(t)) = 2^(3t-5). **

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15:37:07

describe in some detail how we can numerically solve a differential equation dy /dt = f(x), given a point (x0, y0) on its solution curve, an interval (x0, xf) and an increment `dx

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You take the given points and calculate the slope for these points and using whatever time increment dx and multiply the increment by the slope then you add the changes to the current points to get the new points.

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15:37:12

** You start with a point (x0, y0) on the y vs. x graph.

You evaluate the function y' for x=x0 and y=x0, which gives you a slope for your y vs. x graph.

Using the chosen increment `dx you then multiply `dx by the slope to determine how much change there will be in y, and you use this information to obtain a new approximate point on your y vs. x graph by adding the change in y to y0 and the change in x to x0.

You then repeat the process starting with the new point. **

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15:38:04

explain why a numerical solution to differential equation is only an approximate solution in most cases

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We assume that the change happens a little bit each time instead it could happen more rapid or more slow which our assumption is an approximation.

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15:38:06

** You assume the slope at the initial point, but that slope generally changes at least a bit by the time you get to the second point. So you are assuming a constant slope when the slope actually changes.

If your interval is small enough the change in slope will have a small effect. **

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15:42:44

query Problem 1.4.8 Solve 2 * 5^x = 11 * 7^x

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2 * 5^x = 11 * 7^x

log 2 + x (log 5) = log 11 + x(log 7)

x(log 5) - x(log 7) = log 11 - log 2

x(log 5 - log 7) = log 11 - log 2

x = log 11 - log 2 / log 5 - log 7

x = 1.0414 - .3010 / .6990 - .8451

x = .7404 / -.1461

x = -5.07

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15:42:47

** Taking logs of both sides and applying the laws of logarithms we get

log 2 + x log 5 = log 11 + x log 7. Rearranging we obtain

x log 5 - x log 7 = log 11 - log 2 so that

x ( log 5 - log 7) = log 11 - log 2 and

x = (log 11 - log 2) / (log 5 - log 7).

This can be approximated as -5.07. ** DER

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15:43:37

Problem 1.4.28 simplify 2 ln(e^A) + 3 ln(B^e)

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RESPONSE -->

2 ln(e^A) + 3 ln(B^e)

2A + 3B

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15:43:41

** Starting with 2 ln (e^A) + 3 ln (e^B) we use the fact that the natural log and exponential functions are inverses, expressed by the law of logarithms ln(e^x) = x, to get

2 * A + 3 * B or just

2A + 3B. **

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15:48:22

query Problem 1.4.31 (was 1.7.26) P=174 * .9^twhat is the function when converted to exponential form P = P0 e^(kt)?

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RESPONSE -->

P = 174 * e^(kt)

e^(kt) = e^(.9t)

k = -.105

P = 174 e^(-.105t)

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15:48:25

** 174 * .9^t = 174 * e^(kt) if

e^(kt) = .9^t, which is the case if

e^k = .9. Taking the natural log of both sides we get

ln(e^k) = ln(.9) so that

k = ln(.9) = -.105 approx.

So the function is

P = 174 e^(-.105 t). **

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15:58:09

** The population function is exponential and has form P = P0 * e^(kt).

Let t be the time since 1980 and population be in millions. Then we have

40 = P0 e^(k * 0) and

56 = P0 e^(k * 10).

From the first equation we get

40 = P0 so the second equation becomes

56 = 40 * e^(10 k) or

e^(10 k) = 56 / 40 = 1.4. Taking logs we get

10 k = ln(1.4) so that

k = ln(1.4) / 10 = .0336, approx.

Thus our equation is

P = 40 e^(.0336 t).

This doubles when

e^(.0336 t) = 2. Taking the ln of both sides we have

.0336 t = ln(2) so that

t = ln(2) / .0336 = 20.6, approx.

Doubling time is about 20.6 years. **

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I accidentally hit enter before I had a chance to answer the question.

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16:05:37

query Problem 1.4.50 (was 1.7.42 but changed) time to get to 10% of strontium 90 if half-life 29 yrs

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e^(kt)

e^(k(29)) = 1/2

ln e ^ (29k) = ln 1/2

29k = -.6931

k = -.0239

e^(-.0239t) = 10/100

e^(-.0239t) = .1

ln e^(-.0239t) = ln .1

-.0239t = -2.3026

t = 96.3431

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16:05:40

** If the model is Q = Q0 e^(kt) then since half-life is 29 years we know that

e^(k * 29) = 1/2. Taking ln of both sides

29 k = ln(1/2) so that

k = ln(1/2) / 29 = -.0239.

So the model is

Q = Q0 * e^(-.0239 t).

Now if we want to get 10% of the original quantity we have Q = Q0 / 10 so that

Q0 / 10 = Q0 e^(-.0239 t) and

e^(-.0239 t) = 1/10.

Taking logs of both sides and solving for t we get

t = ln(1/10) / -.0239 = 96.3 approx. **

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16:06:21

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

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none so far

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16:07:45

Problem 1.4.31 P=174 * .9^t

What is the function when converted to exponential form P = P0 e^(kt)?

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I just answered this question

P = 174 e^(-.105)

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16:07:48

If P=174(.9)^t is put in the form P = P0 e^(kt) then P0 = 174 and e^(k t) = .9.t so that e^k = .9.

It follows that

e^k = .9 so that

ln(e^k) = ln(.9) or

k = ln(.9) = .105.

The function is therefore

P=174 e^-(.105 t).

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16:09:07

problem 1.4.44 population function for exponential growth.

If 40 meg in 1980 and 56 meg in 1990 give the equation and find the doubling time

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P =

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16:09:11

P=Po b^t is the form of the function. Initial quantity is 40 * 10^6 so Po = 40 * 10^6. Substituting Po = 40 * 10^6:

P=40*10^6 b^t. At t = 10 we have P = 56 * 10^6 so we substitute for P and t:

56*10^6=40*10^6 b^10. We solve for b:

1.4=b^10

b=1.03

P=40*10^6(1.03)^t is our function.

doubling time occurs when the 40^10^6 grows to 80*10^6:

80*10^6=40*10^6(1.03)^t

2=1.03^t

log2=tlog1.03

t=23.4498

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16:09:14

10:32:42

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16:14:21

Problem 1.7.42 percent of original strontium -- 90 after century; 2.47% annual decay rate.

What percent of the original strontium -- 90 would remain after a century?

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e ^ (kt)

e ^ (2.47/100(t))

e ^ (.0247t)

e ^ (kt)

e ^ (k * 1)

e ^ (1. - .0247)

e ^ .9753

2.652

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16:14:24