Assignment 7

course Mth 174

?aw|??H???i??????€assignment #007

?????S??B+???Physics II

03-09-2009

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09:46:26

query problem 7.6.6 approx using n=10 is 2.346; exact is 4.0. What is n = 30 approximation if original approx used LEFT, TRAP, SIMP?

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RESPONSE -->

Error = Actual Value - Approximate Value so,

4.0 - 2.346 = 1.654

n=10, .1654

TRAP = .1654 / 2 = .0827

LEFT= -.0827 / 3

n=30, Error in Left rule = -.02756

SIMP(n) = (2 * MID(n) + TRAP(n) / 3

This is problem 6 in section 7.6 of your text.

LEFT and RIGHT approach the exact value in proportion to the number of steps used.

MID and TRAP approach the exact value in proportion to the square of the number of steps used.

SIMP approachs the exact value in proportion to the fourth power of the number of steps used.

Using these principles we can work out this problem as follows:

** The original 10-step estimate is 2.346, which differs from the actual value 4.000 by -1.654.

If the original estimate was done by LEFT then the error is inversely proportional to the number of steps and the n = 30 error is (10/30) * -1.654 = -.551, approximately. So the estimate for n = 30 would be -.551 + 4.000 = 3.449.

If the original estimate was done by TRAP then the error is inversely proportional to the square of the number of steps and the n = 30 error is (10/30)^2 * -1.654 = -.184, approximately. So the estimate for n = 30 would be -.184 + 4.000 = 3.816.

If the original estimate was done by SIMP then the error is inversely proportional to the fourth power of the number of steps and the n = 30 error is (10/30)^4 * -1.654 = -.020, approximately. So the estimate for n = 30 would be -.02 + 4.000 = 3.98. **

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09:51:45

If the approximation used LEFT then what is your estimate of the n = 30 approximation and how did you get it?

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RESPONSE -->

Error = Actual Value - Approximate Value so,

4.0 - 2.346 = 1.654

n=10, .1654

TRAP = .1654 / 2 = .0827

LEFT= -.0827 / 3

n=30, Error in Left rule = -.02756

I divided .1654 by 2 and then divided that value by 3 because n is increasing by 3.

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09:52:26

If the approximation used TRAP then what is your estimate of the n = 30 approximation and how did you get it?

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RESPONSE -->

Error = Actual Value - Approximate Value so,

4.0 - 2.346 = 1.654

n=10, .1654

TRAP = .1654 / 2 = .0827

I divided the approximate value by 2

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09:52:53

If the approximation used SIMP then what is your estimate of the n = 30 approximation and how did you get it?

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RESPONSE -->

SIMP(n) = (2 * MID(n) + TRAP(n) / 3

I do not understand how to get the MID value

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09:53:17

a < b, m = (a+b)/2. If f quadratic then int(f(x),x,a,b) = h/3 ( f(a) / 2 + 2 f(m) + f(b) / 2).

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09:53:20

How did you show that if f(x) = 1, the equation holds?

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09:53:22

How did you show that if f(x) = x, the equation holds?

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09:53:25

How did you show that if f(x) = x^2, the equation holds?

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09:53:28

How did you use your preceding results to show that if f(x) = A x^2 + B x + c, the equation must therefore hold?

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09:54:24

query problem 7.7.19 integrate 1 / (u^2-16) from 0 to 4 if convergent

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RESPONSE -->

This does not converge.

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09:55:03

does your integral converge, and why or why not?

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RESPONSE -->

It does not converge because it is not a fininte number.

As an example of a function that diverges while its integral converges, 1/x^.5 diverges at 0 but its integral from 0 to 4 converges.

So your reason, as stated, doesn't justify your conclusion.

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09:55:07

If convergent what is your result?

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09:55:52

Why is there a question as to whether the integral does in fact converge?

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RESPONSE -->

Beacause if the intigral is a finite number it converges, otherwise, it diverges.

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10:19:15

Give the steps in your solution.

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RESPONSE -->

integral of 1 / (u^2 - 16) =

lim as b approches 4- is 1/(u^2 - 16)dx =

lim as b approches 4- is (-1)(u^2 - 16)^ -1 =

lim as b approches 4- is (- 1/( (b^2 - 16) - 1/4)

This limit does not exist.

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10:19:29

If you didn't give it, give the expression whose limit showed whether the integral was convergent or divergent.

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RESPONSE -->

integral of 1 / (u^2 - 16) =

lim as b approches 4- is 1/(u^2 - 16)dx =

lim as b approches 4- is (-1)(u^2 - 16)^ -1 =

lim as b approches 4- is (- 1/( (b^2 - 16) - 1/4)

This limit does not exist.

This makes your reasoning more precise. However you have to test the integral, not the function itself.

1 / (u^2-16) = 1 / [(u+4)(u-4)] . Since for 0 < x < 4 we have 1/8 < 1 / (u+4) < 1/4, the integrand is at most 1/4 times 1/(u-4) and at least 1/8 of this quantity, so the original integral is at most 1/4 as great as the integral of 1 / (u-4) and at least 1/8 as great. That is,

1/8 int(1 / (u-4), u, 0, 4) < int(1 / (u^2-4), u, 0, 4) < 1/4 int(1 / (u-4), u, 0, 4).

Thus if the integral of 1 / (u-4) converges or diverges, the original integral does the same. An antiderivative of 1 / (u-4) is ln | u-4 |, which is just ln(4) at the limit u=0 of the integral but which is undefined at the limit u = 4.

We must therefore take the limit of the integral of 1/(u-4) from u=0 to u=x, as x -> 4.

The integral of 1 / (u-4) from 0 to x is equal to ln (4) - ln(x-4) = ln( 4 / (x-4) ).

As x approaches 4 the denominator approaches 0 so the fraction approaches infinity and the natural log approaches infinity. Thus the integral diverges. **

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10:28:57

query problem 7.7.44 (was #39) rate of infection r = 1000 t e^(-.5t)

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RESPONSE -->

b) people are getting sick the fastest, the less amount of days since the start of the epidemic.

c) 1000 people get sick all together?

** This graph increases at first as you move to the right from t = 0. However e^(-.5 t) eventually approaches zero much faster than t increases so the graph has an asymptote at the positive t axis. So it increases for small positive t but eventually returns almost to the t axis, and it can't be strictly increasing. Its concavity changes from downward (negative) for small positive t to upward for larger t; the point at which the concavity changes is important.

We use the standard technique from first-semester calculus to find the point at which this function maximizes.

The first derivative is dr/dt = 1000 e^(-.5 t) - 500 t e^(-.5 t).

Setting this derivative equal to 0 we get

1000 e^(-.5 t) - 500 t e^(-.5 t) = 0;

dividing through by e^-.5 t we get t = 2, which with a first-or second-derivative test confirms that the t = 2 graph point is a relative maximum.

Concavity is determined by the second derivative r'' = e^(-.5 t) [ -1000 + 250 t ], which is 0 when t = 4. This is a point of inflection because the second derivative changes from negative to positive at this point. So the function is concave downward on the interval (-infinity, 4) and concave upward on (4, infinity).

The first derivative has a critical point where the second derivative is zero. This occurs at x = 4, which was identified in the preceding paragraph as the point of inflection for the original function. Since the second derivative goes from negative to positive, this point is a minimum of the first derivative. The first derivative is a decreasing function from t = 0 to t = 4 (2d derivative is negative) and is then an increasing function with asymptote y = 0, the x axis, which it approaches through negative values. Its maximum value for t >= 0 is therefore at t = 0. **

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10:34:52

describe your graph, including asymptotes, concavity, increasing and decreasing behavior, zeros and intercepts

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RESPONSE -->

The graph is exponentially increasing from (1,1000)

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10:36:32

when our people getting sick fastest and how did you obtain this result?

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RESPONSE -->

b) people are getting sick the fastest, the less amount of days since the start of the epidemic. When you multiply each day by -0.5 it decreases at an increasing rate

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10:36:44

How many people get sick and how did you obtain this result?

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RESPONSE -->

Im not sure how to calculate this answer.

** You need to integrate the rate function r = 1000 t e^(-.5t) from t = 0 until forever, i.e., from t = 0 to t = infinity.

An antiderivative of the function is F(t) = 1000 int ( t e^(-.5 t)) = 1000 [ -2 t e^(-.5t) - int ( e^(-.5 t) ) ] = 1000 [ -2 t e^(-.5 t) - 4 e^(-.5 t) ].

Integrating from 0 to x gives F(x) - F(0) = 1000 [ -2 t e^(-.5 x) - 4 e^(-.5 x) ] - 1000 [ -2 * 0 e^(-.5 *0 ) - 4 e^(-.5 * 0 ) ] = 1000 e^-(.5 x) [ -2 t - 4 ] - (-4000).

As x -> infinity, e^-(.5 x) [ -2 t - 4 ] -> 0 since the exponential will go to 0 very much faster than (-2 x - 4) will approach -infinity. This leaves only the -(-4000) = 4000. **

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10:40:58

What improper integral arose in your solution and, if you have not already explained it, explain in detail how you evaluated the integral.

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RESPONSE -->

The improper integral is e^ -.5t

The integral of 3^ -.5tdx =

-1/5e ^ -5t =

-1/5t ^ -5b + 1/5

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10:41:46

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

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RESPONSE -->

I understand the concepts described in 7.6 but am not sure if I am caluclating the errors correctly.

You're welcome to send me some specific examples of problems and your best attempts at solution.

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Your reasoning on the approximations was good.

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