Assignment 7 Query

course Mth 174

oҖazܭassignment #007

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Physics II

10-08-2007

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14:29:58

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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I found the approximations to be:

Left: (10 / 30) * 1.654 = 0.551

Trap: (10 / 30)^2 * 1.654 = 0.184

Simp: (10 / 30)^4 * 1.654 = 0.0204

Good. Those would be your errors. You would subtract the errors from the exact result to get the approximation.

** 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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14:30:44

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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My approximation was:

(10 / 30) * 1.654 = 0.551

I did it by taking the original n value, dividing it by the new n value, then multiplying it by the error. This gave me the new error value.

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14:31:07

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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My approximation was:

(10 / 30)^2 * 1.654 = 0.184

I did it by taking the original n value, dividing it by the new n value, squaring it, then multiplying it by the error. This gave me the new error value.

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14:31:27

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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My approximation was:

(10 / 30)^4 * 1.654 = 0.0204

I did it by taking the original n value, dividing it by the new n value, raising it to the fourth power, then multiplying it by the error. This gave me the new error value.

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14:31:45

This problem has been omitted from the present edition and may be skipped: query problem 7.6.10 If TRAP(10) = 12.676 and TRAP(30) = 10.420, estimate the actual value of the integral.

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14:31:50

What is your estimate of the actual value and how did you get it?

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14:31:52

By what factor should the error in the second approximation be less than that in the first, and how does this allow you to estimate the integral based on the difference in the two approximations?

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14:32:02

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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14:32:05

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

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14:32:08

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

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14:32:10

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

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14:32:25

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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14:33:32

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

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I found that this function does not converge.

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14:37:14

does your integral converge, and why or why not?

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The integral does not converge because I found the antiderivative to be equal to:

(1/8) ln (u - 4) + (1/8) ln (u + 4)

Between 0 and 4 this function does not move towards any value quickly, instead increasing throughout the interval.

You have most of the following. At the end you should be considering limits:

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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14:37:21

If convergent what is your result?

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No result.

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15:07:06

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

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At either end of the interval, the integral acts strange. At the u=4 end:

(1/8) ln(4 - 4) + (1/8) ln (4 + 4)

= (1/8) ln(0) + (1/8) ln (8)

ln(0) is undefined. At the other end, u=0:

(1/8) ln(0 - 4) + (1/8) ln (0 + 4)

= (1/8) ln(-4) + (1/8) ln (4)

ln(-4) is undefined. In this interval there is no way to evaluate the function.

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15:13:37

Give the steps in your solution.

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I used partial fractions to find the antiderivative:

1 / (u^2 - 16) = 1/ (u - 4)(u +4) = A / (u-4) + B (u+4)

Solving that gives:

A = 1/8

B = -(1/8)

Therefore, the derivative is equal to:

(1/8) / (u-4) - (1/8) / (u+4)

The antiderivative is:

(1/8) ln (u-4) - (1/8) ln (u+4)

(Incidentally, I made a mistake with the sign of the second part of the equation. My previous statements still hold, I think).

I then looked at the behavior at the endpoints and found that nothing between 0 and 4 would evaluate. Therefore, it does not converge.

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15:14:20

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

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I assume this would be:

lim u->4 ( (1/8) ln (u-4) - (1/8) ln (u+4) )

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15:30:41

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

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a) I believe was asked to graph the existing function. My graph looked like a mountain with a maximum at t=2, concave up from roughly 0 to 5, concave down from there on.

b) I took this to mean that I needed to find the fastest rate at which the equation changed, meaning that I needed to find the derivative of the original function. I found that to be:

1000 e^(-0.5t) - 500t e^(-0.5t)

To find the highest rate from there, all I should need to do is find the largest value of the derivative, which comes at t=0 and r=1000.

c) I believe this is asking me to find the integral of the original function. Using integration by parts I found that the antiderivative was:

-2000t e^(-0.5t) - 4000 e^(-0.5t)

Then I can just take the limit of this function in the Fundamental theorem to find the integral equals 4000.

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15:35:19

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

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Asymptotes:For x > 0, the x axis is an asymptote.

Concavity: concave down from 0 to roughly 4, then concave up from there on.

Increasing over the interval 2 to infinity.

Decreasing over the interval 0 to 2.

No zeros or intercepts that I could find.

Very good. Compar the following:

** 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.

To get the total number who get sick, 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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15:35:28

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

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See first answer.

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15:35:42

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

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I found that 4000 people got sick. See first answer for reasoning.

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15:45:15

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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I found the improper integral previously recorded then looked at the behavior for large values of t. Since the origin is the same, I only had to look at ( -2000t e^(-0.5t) - 4 e^(-0.5t) ) to find out how it acted (after adding 4000). As the values of t increased, it converged on 4000.

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15:46:07

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

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Diverging and converging are still odd concepts to me, but I think they are making a little sense.

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Excellent work. See my notes.

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Let me know if you have questions. &#