Assignment 7

course Mth 174

ŘÝн~yassignment #007

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

ʣ֗ngymӮg]~w

Physics II

10-08-2007

......!!!!!!!!...................................

15:47:58

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?

......!!!!!!!!...................................

RESPONSE -->

Error = 4.0 - 2.346 = 1.654

The left rule would be around 1.034

the trap rule would be around 1.000037

the simpson rule would be around

Simp(30) = 2*MID(30) + TRAP(30) / 3 = -.000000007

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. **

.................................................

......!!!!!!!!...................................

15:48:17

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

......!!!!!!!!...................................

RESPONSE -->

just showed it in the answer before this one

.................................................

......!!!!!!!!...................................

15:48:27

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

......!!!!!!!!...................................

RESPONSE -->

Just showed it in the first asnwer of the question

.................................................

......!!!!!!!!...................................

15:48:42

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

......!!!!!!!!...................................

RESPONSE -->

showed it in the first question hwere it asked this there too.

.................................................

......!!!!!!!!...................................

15:49:03

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.

......!!!!!!!!...................................

RESPONSE -->

being skipped

.................................................

......!!!!!!!!...................................

15:49:08

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

......!!!!!!!!...................................

RESPONSE -->

skipped

.................................................

......!!!!!!!!...................................

15:49:13

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?

......!!!!!!!!...................................

RESPONSE -->

skipped

.................................................

......!!!!!!!!...................................

15:49:22

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).

......!!!!!!!!...................................

RESPONSE -->

skipped

.................................................

......!!!!!!!!...................................

15:49:27

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

......!!!!!!!!...................................

RESPONSE -->

skipped

.................................................

......!!!!!!!!...................................

15:49:32

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

......!!!!!!!!...................................

RESPONSE -->

skipped

.................................................

......!!!!!!!!...................................

15:49:36

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

......!!!!!!!!...................................

RESPONSE -->

skipped

.................................................

......!!!!!!!!...................................

15:49:42

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

......!!!!!!!!...................................

RESPONSE -->

skipped

.................................................

......!!!!!!!!...................................

15:54:18

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

......!!!!!!!!...................................

RESPONSE -->

integrate 1/(u^2-16) du from 0 to 4

lim b-2- integrate 1/(u^2 - 16) du = lim b-2- (-1) (u^2-16) = (-1/u-16) - 1/2 = lim b-2- (-1/u-16)

You do not have a correct antiderivative. You should be checking your antiderivatives to be sure that when differentiated they give you the original expression.

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. **

.................................................

......!!!!!!!!...................................

15:54:30

does your integral converge, and why or why not?

......!!!!!!!!...................................

RESPONSE -->

no it does not converge

.................................................

......!!!!!!!!...................................

15:54:42

If convergent what is your result?

......!!!!!!!!...................................

RESPONSE -->

my result was on the first part of this question

.................................................

......!!!!!!!!...................................

15:55:06

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

......!!!!!!!!...................................

RESPONSE -->

because it is so close to converge it is hard to determine whether it does or not.

.................................................

......!!!!!!!!...................................

15:55:11

Give the steps in your solution.

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

15:55:25

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

......!!!!!!!!...................................

RESPONSE -->

i showed it on the first part of this question.

.................................................

......!!!!!!!!...................................

15:55:46

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

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

15:56:31

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

......!!!!!!!!...................................

RESPONSE -->

the graph is going down which is concave down and it is decreasing at first and then going back increasing and it goes past zero at 3 and 6 on the x-axis.

** 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. **

** It isn't clear from your responses how you got these results. In case you are relying on a graphing calculator, note the following:

The calculator is fine for checking yourself and verifying conclusions reached by analysis, but you need to use the techniques of calculus to determine inflection points, maxima, minima etc.. **

.................................................

......!!!!!!!!...................................

15:56:53

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

......!!!!!!!!...................................

RESPONSE -->

they are getting sicker faster during the months of winter and fall.

.................................................

......!!!!!!!!...................................

15:57:14

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

......!!!!!!!!...................................

RESPONSE -->

there are about 200 people getting sick over all at this time

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.

.................................................

......!!!!!!!!...................................

15:57:18

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

......!!!!!!!!...................................

RESPONSE -->

.................................................

......!!!!!!!!...................................

15:57:47

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

......!!!!!!!!...................................

RESPONSE -->

i really didnt understand the last question which was number 44 on the how many people were getting sick.

.................................................

See my notes and let me know if you have additional questions. You may still submit a copy of this document with details and self-critique.