assignment 007

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Physics II

11-25-2007

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10:25:36

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 -->

(a) LEFT(10) = 2.346

exact = 4.0

error = 1.654

error will decrease by a factor of about 3 as n increases by a factor of 3.

1.654 / 3 = .5513

4.0 - .5513 =

LEFT(30) = 3.4487

(b) TRAP(10) = 2.346

exact = 4.0

error = 1.654

error will decrease by a factor of about 3^2 as n increases by a factor of 3.

1.654 / 9 = .1838

4.0 - .1838 =

TRAP(30) = 3.8162

(c) SIMP(10) = 2.346

exact = 4.0

error = 1.654

error will decrease by a factor of about 3^4 as n increases by a factor of 3.

1.654 / 81 = .02042

4.0 - .02042 =

SIMP(30) = 3.97958

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10:25:49

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 -->

(a) LEFT(10) = 2.346

exact = 4.0

error = 1.654

error will decrease by a factor of about 3 as n increases by a factor of 3.

1.654 / 3 = .5513

4.0 - .5513 =

LEFT(30) = 3.4487

(b) TRAP(10) = 2.346

exact = 4.0

error = 1.654

error will decrease by a factor of about 3^2 as n increases by a factor of 3.

1.654 / 9 = .1838

4.0 - .1838 =

TRAP(30) = 3.8162

(c) SIMP(10) = 2.346

exact = 4.0

error = 1.654

error will decrease by a factor of about 3^4 as n increases by a factor of 3.

1.654 / 81 = .02042

4.0 - .02042 =

SIMP(30) = 3.97958

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10:26:07

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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RESPONSE -->

(b) TRAP(10) = 2.346

exact = 4.0

error = 1.654

error will decrease by a factor of about 3^2 as n increases by a factor of 3.

1.654 / 9 = .1838

4.0 - .1838 =

TRAP(30) = 3.8162

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10:26:16

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 -->

(c) SIMP(10) = 2.346

exact = 4.0

error = 1.654

error will decrease by a factor of about 3^4 as n increases by a factor of 3.

1.654 / 81 = .02042

4.0 - .02042 =

SIMP(30) = 3.97958

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10:26:27

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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10:26:29

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

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10:26:32

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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10:26:34

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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10:26:36

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

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10:26:38

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

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10:26:40

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

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10:26:44

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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12:33:39

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

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RESPONSE -->

integral of 1 / (u^2-16) from 0 to 4

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

using formula V.26.

integral of 1 / ((u - 4)(u + 4)) = 1/8 (ln |u - 4| - ln |u+4|) + C

from 0 to b =

((ln |b - 4| - ln |b+4|) / 8) - ((ln |0 - 4| - ln |0+4|) / 8) =

((ln |b - 4| - ln |b+4|) / 8)

limit of ((ln |b - 4| - ln |b+4|) / 8) as b approaches 4 does not exist. Integral of 1 / (u^2-16) from 0 to 4 diverges.

The following argument for divergence is more general in its approach, and would apply to many situations in which the antiderivative cannot be expressed in terms of simple functions:

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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12:35:29

does your integral converge, and why or why not?

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RESPONSE -->

No, it diverges. As the integral approaches 4, the area approaches infinity.

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12:35:38

If convergent what is your result?

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12:40:34

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

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RESPONSE -->

If the area approches a finite number as b approches 4, then it is said to converge. The values just don't shrink fast enough for the integral to have a finite value.

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12:43:34

Give the steps in your solution.

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RESPONSE -->

integral of 1 / (u^2-16) from 0 to 4

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

using formula V.26.

integral of 1 / ((u - 4)(u + 4)) = 1/8 (ln |u - 4| - ln |u+4|) + C

from 0 to b =

((ln |b - 4| - ln |b+4|) / 8) - ((ln |0 - 4| - ln |0+4|) / 8) =

((ln |b - 4| - ln |b+4|) / 8)

limit of ((ln |b - 4| - ln |b+4|) / 8) as b approaches 4 does not exist. Integral of 1 / (u^2-16) from 0 to 4 diverges.

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12:44:33

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

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15:02:18

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

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RESPONSE -->

(a)

(b) People are getting sick the fastest on day 2.

The slope is increasing from 0 to 2 and decreasing after 2, and the derivative of the function equals zero at x = 2. This indicates a maximum at x = 2.

Good. Compare your details with 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.

(c) Using integration by parts,

The indefinite integral of 1000 t e^(-.5t) is -2000e^(-0.5t)(t + 2) + C

r = 1000 t e^(-.5t)

t = 19.7854 when r = 1

t = 19.7855 when r = .999955

at t = 19.7855 is approximately the point where less than one person per day is getting sick. I will approximate this to be day 20.

Thus, the integral of 1000 t e^(-.5t) from 0 to 20 is

[-2000e^(-0.5(20))((20) + 2)] - [-2000e^(-0.5(0))((0) + 2)] = 3998

Close, but it's easier to do the integral from 0 to infinity:

** We 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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15:06:39
describe your graph, including asymptotes, concavity, increasing and decreasing behavior, zeros and intercepts

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RESPONSE -->
The graph is concave down. It is increasing at a decreasing rate from 0 to 2 and decreasing at a decreasing rate from 2 to infinity. The graph intercepts the origin.

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15:07:54
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 on day 2.

The slope is increasing from 0 to 2 and decreasing after 2, and the derivative of the function equals zero at x = 2. This indicates a maximum at x = 2.

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15:08:33
How many people get sick and how did you obtain this result?

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RESPONSE -->
(c) Using integration by parts,
The indefinite integral of 1000 t e^(-.5t) is -2000e^(-0.5t)(t + 2) + C

r = 1000 t e^(-.5t)
t = 19.7854 when r = 1
t = 19.7855 when r = .999955
at t = 19.7855 is approximately the point where less than one person per day is getting sick. I will approximate this to be day 20.
Thus, the integral of 1000 t e^(-.5t) from 0 to 20 is

[-2000e^(-0.5(20))((20) + 2)] - [-2000e^(-0.5(0))((0) + 2)] = 3998 People getting sick

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15:09:37
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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15:09:41
Query Add comments on any surprises or insights you experienced as a result of this assignment.

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Good responses. See my notes and let me know if you have questions. &#