Assignment 7

course Mth 174

ڜ͂Y󂌥D^]}assignment #007

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Physics II

02-15-2007

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17:43:20

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 -->

n=30 increases n=1 by factor of 3

Error at n=10 is 2.3462-4.0 = -1.6538

LEFT(30): decrease in error is directly proportional to factor increase in n:

New error = -1.6538/3 = -.5513

New approximation = 4.0 - .5513 = 3.4487

TRAP(30): decrease in error is directly proportional to square of factor increase in n:

New error = -1.6538/3^2 = -.1838

New approximation = 4.0 - .1838 = 3.8162

SIMP(30): decrease in error is directly proportional to 4th power of factor increase in n:

New error = -1.6538/3^4 = -.0204

New approximation = 4.0 - .0204 = 3.9796

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17:43:31

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 -->

See previous response

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17:43:43

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 -->

See previous response

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17:43:48

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 -->

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17:43:56

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 -->

See previous response

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19:14:27

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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RESPONSE -->

TRAP(30) is n*3 of TRAP(10), and the error for TRAP(30) is directly proportional to the factor ratio to the 4th power.

Therefore,

Error[TRAP(30)] is Error[TRAP(10)]/3^4 or Error[TRAP(10)] = 3^4 * Error[TRAP(30)]

Let Actual = A, Error[TRAP(30)] = T(30), Error[TRAP(10)] = T(10)

A + T(10) = 12.676

A + T(30) = 10.420

Substituting 81*T(30) in first equation and solving for T(30) gives:

Error[TRAP(30)] = 0.0282

Substituting into the 2nd equation results in

estimate of actual value of the integral = 10.3918

Good reasoning except that for TRAP the error is inversely proportional to the second power. SIMP is 4th power.

** You need to use the inverse square proportionality of the error with the number of steps.

Trap(30) is approximately (10/30)^2 = 1/9 of TRAP(10). So the difference 10.420 - 12.676 = -2.256 between TRAP(10) and TRAP(30) is approximately 8/9 of the error of TRAP(10).

It follows that the error of TRAP(10) is 9/8 * -2.256 = -2.538. Our best estimate of the integral is therefore -2.538 + 12.676 = 10.138. **

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19:14:38

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

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RESPONSE -->

See previous response

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19:14:57

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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RESPONSE -->

See previous response

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19:17:15

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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19:37:20

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

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RESPONSE -->

If f(x) = 1, we have a quadratic with a and b both equal to 0 and c=1, that is a constant function.

The area under the constant function is a rectangle with length = b-a and height = 1, thus and area of 1*(b-a) = b-a

Using the given equation:

Integral from a to b = [h/3]*[f(a)/2 + 2f(m) + f(b)/2)]

Substituting h=b-a, and f(x) = 1

Integral from a to b = [(b-a)/3]*[1/2 + 2(1) + 1/2]

Integral from a to b = b-a, which equals the result above, therefore the equation is valid for f(x)=1

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19:58:20

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

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RESPONSE -->

If f(x) = x, then the function is a quadratic with a = 0, b=1, and c=0, that is, a linear equation through the origin with a slope of 1.

The area under f(x) from a to b is a trapezoid that can be found by subtracting the area of a triangle with a base from x=0 to x=a from the area of a triangle with a base from x=0 to x=b:

f(a) = a, f(b) = b

Area = 1/2(b-0)*f(b) - 1/2(a-0)*f(a) = 1/2b*f(b) - 1/2a*f(a) = 1/2b^2 - 1/2a^2 = 1/2(b^2-a^2)

Using the equation, with f(m) = [f(b)-f(a)]/2

Integral from a to b = [(b-a)/3][a/2 + 2[(f(b)+f(a)]/2 + b/2)] = [(b-a)/3][a/2 + b + a + b/2] = 1/2(b^2-a^2),

the same result as found geometrically above, therefore the equation is valid for f(x) = x

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20:18:51

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

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RESPONSE -->

If f(x) = x^2, we have a quadratic that is symmetrical about the y-axis and has a minimum at the origin.

f(a) = a^2, f(b) = b^2

The area under the curve from a to b is:

Integral (from a to b) of x^2 dx, which is x^3/3 evaluated from a to b, which = (1/3)*(b^3-a^3).

f(m) = [(b+a)/2}^2

Substituting in the equation:

Area = [(b-a)/3][(a^2)/2 + 2*([(b+a)/2]^2 + (b^2)/2]

= [(b-a)/3][(a^2)/2 + (b^2)/2 + ab + a^2)/2 + (b^2)/2]

= (1/3)*(b^3-a^3), which is the same result as that of the integral evaluated above. Therefore, the equation holds for f(x) = x^2.

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20:26:39

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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RESPONSE -->

The validity of the equations for the three situations shows the validity for separate functions f(x) = 1*x^2, f(x) = 1*x, and f(x) = 1. Since the integral of a function times a constant = the constant time the integral of the function, these three situations are valid for any A, B, and C (i.e., 1 * any constant). Since the integral of the sum of functions is equal to the sum of the integrals of the separate functions, then the equation must hold for any quadratic of the form given, as we have shown that it is valid for each of the three terms (functions).

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20:27:58

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

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RESPONSE -->

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20:46:12

does your integral converge, and why or why not?

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RESPONSE -->

The integral does not converge.

Integral (0 to 4) of 1/(u^2-16). Factoring the denominator into (u-4)(u+4) and using partial fractions results in:

Integral (0 to 4) of 1/[8(u-4)] - 1/[8(u+4)]

Integrating results in:

(1/8) ln |u-4| - (1/8) ln |u+4|, evaluated from 0 to 4

Since ln |u-4| is undefined at u=4, that portion is improper and must be evaluated as a limit:

lim (b>4-) (1/8) ln (u-4), evaluated from 0 to b

= lim (b>4-) [(1/8) ln |b-4| - (1/8) ln |0-4|]

= lim (b>4-) [(1/8) ln |b-4| - (1/8) ln 4]

Since ln |b-4| can increase without limit (although slowly), the integral does not coverge.

That works but you don't have to do the entire integration by parts. The following approach generalized to many integrals for which you cannot find an antiderivative:

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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20:46:36

If convergent what is your result?

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RESPONSE -->

See previous response

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20:56:43

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

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RESPONSE -->

The function 1/(u^2-16) is indeterminate when u is equal to + or - 4, therefore the graph has vertical asymtotes at x=4 and x= -4. In addition, as |x| gets larger and larger, the function approaches 0 as a limit.

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20:57:01

Give the steps in your solution.

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RESPONSE -->

See previous response.

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20:57:20

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

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RESPONSE -->

See previous response.

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21:00:13

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

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21:30:12

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

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RESPONSE -->

The portion of the graph where x < 0 is irrelevant, since t cannot be less than 0.

The graph has a zero/x-intercept/y-intercept at (0,0), has a maximum at (2, approx. 735.76), has a horizontal asymptote at 0 (as x > infinity), is increasing from (0,0) to (2, approx. 735.76) then decreasing as x > infinity, is concave down from (0,0) to (4, approx. 541.34) then concave up as x> infinity.

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21:34:48

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

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RESPONSE -->

People are getting sick fastest at t = 2 days (r = approx. 735.76 people/day.

I found the first derivative of r, which is:

dr/dt = 1000e^(-.5t) - 500t*e^(-.5t) and found the zero of this equation on the calculator

note that this is easy to solve without the calculator; just factor out the e^(-.5 t) and you can solve it in your head

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This is the point at which the slope of r is 0 and is thus a minimum or maximum, in this case a local and global maximum.

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21:45:00

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

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RESPONSE -->

4000 people get sick.

The total number of people who get sick is the integral of r from t=0 to t=infinity.

Integral (0 to infinity) of 1000t*e^(-0.5t)

Using formula #14 p(x) = 1000t, a = -0.5

=[(1/-0.5)(1000t)e^(-0.5t) - (1/(-0.5^2))(1000)e^(-0.5t), evaluated from 0 to infinity

=lim(b>infinity) [-2000te^(-0.5t) - 4000e^(-0.5t)], evaulated from 0 to b

=lim(b>infinity) [-2000be^(-0.5b) - 4000e^(-0.5b) + 2000(0) + 4000(e^0)

The first three terms go to zero as b approaches infinity, leaving the fourth term = 4000.

Excellent solution throughout.

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21:45:13

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 -->

See previous response.

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21:54:06

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

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RESPONSE -->

Problems 7.7-40 and 42 drove me nuts. I tried all kinds of substitutions without success, none of the formulas in the back of the book applied, and I could not see any applicability for the information given that the integral of e^(-x^2/2) = sqrt 2pi. I finally used Derive to calculate the integral for #40, and then had it do step by step, which produced a formula for integrals of x^m*e^(a*x^n) dx that involved the integral of e^(-x^2/2) when applied to this problem. This formula worked for #40, and I used it twice in #42. #39 and #41 were solvable using normal subsitutions.

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Sorry for the delay in posting but you should have included a brief statement of the problems. I don't always have the book handy.

Using integration by parts the integral of x^2 e^-(x^2/2) is found by letting dv = x e^(-x^2/2) and u = x. You get du = dx and v = -1/2 e^(-x^2/2) so that the integral of - v du is just 1/2 * int(e^(-x^2/2)) = sqrt(2 pi) / 2. u v for this integral is -x e^-(x^2/2), which is 0 at both the limits x = 0 and x = infinity.

Repeated integration by parts works for higher even moments.

I'll be posting this with the rest of your work either tonight or tomorrow.