Assignment 20

course Mth 272

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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.

20:20:16

Query problem 6.5.12 (was 6.5.10) trapezoidal and Simpson's rules, n=4, integral 0 to 2 of x `sqrt(x2+1)

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confidence assessment:

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20:20:27

What are your results for the trapezoidal rule and for Simpson's rule, and what is your value for the exact integral?

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Exact is int. = 1/3(2^2+1)^3/2 = 3.7268

Trapexoidal is 1/4[0 + 2(.5sqrt(.5^2+1)) + 2(1sqrt(1^2+1)) + 2(1.5sqrt(1.5^2+1)) + 2(2sqrt(2^2+1)) = 4.574

Simpson is 1/6[[0 + 2(.5sqrt(.5^2+1)) + 2(1sqrt(1^2+1)) + 2(1.5sqrt(1.5^2+1)) + 2(2sqrt(2^2+1)) = 3.392

confidence assessment: 2

You would use x0=0, x1=1/2, x2=1 x3=3/2 x4=2.

The corresponding values of x^2 sqrt(x^2+1) are 0, 0.5590169943, 1.414213562, 2.704163456, 4.472135954.

Using these values:

Trap gives you 3.4567.

Simpson's rule gives you 3.3922.

The exact result, to four significant figures, is int(x sqrt(x^2+1),x,0,2) = 3.393. This is based on the antiderivative 1/3 * (x^2+1)^(3/2)

According to these results Simpson's approximation is within .0012 while trap is within about .064. So the Simpson's approximation is .064 / .0012 = 50 times better, approx..

Theory says that it should be about n^2 = 4^2 = 16 times better. **

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20:20:37

How many times closer is Simpson's rule than the trapezoidal rule?

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20:20:41

Query problem 6.5.21 (was 6.5.19) present value of 6000 + 200 `sqrt(t) at 7% for 4 years by Simpson's rule

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confidence assessment:

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20:20:51

What is your result for the present value?

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present value = $21,831.20

confidence assessment: 2

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20:21:03

What expression did you integrate between what limits to obtain your result?

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int. 6000+200(sqrt(t))(e^(-.07(t)))

4-0/3(8) = 1/6

1/6(130,987.22) = 21,831.20

confidence assessment: 2

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20:21:08

Query distance traveled by pursuer along path y = 1/3 (x^(3/2) - 3 x^(1/2) + 2) over [0, 1].

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20:21:19

What is the distance traveled?

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.2656

confidence assessment: 2

The length of the interval from 0 to 1 is 1. The distance along any continuous curve is at least as long as the interval.

See my note below.

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20:21:29

What expression did you integrate between what limits to obtain your result?

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Used the simpson rule, used n = 4

1-0/3(4) = 1//12

1/12[1/3(0^(3/2) - 3 0^(1/2) + 2) + 1/3(.25^(3/2) - 3 .25^(1/2) + 2) + 1/3(.5^(3/2) - 3 .5^(1/2) + 2) + 1/3(.75^(3/2) - 3 .75^(1/2) + 2) + 1/3(1^(3/2) - 3 1^(1/2) + 2)]

1/12(3.1875) = .2656

confidence assessment: 2

** Integrate to find the arc length of the curve.

If a line with slope m lies above an interval `dx on the x axis then the 'run' of the segment above `dx is just `dx, while the rise is m * `dx. The hypotenuse is therefore `sqrt( `dx^2 + ( m `d)^2 ) = `sqrt( 1 + m^2 ) * `sqrt(`dx^2) = `sqrt(1 + m^2) `dx.

If a curve y = f(x) lies above the interval `dx then the average slope of the curve is a value of y ' for some x in the interval. After a little fancy reasoning we can prove that the arc length is in fact equal to `sqrt( 1 + y'^2) `dx, but proof or not it's easy enough to understand if you understand the thing about m.

This leads to the theorem that for a differentiable function y = f(x) the arc length between x = a and x = b is INT ( `sqrt( 1 + y' ^ 2) dx, x from a to b).

The derivative of the given function is 1/2 x^.5 - 1/2 x^-.5; the square of this expression is 1/4 ( x - 2 + 1/x) so you integrate `sqrt( 1 + 1/4 ( x - 2 + 1/x) ) from 0 to 1.

The integral gives you 1.33327, accurate to 6 significant figures. However you'll probably have to use an approximation technique so you probably won't get this exact result. **

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20:21:37

How did you perform your integration?

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confidence assessment:

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20:21:39

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

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confidence assessment:

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Good answers on some questions; there were also some errors. You didn't provide enough detail for me to spot your exact errors, but I believe my notes will give you the information you need.

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Let me know if you have questions. &#