Assignment 20

course Mth 272

?q???|???C?????assignment #020020.

Applied Calculus II

04-01-2009

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22:23:11

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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RESPONSE -->

first find Reimanns sum

b-a/n = 2-0/4 = 1/2= .5

x0 0 f(x0) = 0

x1 .5 = .55902

x2 1 = 1.4142

x3 1.5 = 2.7042

x4 2 = 4.4721

Then use these values for the Trapezoidal rule

b-a/2n = 1/4 = .25

.25[0+2(.55902)+2(1.4142)+2(2.7042)+4.4721)]

= 3.4567

Then Simpson's Rule

b-a/3n = 1/6

1/6[0+4(.55902)+2(1.4142)+4(2.7062)+4.4721]

= 3.39223

Then the integral u= x^2+1 du=2xdx

1/2 ~ (x^2+1)^1/2dx

= 2/3*1/2(x^2+1)^3/2 from 0 to 2

= 1/3(x^2+1)^3/2 take the integral from 0 to 2

3.7276 - (1/3)

exact value = 3.39345

so Simpson's rule was closer to the actual value.

confidence assessment: 3

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22:23:20

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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RESPONSE -->

answered

confidence assessment: 3

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22:25:58

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

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RESPONSE -->

simps rule = .00122 difference

trap rule = .06325

act value was = 3.39345

confidence assessment: 3

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22:42:27

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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RESPONSE -->

pv= ~from 0 to 4 of (6000+200sqrt(t))e^-.07t dt

distribute..

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

The calculator program's result for this was $21836.98

simpson's rule b-a/3n = 1/6

x0= 0 f(x)= 6000

x1=.5 = 5930.2

x2=1 = 5780.8

x3=1.5 = 5622.5

x4=2 = 5462.0

x5=2.5 = 5302.2

x6=3 = 5144.3

x7=3.5 = 4989.1

x8=4 = 4837.0

1/6[ 6000 + 4(5930.2)+2(5780.8)+4(5622.5)+4(5462)+2(5302.2)+4(5144.3)+2(4989.1)+4837] = $21936.20

There is a $99.22 in the two results

confidence assessment: 3

I don't see anything wrong in your calculation, but the error should be significantly less than that. I have about a $5 difference.

It's not worth worrying about. You're using the method correctly.

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22:42:34

What is your result for the present value?

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RESPONSE -->

answered

confidence assessment: 3

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22:42:44

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

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RESPONSE -->

answered

confidence assessment: 3

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22:46:57

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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RESPONSE -->

This one seem a little ambiguous, did you want the integral of this, because you cant take the anti deriv of y, if it was y prime then yes? Anyways... i did both.

Integral

(2/5)(1/3)x^5/2- (2/3)3X^3/2 + 2

2/15x^5/2- 2x^3/2+2x = 2/15- 2 +2 = 2/15

Getting the answer just by plugging in the numbers to the original

= 4/3

confidence assessment: 3

This is going to be an arc length integral.

The x = 0 point of the curve is (0, 2/3) and the x = 1 point is (1, 0). The straight-line distance between these points is thus sqrt(1^2 + (2/3)^2) = sqrt(13/9) = 1.2 or so. The straight line being the shortest distance between two points, we expect the arc distance to be more than 1.2. However unless the curve gets pretty steep the distance won't be drastically greater than this.

** 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 your result will probably differ after a few significant figures from this result. **

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22:47:05

What is the distance traveled?

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RESPONSE -->

answered

confidence assessment: 3

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22:47:16

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

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RESPONSE -->

answered

confidence assessment: 3

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22:47:22

How did you perform your integration?

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RESPONSE -->

answered

confidence assessment: 3

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22:48:32

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

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RESPONSE -->

The last question wasn't listed as being in the book, and confusing so i posted a question on it, other than that it was pretty easy to understand. Thanks.

confidence assessment: 3

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