Assignment 8

course MTH 174

㻪P|x̥assignment #008

|hrW

Physics II

10-07-2008

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17:10:44

query explain the convergence or divergence of series (no summary needed)

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RESPONSE -->

We start with a definition of an improper integral.

An integral is improper, that is to say that one of the limits of integration is infinite in either direction or the integrand becomes infinite across the interval of the limits of integration.

When one of the limits of integration is infinite, we can take a limit as b approaches infinity or negative infinity to see if the integral has a finite value. If it does, it is said to converge. If it does not have a finite value, it is said to diverge.

In the case of an improper integral where the integrand becomes infinite over the interval of the limits of integration, The limit is taken as b approaches the point where the function is infinite. This can be done by splitting the integrand into parts in the case of the infinite portion of the function is in the middle of the inter us val.

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18:06:26

explain how we know that the integral from 1 to infinity of 1 / x^p converges if p > 1, diverges for p < 1, diverges for p = 1

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RESPONSE -->

When p > 1 integral from 1 to inifinty is is going to be

lim as b -> inf -1/p+1 * 1/b^(p+1) - -1/p+1 * 1/1^(p+1)

lim as b -> inf -1/p+1 * 1/b^(p+1) + 1/p+1 * 1

as b gets closer to infinity the value of 1/b^(p+1) gets closer to 0 leaving just a value of 1/p+1 * 1

When p < 1 the integral evaluates after taking the limit to be bigger and bigger as b gets larger, so the integral converges.

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19:38:52

explain how we know that the integral from 0 to 1 of 1 / x^p diverges if p > 1, converges for p < 1, diverges for p = 1

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RESPONSE -->

from 0 to 1, the integral evaluates to:

lim as b -> 0+ -1/p+1 * 1/1^(p+1) - -1/p+1 * 1/b^(p+1)

When p < 1 in the interval from 0 to 1, as b gets really small and approaches 0, -1/p+1 * 1/b^(p+1) becomes 0, leaving a finite result for the integral.

If p > 1 , the integral evaluates larger and larger as b gets closer to 0, so the interval diverges.

If p = 1, the integral evaluates to

lim as b -> 0+ ln |1| - ln |b|

As b becomes smaller, ln |b| becomes more and more negative without bound, so the integral diverges

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19:56:51

explain how we know that the integral from 0 to infinity of e^(-a x) converges for a > 0.

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RESPONSE -->

The integral of e^(-a x) dx from 0 to infinity evaluates to:

lim b->infinity 1/-a [e^(-ab) - e^(-a*0)]

lim b-> infinity 1/-a [e^(-ab) - 1]

When a > 0 and b increases in size, e^(-ab) becomes 0 leaving the integral to evaluate to 1/a

When a < 0 the integral grows without bound.

When a = 0, lim b->infinity 1/-a [e^(-0*b) - e^(-0*0)]

which will not evaluated due to 1/-0 being undefined

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20:00:37

query problem 7.8.18 integral of 1 / (`theta^2+1) from 1 to infinity

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RESPONSE -->

This appears to be problem 12 in the second edition.

When comparing 1/`theta^(2/2) to 1/x^p from 1 to inifinity, p = 1, so this one diverges because 1/x^p from 1 to inifinity diverges when p = 1 or p < 1

This solution is for the correct function, which was misstated above. Should have been 1 / sqrt(theta^2 + 1). You basically have it right, but consider the subtleties in the following:

** If it wasn't for the 1 in the denominator this would be 1 / `sqrt(`theta^2) or just 1 / `theta, which is just 1 / `theta^p for p = 1. As we know this series would diverge (antiderivative of 1 / `theta in ln(`theta), which appropaches infinity as `theta -> infinity).

As `theta gets large the + 1 shouldn't matter much, and we expect that 1 /`sqrt (`theta^2 + 1) diverges. If this expression is shown to be greater than something that diverges, then it must diverge.

However, 1 / `sqrt ( `theta^2 + 1) < 1 / `theta, and being less than a divergent function that diverges does not prove divergence.

We can adjust our comparison slightly. Since 1 / `theta gives a divergent integral, half of this quantity will yield half the integral and will still diverge. i.e., 1 / (2 `theta) diverges. So if 1 / (2 `theta) < 1 / `sqrt( `theta^2 + 1) we will prove the divergence of 1 / `sqrt(`theta^2 + 1).

We prove this. Starting with

1 / (2 `theta) < 1 / `sqrt( `theta^2 + 1) we square both sides to get

1 / (4 `theta^2) < 1 / (`theta^2+1). Multiplying by common denominator we get

`theta^2 + 1 < 4 `theta^2. Solving for `theta^2 we get

1 < 3 `theta^2

`sqrt(3) / 3 < `theta.

This shows that for all values of `theta > `sqrt(3) / 3, or about `theta > .7, the function 1 / `sqrt(`theta^2+1) exceeds the value of the divergent function 1 / (2 `theta). Our function 1 / `sqrt(`theta^2 + 1) is therefore divergent. **

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20:01:12

does the integral converge or diverge, and why?

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RESPONSE -->

I explained this in my previous response.

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20:01:18

If you have not already stated it, with what convergent or divergent integral did you compare the given integral?

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20:03:34

query problem 7.8.19 (3d edition #20) convergence of integral from 0 to 1 of 1 / `sqrt(`theta^3 + `theta)

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RESPONSE -->

This appears to be problem 13 in the second edition.

In this case comparison is to 1/`theta^(3/2).

Comparing this to 1/x^p from 0 to 1, when p > 1, the integral converges, so this one converges as well.

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20:03:50

does the integral converge or diverge, and why?

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RESPONSE -->

I explained this one previously.

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20:03:53

If you have not already stated it, with what convergent or divergent integral did you compare the given integral?

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RESPONSE -->

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20:23:59

Query problem 8.1.5. Riemann sum and integral inside x^2 + y^2 = 10 within 1st quadrant, using horizontal strip of width `dy.

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RESPONSE -->

This one does not appear to be in the second edition.

Since we are stacking the slices vertically, the integral will be with respect to y, so we need to rearrange the equation.

x^2 + y^2 = 10

x = `sqrt(10 - y^2)

The Reimann Sum will be `sigma `sqrt(10 - y^2) * `dy with slices from y = 0 to `sqrt(10)

The integral will be integral of `sqrt(10 - y^2) * dy from 0 to `sqrt(10)

This looks like # 30 in the tables

1/2 [ y[`sqrt(10 - y^2) + 10 integral 1/(`sqrt(10 - y^2)) dy] ]

1/2 [ y[`sqrt(10 - y^2) + 10 arcsin y/`sqrt(10)] ] from 0 to `sqrt(10)

1/2 [ `sqrt(10)[0 + 10 arcsin 1 ] ] -

1/2 [ 0 [`sqrt(10 - 0^2) + 10 arcsin 0/`sqrt(10)] ]

1/2 * [0 + 10/2 * `pi] = 5/2 * `pi

Of course we could have found the area of this quarter circle by 1/4 * `pi * [`sqrt(10)]^2 = 5/2 * `pi

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20:24:51

Give the Riemann sum and the definite integral it approaches.

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RESPONSE -->

The Reimann Sum will be `sigma `sqrt(10 - y^2) * `dy with slices from y = 0 to `sqrt(10)

The integral will be integral of `sqrt(10 - y^2) * dy from 0 to `sqrt(10)

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20:25:18

Give the exact value of your integral.

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RESPONSE -->

The integral will be integral of `sqrt(10 - y^2) * dy from 0 to `sqrt(10)

This looks like # 30 in the tables

1/2 [ y[`sqrt(10 - y^2) + 10 integral 1/(`sqrt(10 - y^2)) dy] ]

1/2 [ y[`sqrt(10 - y^2) + 10 arcsin y/`sqrt(10)] ] from 0 to `sqrt(10)

1/2 [ `sqrt(10)[0 + 10 arcsin 1 ] ] -

1/2 [ 0 [`sqrt(10 - 0^2) + 10 arcsin 0/`sqrt(10)] ]

1/2 * [0 + 10/2 * `pi] = 5/2 * `pi

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21:03:07

Query problem 8.1.12. Half disk radius 7 m thickness 10 m, `dy slice parallel to rectangular base.

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RESPONSE -->

This time, the integral will be slightly different.

The equation of the circle will be x = `sqrt( 7^2 - y^2)

and will be integrating from 0 to 7. We need twice the length of x this time because we are doing a half circle. In addition, the disk is 10m thick.

The integral will be 2 * 10 * `sqrt(7^2 -y^2) from 0 to y

Again, utilizing table entry #30,

20 * [ 1/2 [ 7 * `sqrt(7^2 - 7^2) + 7^2 arcsin 7/7] -

1/2 [ 0 * `sqrt(7^2 - 0^2) + 7^2 arcsin 0/7] ]

This evaluates to 20 * 49/2 * `pi = 490 `pi.

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22:39:32

Give the Riemann sum and the definite integral it approaches.

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RESPONSE -->

The Riemann sum will be:

`sigma 20 * sqrt(7^2-y^2) * `dy from y = 0 to y =7

The integral will be 2 * 10 * `sqrt(7^2 -y^2) from 0 to 7

I believe I may have mistated the integral earlier saying from 0 to y when it should be 0 to 7.

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22:39:56

Give the exact value of your integral.

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RESPONSE -->

The exact value of the integral is 490 `pi.

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17:13:58

query problem 8.2.11 arc length x^(3/2) from 0 to 2

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RESPONSE -->

The integral for this arc length x^(3/2) is

integral from a to b of `sqrt(1 + (f'(x))^2) dx

f'(x) = 3/2 * x ^*(-1/2) = 3/2 * 1/`sqrt(x)

So the integral is `sqrt(1 + ( 3/2 * 1/sqrt(x) )^2) dx from 0 to 2

I approximated the arc length using the simpson's rule to approximate using 10 slices. The result is approx. 2.374

The arc starts at (0, 0) and ends at (2, 2.828), the latter being correct to at least 3 significant figures. The straight-line distance between the points is greater than 2.374. Your derivative should be 3/2 sqrt(x), not 3/2 * 1 / sqrt(x).

On an interval of length `dx, containing x coordinate c_i, the slope triangle at the top of the approximating trapezoid has slope approximately equal to f (c_i). The hypotenuse of this triangle corresponds to the arc length.

A triangle with run `dx and slope m has rise equal to m * `dx. So its hypotenuse is sqrt((`dx)^2 + (m `dx)^2) = sqrt( (1 + m^2) * (`dx)^2) = sqrt( 1 + m^2) * `dx. For small `dx, the hypotenuse is very close to the curve so its length is very near the arc length of the curve on the given interval.

Since the slope here is f (c_i), we substitute f (c_i) for m and find that the contribution to arc length is

`dL_i = sqrt(1 + f ^2 (c_i) ) * `dx

So that the Riemann sum is

Sum(`dL_i) = sum ( sqrt(1 + f ^2 (c_i) ) * `dx ),

where the sum runs from i = 1 to i = n, with n = (b a) / `dx = (2 0) / `dx. In other words, n is the number of subintervals into which the interval of integration is broken.

This sum approaches the integral of sqrt(1 + (f ' (x)) ^2, over the interval of integration.

In general, then, the arc length is

arc length = integral(sqrt(1 + (f ' (x)) ^2) dx, x from a to b).

In this case

f(x) = x^(3/2) so f ' (x) = 3/2 x^(1/2), and (f ' (x))^2 = (3/2 x^(1/2))^2 = 9/4 x.

Thus sqrt( 1 + (f (x))^2) = sqrt(1 + 9/4 x) and we find the integral of this function from x = 0 to x = 2.

The integral is found by letting u = 1 + 9/4 x, so that u = 9/4 and dx = 4/9 du, so that our integral becomes

Integral ( sqrt( 1 + 9/4 x) dx, x from 0 to 2)

= Integral ( sqrt(u) * 4/9 du, x from 0 to 2)

= 4/9 Integral ( sqrt(u) du, x from 0 to 2)

Our antiderivative is 4/9 * 2/3 u^(3/2), which is the same as 8/27 (1 + 9/4 x) ^(3/2). Between x = 0 and x = 2, the change in this antiderivative is

8/27 ( 1 + 9/4 * 2) ^(3/2) 8/27 ( 1 + 9/4 * 0) ^(3/2)

= 8/27 ( ( 11/2 )^(3/2) 1)

= 3.526, approximately.

Thus the arc length is

integral(sqrt(1 + (f ' (x)) ^2) dx, x from a to b

= Integral ( sqrt( 1 + 9/4 x) dx, x from 0 to 2)

= 3.526.

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17:14:51

what is the arc length?

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RESPONSE -->

The arc length is approx. 2.374

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17:15:05

What integral do you evaluate obtain the arc length?

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RESPONSE -->

`sqrt(1 + ( 3/2 * 1/sqrt(x) )^2) dx from 0 to 2

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17:17:09

What is the approximate arc length of a section corresponding to an increment `dx near coordinate x on the x axis?

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RESPONSE -->

The approximate arc length at x is `sqrt(1 + (f'(x))^2) `dx

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17:19:43

What is the slope of the graph near the graph point with x coordinate x?

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RESPONSE -->

The slope at point x will be f'(x)

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17:26:14

How is this slope related to the approximate arc length of the section?

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RESPONSE -->

Since `dy/`dx is the slope, `dy/`dx = f'(x)

`dy = f'(x) `dx

using 'dx for x and `dy for y in the pythageorean theorem to find the length of the hypotenuse we factor out `dx to become `sqrt(1 + (f'(x))^2) 'dx which is the approximate arc length of the section.

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19:29:58

query problem 8.2.31 volume of region bounded by y = e^x, x axis, lines x=0 and x=1, cross-sections perpendicular to the x axis are squares

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RESPONSE -->

The volume will be the integral of (e^x)^2 dx from 0 to 1

It becomes integral of e^2x dx

Using u = 2x, 1/2du = dx

integral of 1/2 e^u du

1/2 [e^(2*1) - e^(2*0)]

1/2 (e^2 - 1) = approx 3.195

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19:30:25

what is the volume of the region?

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RESPONSE -->

The volume is approx. 3.195

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19:30:47

What integral did you evaluate to get the volume?

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RESPONSE -->

The integral of (e^x)^2 dx from 0 to 1

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19:31:53

What is the cross-sectional area of a slice perpendicular to the x axis at coordinate x?

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RESPONSE -->

(e^x)^2

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19:32:24

What is the approximate volume of a thin slice of width `dx at coordinate x?

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RESPONSE -->

(e^x)^2 * `dx

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19:34:03

How the you obtain the integral from the expression for the volume of the thin slice?

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RESPONSE -->

The integral is simply a re-expression of the Riemann sum of the volumes of thin slices

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19:34:12

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

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RESPONSE -->

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"

Very good work. Just a couple of errors in detail. See my notes.