Assignment 8

course Mth 174

whɒNz݌Tz˶qassignment #008

د`SB

Physics II

06-13-2009

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17:07:15

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

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An integral converges if the answer to the integral is a finite number. If the answer is undefined or it approaches a without bound it diverges. These integrals measure the area under the curve, and many times there is an asymptote that has an effect on whether it converges or diverges.

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17:13:40

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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It is important to be able to tell if the function converges or diverges by taking the integral and looking at the exponent because sometimes two functions can appear to be very similar before you take there integrals. When an integral as an exponent that is less than or equal to 1 it diverges, and greater than 1 converges.

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17:13:50

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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** There is very little difference in the rapidity with which 1/x^1.01 approaches zero and the rapidity of the approach of 1 / x^.99. You would never be able to tell by just looking at the partial sums which converges and which diverges. In fact just looking at partial sums you would expect both to converge, because they diverge so very slowly.

If you graph 1 / x^.99, 1 / x and 1 / x^1.01 you won't notice any significant difference between them. The first is a little higher and the last a little lower past x = 1. You will never tell by looking at the graphs that any of these functions either converge or diverge.

However if you find the antiderivatives you can tell pretty easily. The antiderivatives are 100 x^.01, ln(x) and -100 x^-.01.

On the interval from, say, x = 1 to infinity, the first antiderivative 100 x^.01 will be nice and finite at x = 1, but as x -> infinity is .01 power will also go to infinity. This is because for any N, no matter how great, we can find a solution to 100 x^.01 = N: the solution is x = (N / 100) ^ 100. For this value of x and for every greater value of x, 100 x^.01 will exceed N. The antiderivative function grows without bound as x -> infinity. Thus the integral of 1 / x^.99 diverges.

On this same interval, the second antiderivative ln(x) is 0 when x = 0, but as x -> infinity its value also approaches infinity. This is because for any N, no matter how large, ln(x) > N when x > e^N. This shows that there is no upper bound on the natural log, and since the natural log is the antiderivative of 1 / x this integral diverges.

On the same interval the third antiderivative -100 x^-.01 behaves very much differently. At x = 1 this function is finite but negative, taking the value -100. As x -> infinity the negative power makes x-.01 = 1 / x^.01 approach 0, since as we solve before the denominator x^.01 approaches infinity. Those if we integrate this function from x = 1 to larger and larger values of x, say from x = 1 to x = b with b -> infinity, we will get F(b) - F(1) = -100 / b^.01 - (-100 / 1) = -100 / b^.01 + 100. The first term approaches 0 and b approaches infinity and in the limit we have only the value 100. This integral therefore converges to 100.

We could look at the same functions evaluated on the finite interval from 0 to 1. None of the functions is defined at x = 0 so the lower limit of our integral has to be regarded as a number which approaches 0. The first function gives us no problem, since its antiderivative 100 x^.01 remains finite over the entire interval [0, 1], and we find that the integral will be 100. The other two functions, however, have antiderivatives ln(x) and -100 x^-.01 which are not only undefined at x = 0 but which both approach -infinity as x -> 0. This causes their integrals to diverge.

These three functions are the p functions y = 1 / x^p, for p = .99, 1 and 1.01. This example demonstrates how p = 1 is the 'watershed' for convergence of these series either on an infinite interval [ a, infinity) or a finite interval [0, a]. **

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17:16:06

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

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RESPONSE -->

I this case, the integral has a negative exponent and therefore as x approaches infinity, the limit approaches 0 and you end up with a finite result.

The convergence of the integral of e^-(ax) from 0 to infinity is plausible because this is a decreasing function. However 1 / x^.99 is also decreasing on the same interval but it doesn't converge.

The reason e^(- a x) converges is that it antiderivative is -1/a e^(-a x), and as x -> infinity this expression approaches zero.

Thus the integral of this function from 0 to b is -1/a e^(-a b) - (-1/a e^0) = -1/a e^(-ab) + 1.

If we allow the upper limit b to approach infinity, e^-(a b) will approach 0 and we'll get the finite result 1. **

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17:24:00

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

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RESPONSE -->

For this problem you can compare 1 / (`theta^2+1) to simply 1/(theta)^2 because as theta gets larger the 1 becomes less significant. The integral of 1/(theta)^2 is ln(theta^2) which is infinite so it diverges.

** 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 we can show that 1 / (2 `theta) < 1 / `sqrt( `theta^2 + 1) we will have proved 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.

STUDENT ERROR:

This integral converges because 1/sq rt(theta^2) approaches 0 rapidly.

It will indeed converge but your argument essentially says that it converges because it converges. Way too vague. You have to use a comparison test of some kind. **

You have the right idea, but being less than a converging comparison function would prove convergence; being greater than a diverging function would prove divergence.

However being greater than a converging function, or being less than a diverging function, does not prove anything at all about the convergence of the original function.

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17:24:11

does the integral converge or diverge, and why?

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It diverges because the limit is infinite.

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17:24:48

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

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I compared it with 1/theta^2 which is divergent.

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17:30:08

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 -->

For this one you can also compare the original function to 1/sqrt(theta^3) because as this theta increases the theta on the right becomes less significant. You can then find that the limit of 1/sqrt(theta^3) is greater than that of the original problem. 1/sqrt(theta^3) = 1/ theta^3/2. And the integral causes the exponent to remain positive, and the p test states that this means it converges.

1 / `sqrt(theta^3 + `theta) < 1 / `sqrt(`theta^3) = 1 / `theta^(3/2).

This is an instance of 1 / x^p for x =`theta and p = 3/2.

The integral of 1 / theta^(3/2) converges by the p test.

The given integrand is less than 1 / theta^(3/2), so the original integral also converges, by the comparison test.

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17:30:24

does the integral converge or diverge, and why?

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RESPONSE -->

It converges because the exponent is positive.

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17:30:30

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

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17:33:59

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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You first use the Pythagorean Theorem to find Width so: h2 + (w/2)^2 = r^2

2= sqrt' (4r^2 -h^2)=

2 sqrt' r^2 - h^2=

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The Riemann sum approximation is the sum of w times the chage in h which =

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um of 2 sqrt' r^2 - h times the change in h

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I was not sure what to do from here

FOR HORIZONTAL STRIPS

The solution for the x of the equation x^2 + y^2 = 10 is x = +- sqrt(10 y^2). In the first quadrant we have x > = 0 so the first-quadrant solution is x = +sqrt(10 y^2).

A vertical strip at position y extends from the y axis to the point on the curve at which x = sqrt(10 y^2), so the altitude of the strip is sqrt(10 y^2). If the width of the strip is `dy, then the strip has area

`dA = sqrt(10 y^2) `dy.

The curve extends along the y axis from y = 0 to the x = 0 point y^2= 10, or for first-quadrant y values, to y = sqrt(10). If the y axis from y = 0 to y = sqrt(10) is partitioned into subintervals the the total area of the subintervals is approximated by

A = sum(`dA) = sum ( sqrt(10 y^2) `dy), and as interval width approaches zero we obtain the area

A = integral ( sqrt(10 y^2) dy, y, 0, sqrt(10)).

The integral is performed by letting y = sqrt(10) sin(theta) so that dy = sqrt(10) cos(theta) * dTheta; 10 y^2 will then equal 10 10 sin^2(theta) = 10 ( 1 sin^2(theta)) = 10 sin^2(theta) and sqrt(10 y^2) becomes sqrt(10) cos(theta); y = 0 becomes sin(theta) = 0 so that theta = 0; y = sqrt(10) becomes sqrt(10) sin(theta) = sqrt(10) or sin(theta) = 0, giving us theta = pi / 2.

The integral is therefore transformed to

Int(sqrt(10) cos(theta) * sqrt(10) cos(theta), theta, 0, pi/2) =

Int(10 cos^2(theta) dTheta, theta, 0, pi/2).

The integral of cos^2(theta) from 0 to pi/2 is pi/4, so that the integral of our function is 10 * pi/4 = 5/2 pi.

Note that this is the area of the circle x^2 + y^2 = 10.

FOR VERTICAL STRIPS

The solution for the y of the equation x^2 + y^2 = 10 is y = +- sqrt(10 x^2). In the first quadrant we have y > = 0 so the first-quadrant solution is y = +sqrt(10 x^2).

A vertical strip at position x extends from the x axis to the point on the curve at which y = sqrt(10 x^2), so the altitude of the strip is sqrt(10 x^2). If the width of the strip is `dx, then the strip has area

`dA = sqrt(10 x^2) `dx.

The curve extends along the x axis from x = 0 to the y = 0 point x^2= 10, or for first-quadrant x values, to x = sqrt(10). If the x axis from x = 0 to x = sqrt(10) is partitioned into subintervals the the total area of the subintervals is approximated by

A = sum(`dA) = sum ( sqrt(10 x^2) `dx), and as interval width approaches zero we obtain the area

A = integral ( sqrt(10 x^2) dx, x, 0, sqrt(10)).

The integral is performed by letting x = sqrt(10) sin(theta) so that dx = sqrt(10) cos(theta) * dTheta; 10 x^2 will then equal 10 10 sin^2(theta) = 10 ( 1 sin^2(theta)) = 10 sin^2(theta) and sqrt(10 x^2) becomes sqrt(10) cos(theta); x = 0 becomes sin(theta) = 0 so that theta = 0; x = sqrt(10) becomes sqrt(10) sin(theta) = sqrt(10) or sin(theta) = 0, giving us theta = pi / 2.

The integral is therefore transformed to

Int(sqrt(10) cos(theta) * sqrt(10) cos(theta), theta, 0, pi/2) =

Int(10 cos^2(theta) dTheta, theta, 0, pi/2).

The integral of cos^2(theta) from 0 to pi/2 is pi/4, so that the integral of our function is 10 * pi/4 = 5/2 pi.

Note that this is the area of the circle x^2 + y^2 = 10.

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17:34:39

Give the Riemann sum and the definite integral it approaches.

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RESPONSE -->

Riemann sum approximation = the sum of w * change in h

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17:34:50

Give the exact value of your integral.

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17:37:54

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

Give the Riemann sum and the definite integral it approaches.

You have to do the Riemann sum, get the integral then perform the integration.

A slice of the region parallel to the y axis is a rectangle. If the y coordinate is y = c_i then the slice extends to an x coordinate such that x^2 + (c_i)^2 = 7^2, or x = sqrt(49 - (c_i)^2).

So if the y axis, from y = -7 to y = 7, is partitioned into subintervals y_0 = -7, y1, y2, ..., y_i, ..., y_n = 7, with sample point c_i in subinterval number i, the corresponding 'slice' of the solid has thickness `dy, width sqrt(49 - (c_i)^2) and length 10 so its volume is 10 * sqrt(49 - (c_i)^2) * `dy and the Riemann sum is

sum(10 * sqrt(49 - (c_i)^2) * `dy, i from 1 to n).

The limit of this sum, as x approaches infinity, is then

integral ( 10 * sqrt(49 - y^2) dy, y from 0 to 7) =

10 integral (sqrt(49 - y^2) dy, y from 0 to 7).

Using the trigonometric substitution y = 7 sin(theta), analogous to the substitution used in the preceding solution, we find that the integral is 49 pi / 2 and the result is 10 times this integral so the volume is

10 * 49 pi / 2 = 490 pi / 2 = 245 pi / 2.

A decimal approximation to this result is between 700 and 800. Compare this with the volume of a 'box' containing the figure:

The 'box' would be 14 units wide and 7 units high, and 10 units long, so would have volume 14 * 7 * 10 = 980.

The solid clearly fills well over half the box, so a volume between 700 and 800 appears very reasonable.

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

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RESPONSE -->

Arc length is approximate to the sum of sqrt (1 + f' (x))^2 change in x

y= sqrt' x^3

f(x)= (3x^.5)/2

1 + sqrt' (3^.5)/2 dx = 3.2

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.

what is the arc length?

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RESPONSE -->

3.2

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17:41:25

What integral do you evaluate obtain the arc length?

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RESPONSE -->

you evaluate the integral from 0 to 2

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17:41:59

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 -->

sqrt' (1 + f'(x)^2) dx

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17:43:18

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

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I'm not sure how to figure this out.

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17:44:08

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

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RESPONSE -->

increases in the arc length creates changes in the slope.

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17:45:02

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 -->

V= s^2 change in x

= sum of (e^x)^2 change in x

V= (e^2 - .5) is approximate to 3.195

x runs from 0 to 1.

At a given value of x the 'altitude' of the y vs. x graph is e^x, and a slice of the solid figure is a square with sides e^x so its area is (e^x)^2 = e^(2x).

If we partition the interval from x = 0 to x = 1 into subintervals defined by x_0 = 0, x_1, x_2, ..., x_i, ..., x_n, with interval number i containing sample point c_i, then the slice corresponding to interval number i has the following characteristics:

the thickness of the 'slice' is `dx

the cross-sectional area of the 'slice' at the sample point x = c_i is e^(2x) = e^(2 * c_i)

so the volume of the 'slice' is e^(2 * c_i) * `dx.

The Riemann sum is therefore

sum(e^(2 * c_i * `dx) and its limit is

integral(e^(2 x) dx, x from 0 to 1).

Our antiderivative is easily found to be 1/2 e^(2 x) and the change in our antiderivative is

1/2 e^(2 * 1) - 1/2 e^(2 * 0) = 1/2 e^2 - 1/2 = 1/2 (e^2 - 1).

The approximate value of this result about 3.19.

A 'box' containing this region would have dimensions 1 by e by e, with volume e^2 = 7.3 or so. The figure fills perhaps half this box, so our results are reasonably consistent.

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17:45:11

what is the volume of the region?

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RESPONSE -->

V= 3.195

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17:45:53

What integral did you evaluate to get the volume?

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RESPONSE -->

the integral from 0 to 1 (e^x)^2 change in x

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17:46:26

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

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I think the area of the slice is the integral of that coordinate dx.

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17:47:07

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

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RESPONSE -->

s^2 times the change in x

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17:47:21

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

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17:47:34

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

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