Assignment 8 Query

course Mth 174

??w????O?^?????assignment #008

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.

?????????????

Physics II

10-15-2007

......!!!!!!!!...................................

15:14:26

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

......!!!!!!!!...................................

RESPONSE -->

Both of these terms look at functions as they tend towards infinity, checking to see if there is a limit to the function.

In the case of convergence, if the limit exists, then the value function is said to converge on the limit.

In the case of divergence, the limit does not exist and the value of function increases (or decreases) continually.

** 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]. **

.................................................

......!!!!!!!!...................................

15:32:02

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

......!!!!!!!!...................................

RESPONSE -->

In the case of p > 1, the value of the function quickly moves to a very small number (1 being divided by a large number). Since it quickly moves to a finite number, it then converges.

In the case of p < 1, the value of the function grows without bound, tending towards infinity itself as the upper limit of the integral goes towards infinity. Since there is no finite value, the function is said to diverge.

In the case of p = 1, the value of the function slowly moves towards a finite value. Since it does not quickly move towards a finite number, the function is said to diverge.

The numbers don't tell you much. The key is the antiderivative. If p < 1 then the antiderivative is a positive power of x and hence, since positive powers approach infinity as x approaches infinity, the integral diverges. If p > 1 then the antiderivative is a negative power of x and, since negative power approach 0 as x approaches infinity, the integral converges.

.................................................

......!!!!!!!!...................................

15:45:40

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

......!!!!!!!!...................................

RESPONSE -->

In the case of p > 1, the value of function is 1 divided by a small number. This returns a larger number which shrinks over the interval and does not tend towards a fixed value, therefore diverging.

In the case of p < 1, the function grows slowly towards a fixed value, thereby converging.

In the case of p = 1, the value of the function shrinks over the interval, but does not tend towards a fixed value, therefore diverging.

When p> or = 1, the integral diverges because it approaches infinity.

When p<1, the integral converges because it approaches 1

.................................................

......!!!!!!!!...................................

15:49:15

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

......!!!!!!!!...................................

RESPONSE -->

Because for successively higher values of x, the value of the function tends towards 0 and moves quickly towards it.

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. **

.................................................

......!!!!!!!!...................................

15:58:36

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

......!!!!!!!!...................................

RESPONSE -->

The function as listed in the book is:

1 / 'sqrt(`theta^2+1)

This can be compared to the function 1 / 'sqrt('theta^2) = 1 / 'theta. This function can be considered as a member of the family 1 / x^p, which we know how that acts for different values of p.

Since the value of p in this case is 1, then all we have to do is consider the interval of integration, which in this case is 1 to infinity. In this case it diverges.

Since the square root of any number plus one is greater than the square root of that number, then our comparison function is less than the original in value. Therefore since the comparison function diverges, the original function diverges as well.

Right idea, but being less than a converging comparison function would prove convergence; being greater than a converging function would prove divergence.

However being greater than a converging function, or being less than a diverging function, does not prove convergence. Your argument corresponds to the latter, so you need to do a little more work.

** 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 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. **

.................................................

......!!!!!!!!...................................

15:58:41

does the integral converge or diverge, and why?

......!!!!!!!!...................................

RESPONSE -->

See prior answer.

.................................................

......!!!!!!!!...................................

15:58:51

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

......!!!!!!!!...................................

RESPONSE -->

1 / 'theta

.................................................

......!!!!!!!!...................................

16:10:00

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

......!!!!!!!!...................................

RESPONSE -->

Here we can compare the original function with 1 / 'sqrt( 'theta^3 ), which is the same as 1 / 'theta^( -3/2 ). Since the value of p in this case is less than one and the interval is from 0 to 1, the comparison function converges.

This would be theta^(-3/2) or 1 / theta^(3/2). In the latter form we see that this is indeed a p series, but p > 1, not < 1.

However, I just realized that I made a mistake on the prior answer. 1 divided by a large number equals a small number, and 1 divided by a larger number equals a smaller number. In other words, my last answer was wrong.

Anyway, in this one, the value of the comparison function is greater than the original, and the comparison diverges. Therefore, the original diverges.

being less than a diverging function does not imply divergence; however thsi is in fact less than a converging function, so is convergent.

.................................................

......!!!!!!!!...................................

16:10:05

does the integral converge or diverge, and why?

......!!!!!!!!...................................

RESPONSE -->

See prior.

.................................................

......!!!!!!!!...................................

16:10:16

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

......!!!!!!!!...................................

RESPONSE -->

See prior.

.................................................

......!!!!!!!!...................................

16:46:43

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

......!!!!!!!!...................................

RESPONSE -->

Riemann sum: 'sqrt( 10 - y^2 ) 'dy

Integral: int( 'sqrt( 10 - y^2 ) 'dy, y, 0, 'pi ).

Good, but y values range from 0 to sqrt(10), not from 0 to pi.

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 horizontal 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 Ό the area of the circle x^2 + y^2 = 10.

.................................................

......!!!!!!!!...................................

16:46:52

Give the Riemann sum and the definite integral it approaches.

......!!!!!!!!...................................

RESPONSE -->

See prior answer.

.................................................

......!!!!!!!!...................................

16:47:42

Give the exact value of your integral.

......!!!!!!!!...................................

RESPONSE -->

Exact value: 5 pi / 2

.................................................

......!!!!!!!!...................................

16:52:56

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

......!!!!!!!!...................................

RESPONSE -->

Riemann: 10 ( 2 'sqrt(49 - y^2) ) 'dy

Integral: int(10 ( 2 'sqrt(49 - y^2) ) 'dy, y, 0, 7)

.................................................

......!!!!!!!!...................................

16:53:01

Give the Riemann sum and the definite integral it approaches.

......!!!!!!!!...................................

RESPONSE -->

See above.

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.

.................................................

......!!!!!!!!...................................

16:53:53

Give the exact value of your integral.

......!!!!!!!!...................................

RESPONSE -->

490 'pi / 2 m^3

.................................................

......!!!!!!!!...................................

16:57:35

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

......!!!!!!!!...................................

RESPONSE -->

I found the integral to be:

int( 'sqrt( 1 + (-3/2 'sqrt(x) )^2 ) 'dx, x, 0, 2 )

The approximated value: 3.526

Good. The details:

The arc length is

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

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.

So our integral is

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

We easily integrate this letting u = 1 + 9/4 x so that du = 9/4 dx; the limits x = 0 and x = 2 become u = 1 + 9/4 * 0 = 1 and u = 1 + 9/4 * 2 = 11/2 so we find

4/9 integral(sqrt(u), u from 1 to 11/2).

Our antiderivative of sqrt(u) is 2/3 u^(3/2) and the change in the antiderivative is

2/3 ( 11/2 ) ^ (3/2) - 3/2 ( 1 ) ^ (3/2) = 2/3 ( 11/2)^(3/2) - 1)

and the arc length is

4/9 * 2/3 ( 11/2)^(3/2) - 1) = 8/27 ( (11/2)^(3/2) - 1).

This expression can be simplified or approximated. We get approximately 3.53.

The result should also be compared to the estimated arc length of a sketch of the graph of the original function. The original function starts at (0, 0) and ends at (2, 2 sqrt(2)). A straight line segment between these points has length about 3.46, and the arc will be a little longer than this straight-line approximation. This is consistent with the decimal approximation of the above expression.

.................................................

......!!!!!!!!...................................

16:57:42

what is the arc length?

......!!!!!!!!...................................

RESPONSE -->

See approximated value.

.................................................

......!!!!!!!!...................................

16:57:49

What integral do you evaluate obtain the arc length?

......!!!!!!!!...................................

RESPONSE -->

See above.

.................................................

......!!!!!!!!...................................

17:00:49

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

......!!!!!!!!...................................

RESPONSE -->

Wouldn't this just be:

'sqrt( 1 + (-3/2 'sqrt(x) )^2 ) 'dx

.................................................

......!!!!!!!!...................................

17:04:23

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

......!!!!!!!!...................................

RESPONSE -->

For a general approximation in a small area, I think all that we need to do is divide the rise by the run:

f' (x) 'dx / 'dx = f' (x)

.................................................

......!!!!!!!!...................................

17:07:04

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

......!!!!!!!!...................................

RESPONSE -->

If I got it right, then the approximate arc length of each tiny section is simply the slope of that section multiplied by the length of the section.

.................................................

......!!!!!!!!...................................

17:13:22

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

......!!!!!!!!...................................

RESPONSE -->

I found the volume integral to be:

int( e^(2x), x, 0, 1 )

This evaluated to exactly: (1/2) e^2 - 1/2

Approximation: 3.1945

In the answer section in the back of the book, the answer to this problem is listed as:

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

Did they just leave out the 1/2 on the e^2? When I punched their numbers into the calculator, it didn't match their approximate integral.

(e^2 - 1) / 2 would be right and would agree with both this evaluation and with your result.

It's possible that you misread the grouping in the book's answer. It's also possible that the book messed up the grouping. I wouldn't bet against you or against them, but it's probably one or the other.

Details of the solution:

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 is between 3 and 3.5.

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.

.................................................

......!!!!!!!!...................................

17:13:42

what is the volume of the region?

......!!!!!!!!...................................

RESPONSE -->

V = 3.195

.................................................

......!!!!!!!!...................................

17:13:48

What integral did you evaluate to get the volume?

......!!!!!!!!...................................

RESPONSE -->

See above.

.................................................

......!!!!!!!!...................................

17:14:23

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

......!!!!!!!!...................................

RESPONSE -->

The cross sectional area would be:

(e^x)^2 or just e^(2x)

.................................................

......!!!!!!!!...................................

17:14:48

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

......!!!!!!!!...................................

RESPONSE -->

Simply enough, area multiplied by depth ('dx):

e^(2x) 'dx

.................................................

......!!!!!!!!...................................

17:18:39

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

......!!!!!!!!...................................

RESPONSE -->

By taking the Riemann sum of the overall area and converting it to an integral with the limits of 0 and 1 since 0 and 1 are the bounds on the left and right.

Riemann: e^(2x) 'dx

Integral: int(e^(2x) 'dx, x, 0, 1)

.................................................

......!!!!!!!!...................................

17:19:15

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

......!!!!!!!!...................................

RESPONSE -->

Interesting stuff. Downright cool at times. So that's how they figured out those formulas for volume!

.................................................

"

Good work, but there are a couple of discrpancies and you didn't show the detailed development of the Riemann Sums, so be sure to see my notes.