Assignment 8

course PHY 174

dRYwfŎassignment #008

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Physics II

11-26-2007

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15:30:12

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

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RESPONSE -->

A function converges if the area underneath approaches a finite value.

A function diverges if the are underneath approaches infinity.

** An integral converges if the limit of its Riemann sums can be found and is finite. Roughly this corresponds to the area under the curve (positive for area above and negative for area below the axis) being finite and well-defined (the area under the graph of the sine function, from 0 to infinity, could be regarded as finite but it keeps fluctuating as the sine goes positive then negative, so it isn't well-defined). **

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16:47:51

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 -->

For values of p less than 1, the function grows smaller as x gets larger, but at a rate that does not approach a finite value for the function as x gets arbitrarily large. For p = 1, the function grows smaller as x gets larger, approaching zero. For all p greater than 1, the function gets smaller at a greater rate, approaching zero.

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19:56:11

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 -->

diverges for p >= 1

converges for p < 1

For p > 1

integral of 1/x^2 from 0 to 1

as x gets arbitrarily small, the integral does not approach a finite number.

For p = 1

integral of 1/x from 0 to 1

as x gets arbitrarily small, the integral does not approach a finite number.

For p = .5

integral of 1/x^.5 from 0 to 1

as x gets arbitrarily small, the integral approaches 2.

Good answer. The following is included simply to amplify the question a bit

** 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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20:28:27

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

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RESPONSE -->

For positive values of a, as x approaches infinity, the function does not approach a finite value.

For negative values of a, as x approaches infinity, the function approaches the finite value 0.

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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20:38:51

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

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RESPONSE -->

integral of 1 / (`theta^2+1) from 1 to infinity =

integral of 1 / (`theta^2+1^2) from 1 to infinity =

(This fits formula V.24.)

limit of arctan `theta from 1 to b as b approaches infinity =

arctan(9*10^99) - acrtan(1) = 0.13389

Good, but note that the arctangent function approaches pi/2 as the argument approaches infinity. arctan(1) = pi/4. So the integral approaches pi/2 - pi/4 = pi/4.

arctan(9 * 10^99) - arctan(1) should have given you the same result, to many significant figures; however that close to the calculator's limits things might have gone haywire.

The problem as given in the text was actually 1 / sqrt(theta^2 + 1), and is worth a look:

** 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. **

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20:39:58

does the integral converge or diverge, and why?

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RESPONSE -->

The integral converges. It approaches the finite value 0.13389 as x approaches infinity.

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20:57:27

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

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RESPONSE -->

It could have been compared to the integral of 1/x^p from 1 to infinity for p > 1.

The function, `sqrt (`theta^2 + 1), will always be slightly larger than `theta itself. In other words, `theta raised to a power greater than 1.

Your instinct is correct, but adding 1 to theta^2 won't increase the basic power of the result. For example sqrt(theta^2+1) is always less than 2 theta, so the integrand is always less than 1/ 2 * 1 / theta, and half of a divergent integral is still divergent.

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19:03:21

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 -->

The problem you have there is #20 in the 4th edition.

The `sqrt(`theta^3 + `theta) = `theta^p where p is greater than 1. The integral from 0 to 1 of 1 / `sqrt(`theta^3 + `theta) is similar to the integral from 0 to 1 of 1 / `x^p, where p is greater than 1. It follows that the integral diverges.

You are correct. A little more specifically:

** 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.

So the integral converges by the p test. **

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19:03:45

11-27-2007 19:03:45

does the integral converge or diverge, and why?

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NOTES ------->

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19:04:03

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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21:07:46

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 -->

x^2 + y^2 = 10

s = is the length of each slice

area of each slice = s`dy

s is the length of the adjacent side of a triangle with hypotenuse `sqrt(10) and height y.

s in terms of y is

s = `sqrt(10 - y^2)

The sum of `sqrt(10 - y^2) 'dy

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

5`pi / 2 = area of the region

Very good. Compare with the following:

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.

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21:12:09

Give the Riemann sum and the definite integral it approaches.

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RESPONSE -->

The limit as n approaches infinity, so `dy approaches 0 of the sum from i = 1 to n of `sqrt(10 - y^2) 'dy

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

5`pi / 2 = area of the region

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21:12:23

Give the exact value of your integral.

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RESPONSE -->

5`pi / 2

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21:31:13

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

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RESPONSE -->

1/2 `pi r^2 * L = area of the half cylinder

`dy * s * 10 approximates the volume of the slice

s is twice the length of the adjacent side of a triangle with hypotenuse 7 and height y.

s = 2 * `sqrt(49 - y^2)

The limit as n approaches infinity, so `dy approaches 0 of the sum from i = 1 to n of 10 * 2 * `sqrt(49 - y^2)`dy =

The integral from 0 to 7 of 20 * `sqrt(49 - y^2) dy =

245 `pi m^3 = volume of the region

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21:31:53

Give the Riemann sum and the definite integral it approaches.

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RESPONSE -->

The limit as n approaches infinity, so `dy approaches 0 of the sum from i = 1 to n of 10 * 2 * `sqrt(49 - y^2)`dy =

The integral from 0 to 7 of 20 * `sqrt(49 - y^2) dy =

245 `pi m^3 = volume of the region

Also very good. For comparison:

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.

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21:32:05

Give the exact value of your integral.

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RESPONSE -->

245 `pi m^3

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22:41:01

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

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RESPONSE -->

f(x) = `sqrt(x^3) = x^(3/2)

f'(x) = 3x^(1/2) / 2

arc length of the graph of f(x) from 0 to 2 =

the integral from 0 to 2 of `sqrt(1+(3x^(1/2) / 2)^2) dx = approximately 3.53

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22:41:11

what is the arc length?

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RESPONSE -->

approximately 3.53

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22:41:22

What integral do you evaluate obtain the arc length?

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RESPONSE -->

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

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22:50:24

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

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22:51:53

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

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RESPONSE -->

(3x^(1/2) / 2) `dx / `dx

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22:54:53

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

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RESPONSE -->

(3x^(1/2) / 2) 'dx / `dx

sqrt(1+(3x^(1/2) / 2)^2) `dx

The slope is the quotient of side opp. and side adj. of a triangle with hypotenuse = approx. arc length.

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23:15:30

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 approx. area of a slice = `dx * y

y = 1 - x^2

total volume = the sum of (1 - x^2) `dx

The limit as `dx approaches 0 gives the definite integral from 0 to 1 of (1 - x^2) dx

the definite integral from 0 to 1 of (1 - x^2) dx = 2/3

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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23:15:39

what is the volume of the region?

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RESPONSE -->

2 / 3

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23:16:40

What integral did you evaluate to get the volume?

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RESPONSE -->

The definite integral from 0 to 1 of (1 - x^2) dx = 2/3

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23:17:11

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

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RESPONSE -->

the approx. area of a slice = `dx * y

y = 1 - x^2

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23:17:46

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

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RESPONSE -->

total volume = the sum of (1 - x^2) `dx

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23:18:01

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

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RESPONSE -->

the approx. area of a slice = `dx * y

y = 1 - x^2

total volume = the sum of (1 - x^2) `dx

The limit as `dx approaches 0 gives the definite integral from 0 to 1 of (1 - x^2) dx

the definite integral from 0 to 1 of (1 - x^2) dx = 2/3

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23:18:13

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

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RESPONSE -->

fun

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Really good work here. See inserted notes for comparison and amplification of some ideas.