Assignment 8

course Mth 174

 ãŽÞâºßëvúØòáÓRã¤Ö¡®³ª¶öçassignment #008

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Physics II

02-18-2007

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19:07:50

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

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RESPONSE -->

A series converges if the limit of the series as the number of terms approaches infinity (or negative infinity) approaches a finite number. A series diverges if the limit of the series as the number of terms approaches infinity (or negative infinity) does not approach a finite number.

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19:21:30

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 -->

If p > 1, the integral of 1/ x^p from 1 to infinity is equal to 1/(p-1), which is finite.

If p < 1, the integral (shown on page 350, example 3) grows without bound.

If p = 1, the integral is lim (b>inifinity) of ln x from 1 to b, which is lim(b>infinity) of ln b - ln 1, and ln b grows without bound as b>infinity.

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19:59:24

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 -->

The integral of 1/x^p, evaluated from 0 to b is (1/-p+1)*b^(-p+1). If the exponent of b is positive, as b>0, b^(-p+1) approaches 0. If the exponent is negative, b^(-p+1) grows without limit. The exponent is positive when p<1 and negative when p>1. When p=1, the first part of the expression (1/(-p+1)) becomes indeterminate.

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20:18:22

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), for any a>0, is (-1/a)e^(-a x), evaluated from 0 to b, which is (-1/a)e^(-ab) - (-1/a)e^(0). The first term approaches 0 rapidly as b grows rapidly, while the second term is a constant, thus the integral will converge to the constant.

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20:19:15

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

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20:27:38

does the integral converge or diverge, and why?

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RESPONSE -->

The integral diverges.

Since sqrt(theta^2+1) is >= sqrt(theta^2), 1/sqrt(theta^2+1) <= 1/sqrt(theta^2). Thus the integral of 1/sqrt(theta^2+1) <= integral of 1/sqrt(theta^2) (both integrals from 1 to infinity). The integral of 1/sqrt(theta^2) is the integral of 1/theta from 1 to infinity, which by rule diverges (p=1).

This says that the given integral is less than a divergent integral. You can't draw a conclusion from that.

You're on the right track, but there's a little more work to do.

Consider 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 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. **

This integral is an elementary function and can be computed and evaluated. This results in two terms, one of which (ln(b+sqrt (b^2+1)) diverges and the other (ln(1+sqrt 2) is a constant. Thus the integral diverges.

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20:27:53

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

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RESPONSE -->

See previous response.

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20:29:55

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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20:35:30

does the integral converge or diverge, and why?

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RESPONSE -->

The integral diverges. For theta>=0, theta^3+theta is >=theta^3, so sqrt theta^3+theta >= sqrt theta^3. Thus 1/sqrt theta^3+theta <= 1/sqrt theta^3, and integral of 1/sqrt theta^3+theta <= integral of 1/sqrt theta^3 (both integrals from 0 to 1). Integral of 1/sqrt theta^3 = integral of 1/theta^(3/2), which by rule diverges (p>1).

For p > 1 the integral converges. So the given integrand is less than that of a convergent integral, so your integral converges.

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20:35:41

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

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RESPONSE -->

See previous response.

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20:35:58

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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20:42:13

Give the Riemann sum and the definite integral it approaches.

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RESPONSE -->

Area of slice = xsubi*delta y.

xsubi^2+ysubi^2 = 10, solved for xsubi = sqrt(10-ysubi^2).

The Riemann sum is sum(i=1 to n) xsubi * delta y

= sum(i=1 to n) sqrt(10-ysubi^2)*delta y.

The definite integral it approaches is:

Integral (0 to sqrt 10) sqrt(10-y^2) dy

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20:45:43

Give the exact value of your integral.

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RESPONSE -->

Using formula #30 and formula #28, the integral is

(y/2)*sqrt(10-y^2) + 5 arcsin (y/sqrt 10), evaulated from 0 to sqrt 10. This simplifies to

5 arcsin 1 - 5 arcsin 0 = 5*(pi/2) - 0 = (5pi)/2

Good. This is one quadrant of a circle whose squared radius is 10. Your result is easy to verify indepencently: the total area of the circle is pi r^2 = pi * 10; the area in one quadrant is 1/4 of this, or 5 pi / 4.

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20:50:17

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

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20:59:07

Give the Riemann sum and the definite integral it approaches.

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RESPONSE -->

Volume(slice) = lengthsubi*10*delta y

By Pythagorean theorem: lengthsub1/2 = sqrt(7^2 - ysubi^2), so lengthsubi = 2*sqrt(7^2 - ysubi^2)

Volume(slice) = 2*sqrt(7^2 - ysubi^2) * 10 * delta y = 20*sqrt(7^2 - ysubi^2) delta y

Riemann sum (volume) = sum (i=1 to n) 20*sqrt(7^2 - ysubi^2) delta y

Integral (from 0 to 7) 20*sqrt(7^2-y^2) dy

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21:04:25

Give the exact value of your integral.

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RESPONSE -->

Using formulas #30 and #28, the integral is:

10y*sqrt(7^2-y^2) + 490 arcsin (y/7), evaluated from 0 to 7

Simplifies to 0 + 490 arcsin 1 - 490 arcsin 0

= 490pi/2 - 0

= 245pi m^3

Good. Again not hard to verify independently. This region is half of a cylinder with cross-sectional radius 7 m. Cylinder cross-sectional area is 49 pi m^2; length is 10 m so total area is 490 pi m^3. Half the volume is 245 pi m^3.

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21:05:00

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

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21:12:24

what is the arc length?

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RESPONSE -->

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

Substituting in the arc length formula:

Integral (from 0 to 2) of sqrt[1+((3/2)x^(1/2))^2] dx

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

= 3.53 (approx.) (using calculator to compute)

If you let u = 9/4 x then du = 9/4 dx and your integrand becomes 4/9 sqrt(u); antiderivative is 8/27 u^(3/2), which converts to 8/27 ( 1 + 9/4 x)^(3/2).

The change in this antiderivative from x = 0 to x = 2 is from 8/27 to 8/27(1 + 9/2)^(3/2). Approximated to 3 significant figures this is 3.53, in agreement with your result.

Note that the calculator wasn't necessary except to evaluate the expression. It's not clear whether you used the calculator to do the integration or to approximate the change in the antiderivative function.

However most arc length integrals cannot be computed exactly and approximate methods or calculators are generally necessary.

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21:12:37

What integral do you evaluate obtain the arc length?

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RESPONSE -->

See previous response.

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21:23:11

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 -->

If dx is sufficiently small for f(x) to approximate a straight line at x, then the arc length can be computed using the distance formula:

Arc length = sqrt(dx^2 + dy^2)

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21:24:32

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

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RESPONSE -->

The slope of the graph is f'(x) = (3/2)x^(1/2).

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21:31:11

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

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Since the slope represents dy/dx, if we know the slope and dx, we can find dy to substitute into the distance formula in the previous response.

The greater the slope, the longer the arc length for a given dx.

Geometric reasoning, which I believe you understand:

Think of the top of a trapezoid running from x to x + `dx. The width of the trapezoid is `dx. Its slope is f'(x). The little triangle at the top therefore has run `dx and slope f'(x), so its rise is rise=slope*run = f'(x) * `dx. Its hypotenuse, which approximates the arc distance, is thus hypotenuse = `sqrt(rise^2 + run^2) = `sqrt( (f'(x) `dx)^2 + `dx^2) = `sqrt( 1 + f'(x)^2) `dx. That leads directly to the formula.

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21:31:36

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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21:40:05

what is the volume of the region?

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RESPONSE -->

Side of square = ssubi = y = e^x

Volume(slice) = ssubi^2 * delta x = (e^x)^2 * delta x = e^(2x) * delta x

Riemann sum (volume of region) = sum (i=1 to n) e^(2x) delta x

Integral (from 0 to 1) e^(2x) dx = (e^(2x))/2, evaluated from 0 to 1

= e^2/2 - e^0/2 = e^2/2 - 1/2 = 3.19 units^3 (approx.)

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21:40:17

What integral did you evaluate to get the volume?

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RESPONSE -->

See previous response.

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21:42:59

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

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RESPONSE -->

The cross-sectional area of the slice is (e^x)^2 or e^(2x).

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21:44:47

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

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RESPONSE -->

The approximate volume of a thin slice of width dx is the cross-sectional area * dx or e^(2x) * dx.

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21:47:40

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

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RESPONSE -->

The Riemann sum for the volume is the sum of the volumes of the thin slices, which is approximated by taking the integral of the expresson for the volume of the thin slice and evaluating it between the two ends of the interval.

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21:50:08

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

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RESPONSE -->

This was a very lengthy assignment!

I had some initial difficulty with 8.2, #26 and 27--particularly #26--visualizing the shape of the solid. I had to do #27 first, which then gave me an idea of how to visualize #26.

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Very good work. See my note for what they're worth.