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Mth 174

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App to Geometry Question 1

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Problem Number 1

Find the volume of the solid obtained by rotating the curve y = x^- 1, between x = 0 and x = .87, about the line x = -1.901.

 

V = 1/2 pi r^2

1/2 pi int((1/x^2) dx x, 0, .87)

= 1/2 pi * -1/x + C

-pi/2(.87) + pi/2(0) = -1.8055 approx.

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Let's first, by way of illustration, set up the integral for the entire solid formed by rotating this curve about the given line.

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If the curve is rotated about the line x = -1.9, the solid obtained can be thought of as a stack of circles extending from y = 0 to y = infinity, or alternatively as a series of cylindrical shells expanding outward from the line x = -1.9.

The typical circle obtained by slicing the solid with a horizontal plane at coordinate y on the vertical axis will have a radius equal to the distance from the line x = -1.9 to the graph of y = x^-1. The point on that line corresponding to vertical coordinate y is found by solving the defining equation y = x^-1 for x. We obtain x = y^-1.

The radius of the circle in question thus extends from x = -1.9 to x = y^-1. The distance, which is equal to the radius, is thus

r = y^-1 - (-1.9) = y^-1 + 1.9.

The area of the circle is

A(y) = pi r^2 = pi * (y^-1 + 1.9)^2.

Now we partition the y axis into subintervals, the typical subinterval, the ith subinterval, having width `dy and containing sample point y_i *.

The approximate volume within the solid, corresponding to this interval, is equal to the area of the circle corresponding to the sample point y_i, multiplied by the width `dy of the interval. We thus get the volume increment

`dV = pi I( y_i*^-1 + 1.9)^2 * `dy.

Summing all such intervals and allowing the width `dy to approach zero, noting that y values run from 0 to infinity, we get the volume integral

V = integral( pi (y^-1+1.9)^2 dy, y from 0 to infinity)

or if we write y^-1 as 1/y

V = integral ( pi (1/y + 1.9)^2 dy, y from 0 to infinity).

As it turns out this integral diverges, proving that the volume of the solid is infinite.

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If we now consider only the portion of the curve between x = 0 and x = .87, we see that the y values do not go from 0 to infinity.

At x = .87, we have y = x^-1 = 1 / .87, which is about 1.15. So our integral would run from y = 1.15 to infinity.

You can see, if you find the antiderivative, than the integral once again diverges and that this part of the solid also has infinite volume.

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Mth 174

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Prob. Distribution Question 1

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Problem Number 9

If p(x) = k x (1-x), 0 <= x <= 1, represents a probability distribution, then

• Find the formula for the cumulative probability distribution P(x), and sketch its graph.

• What is the value of k?

• Use the cumulative distribution function to find the probability that x lies between .47 and .53.

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How do I determine the value of k?

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One defining property of a probability distribution is that the total probability is 1.

There is only one value of k for which the integral of kx ( 1 - x), from x = 0 to x = 1, is equal to 1.

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Mth 174

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Annual Rate of Return Question

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Problem Number 10

In 6 years we wish for an income stream of $ 160,000 per year to grow to $ 272,000.   What annual rate of return, compounded continuously, will be required?

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I understand how to determine future and present value , compounding annually, but I'm confused about how to determine the rate?

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Assuming annual rate r, the income that 'flows into' the account during a short time interval `dt containing the time t_i* will have 6 - t_i* years to grow at the given rate.

What will be the income flow during the interval `dt?

To what amount will that income grow in the remaining 6 - t_i* years?

What sum do you get when you add up the income from all such intervals?

What integral do you get when you allow the interval `dt to approach zero?

You might also want to review qa_10 from the first-semester course, as recommended on your assignments page.

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