Honors AP Calculus / Mr. Hansen
3/17/2003

Name: _______________________

 

Answers and Partial Solutions to HW in §11-6

4.(a)

v has units in./sec., and area has units in.2, so product has units in.3/sec. as required

(b)

Since v = 4 – x2, dv/dx = –2x = 0 only when x = 0. Since dv/dx changes from pos. to neg. there, the point (0, 4) is clearly the global max. The other endpoint (x = 4) gives a velocity value of 0, which cannot compete. [On the AP, it is safer to give an explanation like this than to give the easier (and more concise) reason that v = 4 – x2 is obviously a downward-opening parabola with vertex at (0, 4).]

To summarize, v(0) = 4 – 02 = 4, which means max. velocity at center, and v(4) = 4 – 42 = 0, which means zero velocity at the pipe walls.

(c)

Let F denote total flow rate through pipe, and use cylindrical shells to evaluate. We have a variable-factor product: (velocity for a cylindrical shell) times (cross-sectional area of that shell).
dF = v dA
F = ς02 (4 – x2)2px dx » 25.133 in.3/sec.

(d)

6.528 gal./min.

(e)

4(p22) = 16p » 13.056 gal./min., or more than twice as much

(f)

similar to finding volume by cylindrical shells, with v taking the place of h

 

 

8.(a)

T(0) = 20 sin 0 = 0, and T(.25) = 20 sin (2p · .25) = 20 as required. Note that there is an error in the text; the function T(t) should be defined as temperature above normal as opposed to mere temperature. There was an AP question very similar to this a few years ago; in that one, the function given was for outside temperature, and students were supposed to compute the temperature difference by subtracting a reference temperature from that function. Needless to say, many people forgot to do that and simply integrated temperature with respect to time. Although integrating T with respect to time is the correct procedure in problem #8, you have to remember that T is not temperature but rather the temperature above normal, i.e., the temperature difference.

(b)

This is a variable-factor product of temperature difference (as explained above) and a small slice of time. We get ς0.25 T(t) dt = ς0.25 20 sin 2pt dt = 10/p » 3.183 degree-days.

 

 

15.(a)

We can factor f as (3 + x)(3 – x), giving ±3 as the only roots by inspection. As for g, synthetic division by (x – 3) comes out even, showing that 3 is a root; the quotient, namely –x2/3 – 2x – 3, must have the same roots as the polynomial x2 + 6x + 9 since multiplying by –3 does not alter the roots. By inspection, this latter polynomial factors into (x + 3)(x + 3), giving a double root at –3. You are expected to be able to perform analysis of this difficulty during the AP exam.

(b)

ς–33 f(x) dx = ς–33 (9 – x2) dx = 2ς03 (9 – x2) dx = 2(9x – x3/3)|03 = 36 by properties of even functions over symmetric intervals. Similarly,
ς–33 g(x) dx = ς–33 (–x3/3 – x2 + 3x + 9) dx
                    = ς–33 (– x2 + 9) dx [since odd terms integrate to 0 over symmetric interval]
                    = 2ς03 (– x2 + 9) dx
                    = 36 as before.

(c)

Since f ’(x) = –2x changes from pos. to neg. at 0, x = 0 gives a max. for f at (0, 9). [You cannot apply the second-derivative test.]

For g, we know g ’(x) = –x2 – 2x + 3 = –(x – 1)(x + 3) = 0 iff x = 1 or x = –3. Since g(1) = 32/3 > g(–3), x = 1 is the only candidate for a local max. At the other endpoint, g(3) = 0, so x = 1 is the only global max. candidate. By second deriv. test, g ’’(x) = –2x – 2 ή g ’’(1) = –4 < 0 ή max. at the point (1, 32/3), when x = 1.

(d)

[See definition of moment on p.568.] The first moment of g wrt y-axis is ς–33 xg(x) dx = 21.6 by calc. By setting 36xbar = 21.6, we get xbar = .6.

(e)

Clearly false, since gmax is at x = 1, not .6.

(f)

Also false. Although the assertion is true for symmetric distributions, g is not symmetric. Area to left of centroid = ς–30.6 g(x) dx = 17.1072 Ή area to right of centroid = ς0.63 g(x) dx = 18.8928.

(g)

[Again refer to the definition of moment on p.568.] Skewness of g = ς–33 (x – 0.6)3g(x) dx » –17.774 by calc.

(h)

First moment of f wrt y-axis is ς–33 xf(x) dx = ς–33 (9x – x3) dx = 0 by inspection (odd function integrated over a symmetric interval). Setting 36xbar = 0 gives xbar = 0, which means that our third-moment skewness must be computed about the line x = 0. Skewness of f = ς–33 (x – 0)3f(x) dx = ς–33 (9x3 – x5) dx = 0 by inspection (odd function integrated over a symmetric interval).

Saying that the skewness of f is 0 makes sense because x is not skew at all; f is symmetric. [In statistics, you will learn that a function like g that has negative skewness is said to be skew left. The visual cue to remember this is that the function “dribbles out” toward the left.]

(i)

The easiest solution is to define h(x) = g(–x) = x3/3 – x2 – 3x + 9, which is a “mirror image” of g wrt the y-axis. By replicating the actions of steps (d) and (g), using xbar = –0.6, you can compute the skewness of h to be 17.774. This makes sense, since h is merely a “skew right” version of g.