Honors AP Calculus / Mr. Hansen
2/2/2003

Name: _________________________

§8-9

2.

20p

8.(a)

r = 10/(3 – 3 cos q)

(b)

12p Ö5

(c)

33.074

14.

50 – 6.25p

20.(a)

974.307 million square miles

(b)

Use root finder to get q » 1.890.

(c)

k » 0.1762

(d)

P » 81.9 hr.

(e)

Total area » 8429.778 million square miles. Time (proportional fraction) must be 9.466 hrs.

(f)

From 0 to 0.2 radians, L » 20,200 mi. From .8 to 1.890 radians, L » 56,800 mi.

22.

[The content of #22 will not be on the test. However, the answer to part (a) illustrates material that you should be able to handle if it appears on the test, because we discussed it in class.]

(a)

We learned about the conversion between polar and rectangular coordinates, namely

x = r cos q
y = r sin q

Therefore, at any point (r, q), y/x can be found by substitution from the equations above to be (r sin q)/(r cos q) = tan q. (Q.E.D.)

(b)

Since the function is parameterized by q, we can use the parametric chain rule to get dy/dx = (dy/dq) / (dx/dq). But the LHS of this eqn. is (by geometric def. of derivative) the slope of the tangent line, i.e., the tangent of the angle f as shown. (Q.E.D.)

(c)

Perhaps you don’t recall the given identity for tan(A – B) from your study of trigonometry. However, since you were not required to prove the identity, you can still apply it as follows:

Let x´ and y´ denote dx/dq and dx/dq, respectively.
tan y = tan(f – q)
          = (tan f – tan q)/(1 + tan f tan q)
          = (y´/x´ – y/x) / (1 + yy´/xx´)

Now multiply both numerator and denominator by xx´. (Q.E.D.)

(d)

From part (a) and application of chain rule, we have

x´ = –r sin q
y´ = r cos q

\ numerator = xy´ – yx´
          = (r cos q)(r cos q) – (r sin q)(–r sin q)
          = r²(cos² q + sin² q)
          = r²

(e)

Again applying the equations from part (a), it is obvious that r² = x² + y². Let r´ denote dr/dq, and use implicit differentiation (wrt q) of the equation r² = x² + y² to get

2rr´ = 2xx´ + 2yy´

Divide through by 2 to see that the denominator in the result of part (c) is simply r´. Thus tan y = r/r´ as claimed. (Q.E.D.)

(f)

By our formula in part (e), we have tan y = r/r´ = (a – a cos q)/(a sin q)
= (1 – cos q)/sin q.

There is a trig identity that says that this last expression equals tan(q/2).

Thus q/2 and the original angle y differ by some multiple of p radians.

But since q Î [0, 2p] and y Î [0, p] by the way the angles were defined, clearly q/2 and y are the same angle. (There is no other way they could both have the same value for the tangent function.)

(g)

Since y is constant (given), tan y is a constant. Let k = tan y. By the formula in part (e),
k = r/r´ Þ r´ = r/k = Kr where K = 1/k is also a constant. We are well acquainted with the diffeq. r´ = Kr, and we know that it has the general solution r = Ce^Kq.

If we measure r in inches and hold a ruler next to the photograph as carefully as possible, we get the following two data points in polar coordinates:

(1/8, 2p)
(31/32, 6p)

(The choice of points is not critical; just choose two that are far enough apart to give some semblance of accuracy.) Using these initial conditions and some basic precal-type algebra, we find C = .0449..., K = .16295..., so that when the polar graph r = Ce^Kq is plotted and traced, the spiral matches the book’s photograph very closely. As a check, the radius for q = 4p would be .348 by calc., and that is essentially exactly what a careful ruler measurement would give.

The chambered nautilus is doing some interesting polar graphing without even realizing it. Pretty neat, wouldn’t you say?

Review

R1.(a)

[Use calc.]

(b)

f ´(x) = 3x² – 18x + 30; f ´´(x) = 6x – 18
g ´(x) = 3x² – 18x + 27; g ´´(x) = 6x – 18
h ´(x) = 3x² – 18x + 24; h ´´(x) = 6x – 18

(c)

h ´(x) = 3(x – 2)(x – 4) = 0 when x Î {2, 4}.
h ´´(2) = 6(2) – 18 < 0 Þ x = 2 gives a local max.
h ´´(4) = 6(4) – 18 > 0 Þ x = 4 gives a local min.

Note that f ´ has no zeros, and g ´ has only one zero.

(d)

That would be g.
g ´(3) = 3(x – 3)² = 0 only when x = 3
However, g ´(x) > 0 for all x in a deleted nbhd. of 3, by which we mean the set

(3 – e, 3 + e) \ {3}

Therefore, x = 3 gives neither a local max. nor a local min.

(e)

A pt. of infl., by definition, is a point of continuity at which the second deriv. changes sign. Functions f, g, and h are polynomials, hence cont. everywhere. All three experience a sign change in second deriv. at x = 3 since 6x – 18 changes sign there. Hence all three have a pt. of infl. at x = 3, and 6x – 18 = 0 for all three. (Q.E.D.)

R2.(a)

f

incr.

no max. or min.

incr.

f ´

+

DNE

+

x

2

f

È

pt. of infl.

Ç

f ´´

+

DNE

–

x

2

(b)

Fcn. must have upward concavity (È shape) on (–2, 3), downward concavity (Ç shape) on (3, 5), then upward concavity again to right of x = 5. There could be a cusp at x = –2 or a complete break. (A cusp is simpler to draw.) Other than that, though, the fcn. is assuredly continuous, with a local min. at x = 1. There is also upward concavity to the left of x = –2. At x = 5, there must be a plateau pt. that is also a pt. of inflection. Note that if you draw the fcn. as described, you will also get a pt. of infl. at x = 3.

(c)

i. f ´(x) = 2/3 x^–1/3 – 1, f ´´(x) = –2/9 x^–4/3

ii. Zooming in reveals local min. (cusp) at (0, 0) and local max. at x » .3.
You should find these analytically as follows:
f ´(x) = 0 only when x = 8/27 [solution of eq. omitted, but you should show work]
f ´(x) DNE only when x = 0, and since f ´(x) < 0 for x Î (–.1, 0) and f ´(x) > 0 for x Î (0, .1), x = 0 clearly gives a local min.

There are no other critical values. (You must say that.)

iii. f ´´(x) < 0 for all x ¹ 0, and f ´´(0) DNE; since f ´´ does not change sign anywhere, there can be no pts. of infl.

iv. We must check critical points and interval endpoints.
f(0) = 0, f(8/27) = 4/27, f(5) = 5^2/3 – 5 » –2.076
\ global max. at (8/27, 4/27), global min. at (5, –2.076)

(d)

Global min. at (0, 0), local max. at (2, .541), no global max. since fcn. is unbounded in Quadrant II

Pts. of infl. at (.586, .191) and (3.414, .384)

However, you can’t reason from the graphing calc. except in multiple choice. Here is how you would analyze f in a free-response problem:

Analysis of first deriv.: f ´(x) = x(2 – x)e^–x, which is – for x < 0, + for x Î (0, 2), then – again for x > 2. That is how you can tell there is a local min. for x = 0 and a local max. for x = 2. Since f is (by inspection) always > 0 except at the origin, we know the origin is not only a local min. but a global min.

Analysis of second deriv.: f ´´(x) = (x² – 4x +2)e^–x, which takes its sign according to the quadratic x² – 4x +2 since e^–x is always > 0. By quadratic formula (or factoring and sketching), the quadratic has roots at 2 ± Ö2 and is < 0 in between. Since f is cont. everywhere and f ´´ changes sign at 2 ± Ö2, those x values give pts. of inflection.

R3.(a)

Let x = width of a cell, y = length of a cell.
xy = 10 Þ y = 10/x on domain (0, ¥) = Â⁺, i.e., the positve real numbers
Objective: Minimize L(x) = 12x + 7y = 12x + 70/x
L´(x) = 12 – 70/x² = 0 only when x » 2.415... [work omitted]

Important: We know this is the only critical point on domain Â⁺ since L´ is defined on Â⁺ and has only one root. Since L´ changes sign from – to + there [or you could say since L´´(x) = 140/x³ > 0 everywhere on Â⁺], the value 2.415... gives a global min.

y value for this x is 4.140... [work omitted]

Overall length of battery = 6x = 14.491...

The optimal battery based on the objective fcn. above (approx. 14.5" by 4.1") is much longer and narrower than the typical battery. Thus, minimal wall length does not seem to be a major consideration in battery design.

(b)

Let x = width of rectangle = radius of cylinder
Let y = height of rectangle = height of cylinder

Domain for both x and y is Â⁺
Objective: Maximize V(x) = pr²h = px²y = px²(8 – x³) = 8px² – px⁵

V ´(x) = px(16 – 5x³) = 0 only when x = 0 (not allowed) or x = 1.4736... [work omitted].
Important: We know 1.4736... gives a global max. since V is cont. on its domain, has only one critical pt. on its domain, and V ´´(x) = 16p – 20px³ < 0 at our solution. [The value is about –150.796, but if you didn’t have a calculator, you could simply estimate. You know x > 1, so 16p – 20px³ < –4p , and all we care about is the sign. Or, if you don’t like the second deriv. test, you could analyze the sign of V ´(x) = px(16 – 5x³) in a small deleted nbhd. of 1.4736... as follows: clearly V ´ > 0 to the left of x = 1.4736... and V ´ < 0 to the right, since the factor px does not change sign there, but the factor (16 – 5x³) does. The sign change in V ´ proves that we have a local max., and since we have cont. and no other critical pts. on Â⁺, x = 1.4736... actually gives a global max.]

Answer [work omitted]: rectangle dimensions are 1.4736... (width) by 4.8 (height).

R4.(a)

A = ∫₁² e^y dy = 4.6707... [work omitted]

(b)

Let y = first fcn., z = second fcn.
A = ∫_–1⁸ (y – z) dx = ∫_–1⁸ (x^1/3 – x/3 + 2/3) dx = 6.75 [work omitted]

(c)

The functions switch place (top and bottom) at the origin. Correct method:
A = ∫_–1⁰ (x³ – x) dx + ∫₀¹ (x – x³) dx

Test

T1.

f ´(x) = 3(x – 2.5)(x – 2.7), so f ´ changes from + to – at x = 2.5 and from – back to + at x = 2.7. Thus there is a local max. at x = 2.5 and a local min. at x = 2.7. [You could also compute f ´´(2.5) and f ´´(2.7) to prove these facts, but the factor analysis is certainly faster in this case.]

f ´´(x) exists everywhere, and f ´´(x) = 6x – 15.6 = 6(x – 2.6) = 0 iff x = 2.6
\ x = 2.6 is the only possibly inflection pt.

Since polynomial fcn. f is everywhere cont., and since a sign change in f ´´ clearly occurs at x = 2.6, x = 2.6 gives a pt. of infl.

However, f ´(2.6) = 3(x – 2.5)(x – 2.7) = 3(.1)(–.1) = –.03 ¹ 0, so f is not horizontal at the pt. of infl.

T2.

Here we are given that f is cont. [unlike in R2.(b) above], so there must be a cusp at x = 2. The fcn. has a pt. of infl. at x = 1, an upward-pointing cusp at x = 2, a second pt. of infl. at x = 3, and yet another pt. of infl. (though this time with a vertical tangent) at x = 4. The function should decrease to the right of x = 4. Make sure that f(0) = 3 as required, and you should have a valid sketch.

T3.

L = ∫₀² (1 + 9x⁴)^1/2 dx = 8.6303... [Numeric integration is required unless you have a table of integrals. Nowadays, we assume that you will use the calculator. On the AP exam, you need to show the work leading to the integral (namely the squaring of dy/dx and its insertion into the arc-length formula), the definite integral, and the answer. Of course, if you are in the no-calculator section of the exam, you would not be expected to compute the answer.]

T4.

[Omit, since we did not study §8-8. The answer is 77.3245... in case you are curious.]

T5.

32.746 cm³ [work omitted since this is quite similar to R3.(b)]

T6.

V = ∫₀⁸ p(y^1/3)² dy = 19.2p [work omitted]

T7.

V = ∫₀² 2px(8 – x³) dx = 19.2p [work omitted]

T8.

V_cyl = p r²h = p (2²)8 = 32p
ratio is 19.2p /(32p ) = .6 or 60%

By parameterizing the problem (i.e., by letting r be an adjustable constant instead of 2 as in this problem), you can actually prove the following theorem: A cubic paraboloid occupies 60% of the volume of its circumscribed cylinder.

T9.(a)

[Use calc.]

(b)

Use the parametric arc length formula, showing formula, plug-ins, and solution.
L = ∫₀^2p ((dx/dt)² + (dy/dt)²)^1/2 dt = ∫₀^2p ((–5 sin t)² + (2 cos t)²)^1/2 dt = 23.0131...

(c)

Use disks (circular cross sections) perpendicular to the x-axis.

There are several ways to do this. One way is to leave everything in terms of t and compute V = ∫_p⁰ p(2 sin t)²(–5 sin t dt). Note that the second factor is actually dx, but everything including the limits of integration has to be rewritten in terms of t.

Another way is probably easier, though it won’t necessarily work in general. Analyze the curve algebraically to eliminate the parameter t altogether: Observe that x²/25 + y²/4 = 1 by subst. and use of the familiar identity cos² q + sin² q = 1. Solve for y to get
y = ± 2(1 – x²/25)^1/2, but actually the sign is irrelevant since we will be squaring y (the radius of each slice) anyway. V = ∫_–5⁵ py² dx = ∫_–5⁵ p(4(1 – x²/25)) dx = 80p/3 [work omitted]. But 4/3 p (r_x) (r_y)² = 4/3 p (5) (2²) = 80p/3 also. (Q.E.D.)

A third way, and probably the best, is to eliminate the parameter t and then rework the problem with the x radius and y radius being parameters (adjustable constants). You can prove V = 4/3 p (r_x) (r_y)² directly, which (as in T8) then becomes a theorem in its own right: The volume of an ellipsoid with circular cross sections equals 4/3 p times the radius perpendicular to the slices times the square of the maximum cross section radius.

T10.

Use polar arc length formula [work omitted] to get L = ∫₀^6p (r² + (dr/dq)²)^1/2 dq = 280.6961... The answer can actually be given in an exact closed form as
L = 10(e^0.6p – 1)Ö25.25, which you should be able to find by FTC.

T11.

Use polar area formula twice, once for the area swept out in Quadrant I for q Î [4p, 4.5p] and once for the area in Quadrant I for q Î [2p, 2.5p]. The shaded area is then the first answer minus the second. Using the formula A = ∫ ½ r² dq for each gives

∫_4p ^4.5p ½ (5e^0.1q )² dq – ∫_2p ^2.5p ½ (5e^0.1q )² dq = 203.7405 [work omitted]

As in T10, you should be able to use FTC, if asked, to find the exact answer, namely
62.5(e^0.9p – e^0.8p – e^0.5p + e^0.4p).