CALCULUS III FALL 1999

HOMEWORK 2 – ANSWERS

§9.3 Questions 6,8,12,22,24; §9.4 Questions 26,36,40,42,78

6. We have

\frac{dx}{dθ} = a(− sin θ + sin θ + θ cos θ) = aθ cos θ

and

\frac{dy}{dθ} = a(cos θ − cos θ + θ sin θ) = aθ sin θ.

So the required length is

L = ∫_{0}^{π} \sqrt{a^{2}θ^{2} cos^{2} θ + a^{2}θ^{2} sin^{2} θ} dθ = ∫_{0}^{π} |a|θ dθ = \frac{1}{2}π^{2}|a|.

8. We have

\frac{dx}{dt} = e^{t} − 1 and \frac{dy}{dt} = 2e^{t/2}.

So the required length is

L = ∫_{0}^{1} \sqrt{(e^{t} − 1)^{2} + 4e^{t}} dt = ∫_{0}^{1}(1 + e^{t}) dt = e.

12. We have

\frac{dx}{dθ} = −\frac{2a}{sin^{2} θ} and \frac{dy}{dθ} = 4a sin θ cos θ.

So the required length is

L = ∫_{π/4}^{π/2} \sqrt{\frac{4a^{2}}{sin^{4} θ} + 16a^{2} sin^{2} θ cos^{2} θ} dθ ≃ 2.26|a|.

CALCULUS III  HOMEWORK 2 – ANSWERS  FALL 1999 2.

22. We have dx/dt = 3 − 3t^{2} and dy/dt = 6t, so the required surface area is

S = 2π ∫_{0}^{1} 3t^{2}\sqrt{(3 − 3t^{2})^{2} + 36t^{2}} dt = 6π ∫_{0}^{1} t^{2}\sqrt{9t^{4} + 18t^{2} + 9} dt

= 6π ∫_{0}^{1} t^{2}(3t^{2} + 3) dt = 18π[\frac{t^{5}}{5} + \frac{t^{3}}{3} ]_{0}^{1} = \frac{48}{5}π.

24. Notice that, as θ varies from −π to π the curve is below the x-axis and as it varies from 0 to π it desrcibes the mirror image above the x-axis, this follows from the fact that sin is an odd function. Note that for 0 ⩽ θ ⩽ π, dy/dx = (cos θ − cos 2θ)/(sin 2θ − sin θ) except when θ = 0, π/3, π. In this parameter interval, dy/dθ > 0 for θ < θ_{0} where cos θ_{0} = (1 − \sqrt{5})/4 and dy/dθ < 0 for θ > θ_{0}. Similarly, dx/dθ > 0 for θ < π/3; and dx/dθ < 0 for θ > π/3. So dy/dx > 0 if 0 < θ < π/3; dy/dx < 0 if π/3 < θ < θ_{0}; dy/dx > 0 if θ_{0} < θ < π. From these observations we can deduce that the curve has no self intersections for the parameter interval [0, π]. This can be much more easily observed by asking your calculator to plot the curve. But you should at least know how to derive it without a calculator. The required surface area is obtained by rotating the part of the curve corresponding to 0 < θ < π about the x-axis. This gives

S = 2π ∫_{0}^{π}(2 sin θ − sin 2θ)\sqrt{4(sin θ − sin 2θ)^{2} + 4(cos θ − cos 2θ)^{2}} dθ

= 8π\sqrt{2} ∫_{0}^{π}(1 − cos θ) sin θ\sqrt{1 − sin θ sin 2θ − cos θ cos 2θ} dθ

= 8π\sqrt{2} ∫_{0}^{π}(1 − cos θ)^{3/2} sin θ dθ = 8π\sqrt{2}[2(1 − cos θ)^{5/2}/5]_{0}^{π} = \frac{128}{5}π.

26. We have x = r cos θ = 2 sin θ cos θ = sin 2θ and y = r cos θ = 2 sin^{2} θ = 1 − cos 2θ. Consequently, x^{2} + (y − 1)^{2} = 1; which is the circle center (0, 1) and radius 1.

36. We have r^{2} cos^{2} θ − r^{2} sin^{2} θ = 1 which is equivalent to r^{2} cos^{2} 2θ = 1 or r^{2} = sec 2θ for −π/4 < θ < π/4.

40. We have x = r cos θ = −4 sin θ cos θ = −2 sin 2θ and y = r sin θ = −4 sin^{2} θ = 2(cos 2θ − 1). Consequently, x^{2} + (y + 2)^{2} = 4 and therefore the curve is the circle center (0, −2) and radius 2. Alternatively, we recall from class that the curve r = −4 cos θ is the circle center (−2, 0) and radius 2. Now note that sin θ = cos(θ+π/2) and therefore the required cirve arises from rotating the latter circle through π/2.

42. Note that r = 2(sin θ + cos θ) = 2\sqrt{2} cos(θ − π/4). So we should expect the curve obtained by rotating r = 2\sqrt{2} cos θ through π/4 (counterclockwise) about the origin. The latter is a circle and so we see that the required curve is

and this curve is described once as θ varies from 0 to π.

78.

a). r = sin(θ/2) must be a curve which stays within distance one of the origin. It will pass through the origin when θ = 0, 2π, 4π. Other points where it crosses the y-axis arise when cos θ = 0; these are

θ = π/2, 3π/2, 5π/2, 7π/2.

Calculating the y-coordinate for these values shows that, apart from the origin, the curve crosses the y-axis twice more. The only possible curve is VI.

b). It will pass through the origin when θ = 0, 4π, 8π, so again we have two passes through the origin. Other intersections with the y-axis come from

θ = π/2, 3π/2, 5π/2, 7π/2, 9π/2, 11π/2, 13π/2, 15π/2.

This leads to four different values of y, namely ± sin π/8, ± sin 3π/8, so there are four crossings of the y-axis different from the origin. This now has to be curve III.

c). In this case, we always have |r| ⩾ 1, so the only possibility is IV. d). This curve passes through the origin every time θ is a multiple of π. There

are no other places where it crosses the x-axis, so it must be V. e). This crosses the x-axis at all multiples of π. At the odd multiples r = −3

and at the even multiples r = 5. So the curve crosses the x-axis at the points (3, 0) and (5, 0) (and the origin). The only possibility is II.

f). As θ approaches zero this r approaches ∞; and as θ approaches ∞, r approaches zero from above. So this curve is a spiral and must be I.