CALCULUS III FALL 1999 HOMEWORK 4 – ANSWERS

§9.6 Questions 10,14,44; §10.1 Questions 16,18,22,32,38

CALCULUS III  HOMEWORK 4 – ANSWERS  FALL 1999 2.

10. The y-axis is the major axis of this ellipse and so its focis are on this axis. The eccentricity satisfies 4 = 25(1 − e^{2}) and so e = \sqrt{21}/5. The foci are therefore at (0, ±\sqrt{21}) and the vertices are at (0, ±5). The ellipse looks like 14. The major axis of this hyperbola is the y-axis and so its foci are on this axis. The

eccentricity satisfies 144 = 25(1 − e^{2}) and so e = 13/5. The foci are therefore at (0, ±13) and the vertices at (0, ±5). The asymptotes are y = ±5x/12. The hyperbola looks like

44. First we assume that the foci are at (0, 0) and (−2\sqrt{2}, 0) and that the major axis is horizontal with length 4. This is just a rotation and translation of the required ellipse. We have a = 2 and ae = \sqrt{2} which gives e = 1/\sqrt{2} and ℓ = a(1 − 1/2) = 1. So the polar equation of this ellipse is

r = \frac{1}{1 + (cos θ)/\sqrt{2}}.

Applying a rotation of π/4 about the origin will result in an ellipse with foci at the origin and (−2, −2). This ellipse will have polar equation

r = \frac{1}{1 + (cos(θ − π/4))/\sqrt{2}} = \frac{2}{2 + cos θ + sin θ} .

The cartesian equivalent is obtained by multiplying the right side numerator and denominator by r. This gives 2\sqrt{x^{2} + y^{2}} + x + y = 2. Finally, we translate 1 unit to the right and 1 unit vertically to obtain the desired ellipse. This must have equation

2\sqrt{(x − 1)^{2} + (y − 1)^{2}} + (x − 1) + (y − 1) = 2

which is equivalent to 3x^{2} − 2xy + 3y^{2} = 8.

16.

a_{n} = \frac{4n − 3}{3n + 4} = \frac{4 − 3/n}{3 + 4/n} → \frac{4}{3} as n → ∞.

18.

a_{n} = \frac{n^{1/3} + n^{1/4}}{n^{1/2} + n^{1/5}} = \frac{n^{−1/6} + n^{−1/4}}{1 + n^{−3/10}} → 0 as n → ∞.

22. a_{n} = sin(nπ/2) and so a_{n} = 0 for even n. For odd n, we note that |a_{n}| = 1 and so the sequence diverges.

32.

a_{n} = ln(n + 1) − ln n = ln \frac{n + 1}{n} = ln(1 + \frac{1}{n}) → 0 as n → ∞.

38.

a_{n} = (\sqrt{n + 1} − \sqrt{n})\sqrt{n + \frac{1}{2}} = \frac{\sqrt{n + 1/2}}{\sqrt{n + 1} + \sqrt{n}}

= \frac{\sqrt{1 + 1/2n}}{\sqrt{1 + 1/n} + 1} → \frac{1}{2} as n → ∞.