Problem 1 - Geometry

Answer the following questions about the curve defined by

$$x=a\cos^3 t, \quad y=a\sin^3 t,$$

with a parameter $t$ $(0\leq t\leq 2\pi)$, where $a>0$.

1.

Draw an outline of this curve.

2.

Compute the area of the region enclosed by the curve.

3.

Compute the length of this curve.

1.

2.

$$S = 4\int_{0}^{\frac{\pi}{2}} -ydx$$

$$= -4a^2\int_{0}^{\frac{\pi}{2}} \sin^3 t d(\cos^3 t)$$

$$= 12a^2\int_{0}^{\frac{\pi}{2}} \sin^3 t \sin t \cos^2 t dt$$

$$= 12a^2\int_{0}^{\frac{\pi}{2}} \cos t \cos t \sin^4 t dt$$

$$= 12a^2\int_{0}^{\frac{\pi}{2}} (1-\sin^2 t)\sin^4 t dt$$

$$= 12a^2\int_{0}^{\frac{\pi}{2}} \sin^4 t - \sin^6 t dt$$

$\sin^4 t = \left(\frac{e^{it}-e^{-it}}{2i}\right)^4$
$\cdots$
$= \frac{1}{16}\left(6 + 2\cos 4t - 8\cos 2t \right)$
$\sin^6 t = \left(\frac{e^{it}-e^{-it}}{2i}\right)^6$
$\cdots$
$= -\frac{1}{32}\left(-10 + \cos 6t + 15\cos 2t - 6\cos 4t\right)$
$\cdots$

3.

$$\frac{dx}{dt} = - 3 a \sin t \cos^2 t \quad \frac{dy}{dt} = 3 a \cos t \sin^2 t$$

$$l = 4\int_{0}^{\frac{\pi}{2}} \sqrt{9a^2(\sin^2 t\cos^4 t + \cos^2 t\sin^4 t)} dt$$

$$= 4\int_{0}^{\frac{\pi}{2}} \sqrt{9a^2\sin^2 t\cos^2 t(\cos^2 t + \sin^2 t)} dt$$

$$= 4\int_{0}^{\frac{\pi}{2}} \sqrt{9a^2\sin^2 t\cos^2 t} dt$$

$$= 12a\int_{0}^{\frac{\pi}{2}} \sin t\cos t dt$$

$$= 12a\int_{0}^{\frac{\pi}{2}} \frac{\sin 2t}{2}dt$$

$$= 12a\left[-\frac{1}{4}\cos 2t\right]_{0}^{\frac{\pi}{2}}$$

$$= 12a\left[\frac{1}{4} + \frac{1}{4}\right]$$

$$= 6a$$