# Recitation 27

Example 1: The curve traced out by a point $P$ on the circumference of a circle as the circle rolls along a straight line is called a cycloid. If the circle has radius $r$ and rolls along the $x$-axis and if one position of $P$ is the origin, find parametric equations for the cycloid.

Hint: Choose as parameter the angle of rotation $\theta$ of the circle.

Definition: The curve is concave up when $d^2y/dx^2>0$ and concave down
when $d^2y/dx^2<0$.

Example 2: Find $dy/dx$ and $d^2 y/dx^2$ for $x=t^2+1, y=t^2+t$. For which values of $t$ is the curve concave upward?

Hint: Use $dy/dx = (dy/dt)/(dx/dt)$ and $d^2y/dx^2 = \frac{d(dy/dx)/dt}{dx/dt}$.

Example 3: Graph the curve $x=\cos t+2\cos2t,y=\sin t+2\sin2t$ to discover where it crosses itself. Then find equations of both tangents at that point.

Solution: The curve crosses itself at $(-2, 0)$.

This happens when $$x = \cos t + 2 \cos 2t = \cos t(1+ 4\cos t)-2 = -2, y = \sin t + 2\sin 2t = \sin t(1+4\cos t) = 0,$$ i.e., $\cos t = -1/4$. The tangent line at $(x_0, y_0)$ where $x_0 = x(t_0), y_0 = y(t_0)$ is given by $y'(t_0)(x-x_0) = x'(t_0)(y-y_0)$, i.e., $(\cos t_0 + 4\cos 2t_0)(x - x_0) = (-\sin t_0 - 4\sin 2t_0)(y - y_0)$. Now set $\cos t_0 = -1/4, \sin t_0 = \pm\sqrt{15}/4, x_0 = -2, y_0 = 0$ and get two equations of both tangents $\sqrt{15}(x+2)\pm y=0$.

Example 4: Use the parametric equations of an ellipse, $x = a \cos \theta$,
$y=b\sin\theta$,$0\leq\theta\leq2\pi$, to find the area that it encloses.

Solution: Suppose $x=f(t)$ and $y=g(t)$. Notice that as $\theta$ increases, $dx$ is negative and $y$ is positive when $0 < \theta < \pi$, and then $dx$ positive and $y$ negative when $\pi < \theta < 2\pi$. Therefore $A=\int_0^{2\pi}g(t)f'(t)(-dt)=\int_0^{2\pi}b\sin\theta a\sin\theta d\theta=\pi ab$.

Example 5: Find the exact length of the curve $x=1+3t^2, y=4+2t^3, 0\leq t\leq 1$.

Solution: The length of the curve is $\int_0^1\sqrt{(6t)^2+(6t^2)^2}dt=4\sqrt{2}-2$.

Example 6: Find the exact area of the surface obtained by rotating the given curve $x=a\cos^3\theta, y=a\sin^3\theta, 0\leq \theta\leq\pi/2$ about the $x$-axis.

Solution: The area of the surface is $\int_0^{\pi/2} 2\pi a\sin^3\theta \sqrt{(3a\cos^2\theta\sin\theta)^2+(3a\sin^2\theta\cos\theta)^2}dt=\frac{6\pi}{5}a^2$.