Consider the parametric curve
$$x=5+\cos(t)$$
$$y = 1 + \sin(t)$$(1) Find the speed of a particle whose position is given by this parametric curve.
(2) Rotate the parametric curve about the $y$ – axis. Find the surface area A of the surface obtained.
Here is my work on this problem:
Solving (1)
The speed of a particle is given by:
$$\frac{ds}{dt}= \sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}$$
$\frac{dx}{dt}=-\sin(t)$
$\frac{dy}{dt}=\cos(t)$
$\frac{ds}{dt} = \sqrt{(-\sin(t))^2+(\cos(t))^2} = \sqrt{\sin^2(t)+\cos^2(t)} = 1$
The speed is, $\frac{ds}{dt}=1$
Solving (2)
The differential area for rotating about the $y$ – axis is given by the equation:
$$dA = 2 \pi x \sqrt{f'(t)^2+g'(t)^2}dt$$
$f(t) = x = 5 + \cos(t)$
$f'(t) = -\sin(t)$
$g(t) = y = 1 + \sin(t)$
$g'(t) = \cos(t)$
$dA = 2 \pi (5 + \cos(t)) \sqrt{(-\sin(t))^2+(\cos(t))^2}dt$
$dA = 2 \pi (5 + \cos(t))dt$
$A = \int (10\pi + 2\pi\cos(t))dt$
$A = \int (10\pi + 2\pi\cos(t))dt$
$A = 10\pi t + 2\pi\sin(t) + C$
Could you point out where I have gone wrong?
Best Answer
For (2), note that the surface area is the same for the curve $$ x =5+\cos\theta , \>\>\>\>\>y = \sin\theta, \>\>\>\>\> t\in[0,2 \pi]$$
The area integral is
\begin{align} A & = 2\int_0^6 2\pi x \sqrt{1+ (y’_x)^2}dx\\ &= 4\pi\int_\pi^0 (5+\cos\theta)\sqrt{1+\cot^2\theta}(-\sin\theta)d\theta \\ &= 4\pi\int^\pi_0 (5+\cos\theta)d\theta = 20\pi^2\\ \end{align}