Question

To find the extreme values of a function​ f(x,y) on a curve xequals​x(t), yequals​y(t), treat f as a function of the single variable t and use the chain rule to find where​ df/dt is zero. As in any other​ single-variable case, the extreme values of f are then found among the values at the critical points​ (points where​ df/dt is zero or fails to​ exist), and endpoints of the parameter domain. Find the absolute maximum and minimum values of the following function on the given curves. Use the parametric equations xequals2 cosine t​, yequals2 sine t.

Answer 1

Answer:

Absolute maximum is 2  

Absolute minimum at -2

Step-by-step explanation:

The given parametric functions are:

$x=2\cos t,y=2\sin t$

By the chain rule:

$f'(t)=\frac{\frac{dy}{dt} }{\frac{dx}{dt} }$

$\frac{df}{dt} =\frac{2 \cos t}{-2\sin t} =-\cot(t)$

At fixed points, $f'(t)=0$

$\implies -\cot (t)=0$

This gives $t=\frac{\pi}{2} ,\frac{3\pi}{2}$ on $0\le t\le 2\pi$

This implies that the extreme points are $(2\cos \frac{\pi}{2}, 2\sin \frac{\pi}{2})=(0,2)$ and $(2\cos \frac{3\pi}{2}, 2\sin \frac{3\pi}{2})=(0,-2)$

By eliminating the parameter, we have $x^2+y^2=4$

This is a circle with radius 2, centered at the origin.

Hence (0,2) is an absolute maximum ,at $t=\frac{\pi}{2}$ and (0,-2) is an absolute minimum at  $t=\frac{3\pi}{2}$