To find the extreme values of a function f(x, y) on a curve xequalsx(t), yequalsy(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.

Absolute maximum is 2  

Absolute minimum at -2

Step-by-step explanation:

The given parametric functions are:

By the chain rule:

At fixed points,

This gives on

This implies that the extreme points are and

By eliminating the parameter, we have

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

Hence (0,2) is an absolute maximum ,at and (0,-2) is an absolute minimum at  

The extreme value of a function is the minimum and the maximum values of the function.

The extreme value is at:

The given parameters are:

Differentiate using chain rule. So, we have:

Simplify

Further simplify

Set f'(t) to 0

Divide both sides by -1

Take arccot of both sides

Substitute the above values of t in and

So, we have:

Hence, the extreme value is at:

Read more about extreme values at:

link

step-by-step explanation:

y = | -2(-2) |     →     y = | 4 |     →     y = 4       ⇒   (-2, 4)

y = | -2(-1) |     →     y = | 2 |     →       y =   2       ⇒   (-1, 2)    

y = | -2(0) |     →     y = | 0 |     →       y = 0         ⇒   (0, 0)

y = | -2(1) |   c →     y = | -2 |     →     y =   2         ⇒   (1, 2)    

y = | -2(2) |     →     y = | - 4 |     →     y =   4         ⇒   (2, 4)