00:01
In this question, we want to find the linear approximation of the function below at the indicated point.
00:06
My function f of x, y, is the square root of 41 minus x squared minus 4y squared at the point 4 .2.
00:14
So how do i get my linear approximation? well, my l of x, y is equal to f at the point.
00:26
F of 4 .2 plus the partial of f with respect to x at the point 42, times the quantity of x minus four plus the partial of f with respect to y at the point four two times the quantity of y minus two so let's evaluate so f of four two it is the square root of 41 minus four squared is 16 minus two squared is four times four is 16 again squared of nine three okay so we're getting three there now i need my partial derivatives.
01:09
So my partial with respect to x, 1 over 2 times the square root of 41 minus x squared minus 4y squared times negative 2x.
01:23
Okay? and so this is negative x over the square root of 41 minus x squared minus 4y squared.
01:38
If i take that and i evaluate it at the point 4, up top i'm getting negative 4.
01:45
In the bottom, we just saw that we're getting 3.
01:50
Now i'll get my partial of f with respect to y.
01:54
That'll be 1 over 2 times the square root of 41 minus x squared minus 4y squared times negative 8y by the chain.
02:09
This simplifies to negative 4y over the square root of 4.
02:14
41 minus x squared minus 4 y square.
02:21
If i evaluate at the point 4 2, up top, negative 4 times y, negative 8, over 3.
02:31
And so my l of x, y this time, my f of 4 2, that was 3.
02:39
So 3, plus my partial with respect to x was negative 4 thirds...