Question

Given f(x,y)=x^2+4y^2. using x=1 and y=1 and differentials, find an approximation for (1.01)^2 + 4(1.99)^2

Answer 1

First get the Partial of x and y.
So Fx =2x
Fy = 8y

Answer 2

Fx(1,1) = 2
Fy(1,1) = 8
f(1,1) = 5

Answer 3

\[T(x,y) = Fx(x _{0},y _{0})\Delta x + Fy(x _{0},y _{0})\Delta y+ f(x _{0},y _{0})\]

Answer 4

T= 2(x-1) + 8(y-1) + 5

Answer 5

T= 2x-2 + 8y - 8 + 5
T= 2x +8y -5

Answer 6

T((1.01^2, 4(1.99)^2) = 2((1.01)^2) + 8(4(1.99)^2) -5

Answer 7

\[\approx 123.76\]

Answer 8

If that isnt the correct answer I know it is atleast the right steps.