Answer to Question #255382 in Calculus for Aryaf

Question

Xy^2z^3=8 at (2,2,1)

Answer 1

Let $f(x,y,z): xy^2z^3-8=0$

$\therefore \frac{\delta f}{dx}=y^2z^3; \frac{\delta f}{dy}=2xyz^3; \frac{\delta f}{dz}=3xy^2z^2$

At $(2,2,1):$

$\frac{\delta f}{dx}=4;\frac{\delta f}{dy}=8;\frac{\delta f}{dx}=24$

Therefore, equation of tangent is:

$4(x-2)+8(y-2)+24(z-1)=0\\ \Rightarrow x-2+2y-4+6z-6\\ \Rightarrow x+2y+6z=12$