Question

Consider the surface S=(x,y,z)∈R³|x²+y²+z²=9.

(a)Define an R³−R function f such that S is the contour surface of f at level 9.

(b)Find an equation for the plane V that is tangent to S at the point (x,y,z) = (2,1,2).

(c) Sketch the surface Sin R³,together with a section of the plane V to illustrate that V is tangent to S at the point (2,1,2).

Answer 1

Solution :-

$s=\{(x,y,z)\in R^3 |x^2+y^2+z^2=9 \}$

(a) $f:R^3 \rightarrow R \\ such \ that \ S \ is \ the \ contour \ surface \ of \ f \ at \ level \ 9 \\ so \ \ \ \ \ f(x,y,z)=x^2+y^2+z^2-9=0$

(b)$f(x,y,z)=x^2+y^2+z^2-9 \\$

than $df=2xdx+2ydy+2zdz$

$df|_{(2,1,2)}=4\hat{i}+2\hat{j}+4\hat{k}$

so equation of tangent plane V will be

4(x-2)+2(y-1)+4(z-2)=0

$\implies 4x-8+2y-2+4z-8=0$

$\implies \boxed{4x+2y+4z=18}$

(c)