Answer to Question #93314 in Calculus for Subhasis Padhy

Question

Find the directional derivatives of f(x,y,z)= x²+2y²+3z² at P₀(1,1,1) in the direction of a= i+j+k .

Answer 1

Gradient of f(x,y,z)= x²+2y²+3z² is g(x,y,z)= 2xi+4yj+6zk.

Therefore gradient  of  f at point P₀(1,1,1) is g=2i+4j+6k

Unit vector in the direction of a= i+j+k is e=a/|a|, i.e

$e=\frac{1}{\sqrt 3}i+\frac{1}{\sqrt 3}j+\frac{1}{\sqrt 3}k.$

The directional derivatives of f(x,y,z)= x²+2y²+3z² at P₀(1,1,1) in the direction of a= i+j+k is inner product g and e, i.e.

$g\cdot e=2\frac{1}{\sqrt 3}+4\frac{1}{\sqrt 3}+6\frac{1}{\sqrt 3}=4\sqrt 3.$