Answer to Question #90394 in Calculus for Venkatalakshmi

Question

Find the directional derivative of x2yz+4xz2 at (1,2,1) in the direction vector 2i-j-2k.

Answer 1

$$\varphi=x^2yz+4xz^2, a=2i-j-2k$$

$${\partial \varphi\over \partial a}={\partial \varphi\over \partial x}\cos\alpha+{\partial \varphi\over \partial y}\cos\beta+{\partial \varphi\over \partial z}\cos\gamma$$

$$\cos\alpha={2 \over \sqrt{(2)^2+(-1)^2+(-2)^2}}={2 \over 3}$$

$$\cos\beta={-1 \over \sqrt{(2)^2+(-1)^2+(-2)^2}}=-{1 \over 3}$$

$$\cos\gamma={-2 \over \sqrt{(2)^2+(-1)^2+(-2)^2}}=-{2 \over 3}$$

$${\partial \varphi\over \partial x}=2xyz+4z^2, {\partial \varphi\over \partial y}=x^2z, {\partial \varphi\over \partial z}=x^2y+8xz$$

$${\partial \varphi\over \partial a}={2 \over 3}(2xyz+4z^2)-{1 \over 3}(x^2z)-{2 \over 3}(x^2y+8xz)$$

At point (1,2,1) we get the directional derivative:

$${\partial \varphi\over \partial a}|_{(1,2,1)}={2 \over 3}(2(2)(1)(2)+4(1)^2)-$$

$$-{1 \over 3}(1)^2(1)-{2 \over 3}((1)^2(2)+8(1)(1))=-{5 \over 3}$$

$${\partial \varphi\over \partial a}|_{(1,2,1)}=-{5 \over 3}$$