Answer to Question #223426 in Calculus for RAHUL CHOUBEY

Question

directional derivatives of xy2+yz3 at (1,-1,1) along i+2j+2k

Answer 1

$\text{To compute the directional derivative, we first compute }\\\frac{\partial f}{\partial x}i+\frac{\partial f}{\partial y}j+\frac{\partial f}{\partial z}k \text{ at, where $f=xy^2+yz^3$} \\\frac{\partial f}{\partial x}=y^2 \text{, at point(1,-1,1), $\frac{\partial f}{\partial x}=1$} \\\frac{\partial f}{\partial y}=2xy+z^3 \text{, at point(1,-1,1), $\frac{\partial f}{\partial y}=-1$} \\\frac{\partial f}{\partial z}=3yz^2 \text{, at point(1,-1,1), $\frac{\partial f}{\partial z}=-3$} \\=i-j-3k \\\text{Hence, the directional derivative is (i-j-3k).u, where u is the unit vector in direction }\\\text{i+2j+2k}\\u=\frac{i+2j+2k}{\sqrt{1^2+2^2+2^2}}=\frac{i+2j+2k}{3} \\\implies u = (\frac{1}{3}, \frac{2}{3}, \frac{2}{3}) \\\text{Therefore the directional derivative is}\\1.\frac{1}{3}+-1.\frac{2}{3}+-3.\frac{2}{3}=-\frac{7}{3}$