Answer to Question #181253 in Calculus for Deepika

Question

Find the directional derivative of f(x,y,z)=(x^2+y^2+z^2)^-1/2 at (3,1,2) on the direction of (yzi^ +xzj^ +xyk^)

Answer 1

$\displaystyle f(x,y,z)=\left(x^2+y^2+z^2\right)^{-\frac{1}{2}} \\ \frac{\partial f(x,y,z)}{\partial x} = -x\left(x^2+y^2+z^2\right)^{-\frac{3}{2}}\\ \frac{\partial f(x,y,z)}{\partial x}\biggr\vert_{(3, 1, 2)} = -3(14)^{-\frac{3}{2}}\\ \frac{\partial f(x,y,z)}{\partial y} = -y\left(x^2+y^2+z^2\right)^{-\frac{3}{2}} \\ \frac{\partial f(x,y,z)}{\partial y}\biggr\vert_{(3, 1, 2)} = -(14)^{-\frac{3}{2}}\\ \frac{\partial f(x,y,z)}{\partial z} = -z\left(x^2+y^2+z^2\right)^{-\frac{3}{2}}\\ \frac{\partial f(x,y,z)}{\partial z}\biggr\vert_{(3, 1, 2)} = -2(14)^{-\frac{3}{2}}\\ \nabla f(x, y, z) = -(14)^{-\frac{3}{2}}(3\textbf{i} + \textbf{j} + 2\textbf{k})\\ \textsf{Let}\,\, u\,\, \textsf{be a unit vector}\\ u = yz\textbf{i} + xz\textbf{j} + xy\textbf{k}\\ \hat{u} = \frac{(yz, xz, xy)}{\sqrt{(yz)^2 + (xz)^2 + (xy)^2}}\\ \therefore\textsf{Directional derivative}\,\,(\nabla_u f) = -(14)^{-\frac{3}{2}}(3, 1, 2) \times \frac{(yz, xz, xy)}{\sqrt{(yz)^2 + (xz)^2 + (xy)^2}} \\= \frac{-(14)^{-\frac{3}{2}}\left(3yz + xz + 2xy\right)}{\sqrt{(yz)^2 + (xz)^2 + (xy)^2}}$