Question #181253

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^)


1
Expert's answer
2021-05-02T05:16:53-0400

f(x,y,z)=(x2+y2+z2)12f(x,y,z)x=x(x2+y2+z2)32f(x,y,z)x(3,1,2)=3(14)32f(x,y,z)y=y(x2+y2+z2)32f(x,y,z)y(3,1,2)=(14)32f(x,y,z)z=z(x2+y2+z2)32f(x,y,z)z(3,1,2)=2(14)32f(x,y,z)=(14)32(3i+j+2k)Letube a unit vectoru=yzi+xzj+xyku^=(yz,xz,xy)(yz)2+(xz)2+(xy)2Directional derivative(uf)=(14)32(3,1,2)×(yz,xz,xy)(yz)2+(xz)2+(xy)2=(14)32(3yz+xz+2xy)(yz)2+(xz)2+(xy)2\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}}


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