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Expert-verified Found in: Page 989 ### Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861 # Find the directional derivative of the given function at the specified point P in the direction of the given vector. Note: The given vectors may not be unit vectors. $f\left(x,y,z\right)=\sqrt{\frac{xy}{z}},P=\left(2,3,1\right),\mathbf{v}=⟨2,1,-2⟩$

${D}_{\mathrm{u}}f\left(2,3,1\right)=\frac{10}{\sqrt{6}}$

See the step by step solution

## Step 1. Given Information

$\text{The directional derivative of a function}f\left(x,y,z\right)\text{at a point}P=\left({x}_{0},{y}_{0},{z}_{0}\right)\text{in the direction of a}\phantom{\rule{0ex}{0ex}}\text{unit vector}\mathbf{u}=⟨a,b,c⟩\text{is given by}\phantom{\rule{0ex}{0ex}}{D}_{u}f\left({x}_{0},{y}_{0},{z}_{0}\right)=\underset{h\to 0}{lim} \frac{f\left({x}_{0}+ah,{y}_{0}+bh,{z}_{0}+ch\right)-f\left({x}_{0},{y}_{0},{z}_{0}\right)}{h}\phantom{\rule{0ex}{0ex}}\text{Here}f\left(x,y,z\right)=\sqrt{\frac{xy}{z}}\text{, and}P=\left(2,3,1\right).\phantom{\rule{0ex}{0ex}}Alsothevectorisv=\left(2,1,-2\right)$

## Step 2. Solution

$\text{The unit vector "}\mathbit{u}\text{" in the direction of "}v\text{" is given by}\phantom{\rule{0ex}{0ex}}\begin{array}{l}\mathbf{u}=\frac{1}{\parallel \mathbf{v}\parallel }\mathbf{v}\\ =\frac{1}{\sqrt{{2}^{2}+{1}^{2}+\left(-2{\right)}^{2}}}⟨2,1,-2⟩\\ =\frac{1}{3}⟨2,1,-2⟩\\ Nowbydefinition:\\ \begin{array}{l}{D}_{u}f\left(2,3,1\right)=\underset{h\to 0}{lim} \frac{f\left(2+\frac{2h}{3},3+\frac{h}{3},1-\frac{2h}{3}\right)-f\left(2,3,1\right)}{h}\\ =\underset{h\to 0}{lim} \frac{\sqrt{\left(2+\frac{2h}{3}\right)\left(3+\frac{h}{3}\right)÷\left(1-\frac{2h}{3}\right)}-\sqrt{\frac{6}{1}}}{h}\\ =\underset{h\to 0}{lim} \frac{\sqrt{6+\frac{8h}{3}+\frac{2h}{9}-\sqrt{6-4h}}}{h\sqrt{1-\frac{2h}{3}}}\\ Rationalizethenumeratorgives:\\ \begin{array}{l}{D}_{u}f\left(2,3,1\right)=\underset{h\to 0}{lim} \frac{\sqrt{6+\frac{8h}{3}+\frac{2{h}^{2}}{9}}-\sqrt{6-4h}}{h\sqrt{1-\frac{2h}{3}}}×\frac{\sqrt{6+\frac{8h}{3}+\frac{2{h}^{2}}{9}}+\sqrt{6-4h}}{\sqrt{6+\frac{8h}{3}+\frac{2{h}^{2}}{9}}+\sqrt{6-4h}}\\ =\underset{h\to 0}{lim} \frac{6+\frac{8h}{3}-\frac{2{h}^{2}}{9}-6+4h}{h\sqrt{1-\frac{2h}{3}}\left(\sqrt{6+\frac{8h}{3}+\frac{2{h}^{2}}{9}}+\sqrt{6-4h}\right)}\\ =\underset{h\to 0}{lim} \frac{\sqrt{\frac{20}{3}-\frac{2}{9}h}}{\sqrt{1-\frac{2h}{3}}\left(\sqrt{6+\frac{8h}{3}+\frac{2{h}^{2}}{9}}+\sqrt{6-4h}\right)}\\ Takinglimitash\to 0gives:{D}_{\mathrm{u}}f\left(2,3,1\right)=\frac{10}{\sqrt{6}}\end{array}\end{array}\end{array}$ ### Want to see more solutions like these? 