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Q. 29

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. $f\left(x,y,z\right)=x{y}^{2}{z}^{3},P=\left(0,0,0\right),\mathbf{v}=⟨1,-2,-1⟩$

${D}_{u}f\left(0,0,0\right)=0$

See the step by step solution

## Step 1. Given Information

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

## Step 2. Solution

$\text{Also the vector is}v=⟨1,-2,-1⟩\phantom{\rule{0ex}{0ex}}\text{The unit vector "}u\text{" in the direction of "}v"\text{is given by}\phantom{\rule{0ex}{0ex}}\begin{array}{l}\mathbf{u}=\frac{1}{\parallel \mathbf{v}\parallel }\mathbf{v}\\ =\frac{1}{\sqrt{{1}^{2}+\left(-2{\right)}^{2}+\left(-1{\right)}^{2}}}⟨1,-2,-1⟩\\ =\frac{1}{\sqrt{6}}⟨1,-2,-1⟩\\ \text{Hence the directional derivative is:}\\ \begin{array}{l}{D}_{\mathrm{u}}f\left(0,0,0\right)=\underset{h\to 0}{lim} \frac{f\left(\frac{h}{\sqrt{6}},-\frac{2h}{\sqrt{6}},-\frac{h}{\sqrt{6}}\right)-f\left(0,0,0\right)}{h}\\ =\underset{h\to 0}{lim} \frac{\frac{h}{\sqrt{6}}{\left(\frac{-2h}{\sqrt{6}}\right)}^{2}{\left(\frac{-h}{\sqrt{6}}\right)}^{3}-0}{h}\\ =\underset{h\to 0}{lim} \frac{-4{h}^{5}}{{6}^{3}}\\ =0\\ Hence{D}_{u}f\left(0,0,0\right)=0\end{array}\end{array}$