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Q. 29

Expert-verifiedFound in: Page 989

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.

$f(x,y,z)=x{y}^{2}{z}^{3},P=(0,0,0),\mathbf{v}=\u27e81,-2,-1\u27e9$

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

$\text{The directional derivative of a function}f(x,y,z)\text{at a point}P=\left({x}_{0},{y}_{0},{z}_{0}\right)\text{in the direction of a}\phantom{\rule{0ex}{0ex}}\text{unit vector}\mathbf{u}=\u27e8a,b,c\u27e9\text{is given by}\phantom{\rule{0ex}{0ex}}{D}_{u}f\left({x}_{0},{y}_{0},{z}_{0}\right)=\underset{h\to 0}{lim}\u200a\frac{f\left({x}_{0}+ah,{y}_{0}+bh,{z}_{0}+ch\right)-f\left({x}_{0},{y}_{0},{z}_{0}\right)}{h}\phantom{\rule{0ex}{0ex}}\text{Here}f(x,y,z)=x{y}^{2}{z}^{3}\text{, and}P=(0,0,0).$

$\text{Also the vector is}v=\u27e81,-2,-1\u27e9\phantom{\rule{0ex}{0ex}}\text{The unit vector "}u\text{" in the direction of "}v"\text{is given by}\phantom{\rule{0ex}{0ex}}\begin{array}{l}\mathbf{u}=\frac{1}{\parallel \mathbf{v}\parallel}\mathbf{v}\\ =\frac{1}{\sqrt{{1}^{2}+(-2{)}^{2}+(-1{)}^{2}}}\u27e81,-2,-1\u27e9\\ =\frac{1}{\sqrt{6}}\u27e81,-2,-1\u27e9\\ \text{Hence the directional derivative is:}\\ \begin{array}{l}{D}_{\mathrm{u}}f(0,0,0)=\underset{h\to 0}{lim}\u200a\frac{f\left(\frac{h}{\sqrt{6}},-\frac{2h}{\sqrt{6}},-\frac{h}{\sqrt{6}}\right)-f(0,0,0)}{h}\\ =\underset{h\to 0}{lim}\u200a\frac{\frac{h}{\sqrt{6}}{\left(\frac{-2h}{\sqrt{6}}\right)}^{2}{\left(\frac{-h}{\sqrt{6}}\right)}^{3}-0}{h}\\ =\underset{h\to 0}{lim}\u200a\frac{-4{h}^{5}}{{6}^{3}}\\ =0\\ Hence{D}_{u}f(0,0,0)=0\end{array}\end{array}$

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