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Question Number 204499 by DeArtist last updated on 19/Feb/24

$$1.\mathrm{Find}\mathrm{the}\mathrm{directional}\mathrm{derivative}\mathrm{of}$$$$F(x,y,z)=2xy-z^2\mathrm{at}\mathrm{the}\mathrm{point}(2,-1,1)\mathrm{in}$$$$\mathrm{a}\mathrm{direction}\mathrm{towards}(3,1,-1)$$$$\mathrm{in}\mathrm{what}\mathrm{direction}\mathrm{is}\mathrm{the}\mathrm{directional}\mathrm{derivative}$$$$\mathrm{maximum}?\mathrm{what}\mathrm{is}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{this}\mathrm{maximum}?$$

Answered by TonyCWX08 last updated on 19/Feb/24

$$$$$$▽F$$$$=(\frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z})$$$$=(2y,2x,-2z)$$$$$$$$▽F(2,-1,1)=<-2,4,-2>$$$$$$$$letP=(2,-1,1)$$$$letQ=(3,1,-1)$$$$$$$$P\stackrel{\rightharpoonup }{Q}=<1,2,-2>$$$$\hat{u}$$$$=\frac{<1,2,-2>}{\sqrt{1^2+2^2+{(-2)}^2}}$$$$=\frac{<1,2,-2>}{3}$$$$D_{\hat{u}}F=DirectionalDerivative$$$$=<-2,4,-2>\times \frac{1}{3}<1,2,-2>$$$$=\frac{1}{3}(-2+8+4)$$$$=\frac{10}{3}$$$$$$$$DirectionofDirectionalDerivative$$$$=\frac{<-2,4,-2>}{\sqrt{4+16+4}}$$$$=\frac{<-2,4,-2>}{\sqrt{24}}$$$$=\frac{<-2,4,-2>}{2\sqrt{6}}$$$$=<-\frac{1}{\sqrt{6}},\frac{2}{\sqrt{6}},-\frac{1}{\sqrt{6}}>$$$$$$$$MaximumValue$$$$=\sqrt{2{(-\frac{1}{\sqrt{6}})}^2+{\left(\frac{2}{\sqrt{6}}\right)}^2}$$$$=\sqrt{\frac{1}{3}+\frac{2}{3}}$$$$=\sqrt{1}$$$$=1$$

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