Chris Cruz

2021-12-27

Let $f\left(x,y\right)=x{e}^{{x}^{2}-y}$ and $P=\left(1,1\right)$.
(a) Calculate $‖\mathrm{\nabla }{f}_{p}‖$.
(b) Find the rate of change of f in the direction $\mathrm{\nabla }{f}_{p}$
(c) Find the rate of change of f in the direction of a vector making an angle of ${45}^{\circ }$ with $\mathrm{\nabla }{f}_{p}$

Stuart Rountree

Step 1
As per given by the question,
There are given that the $f\left(x,y\right)=x{e}^{{x}^{2}-y},P\left(1,1\right)$
Now,
(a). Calculate $‖\mathrm{\nabla }fp‖$.
$\mathrm{\nabla }f=<{f}_{x}\left(x,y\right),{f}_{y}\left(x,y\right)>$
$\mathrm{\nabla }f=$
$\mathrm{\nabla }f=<\left(2{x}^{2}{e}^{{x}^{2}-y}\right)+{e}^{{x}^{2}-y},-x{e}^{{x}^{2}-y}>$
Now,
Evaluate $\mathrm{\nabla }f$ at the point $P\left(1,1\right)$.
$\mathrm{\nabla }{f}_{\left(1,1\right)}=<2\left(1\right){e}^{\left(1\right)-1}+{e}^{\left(1\right)-1},-\left(1\right){e}^{1-1}>$
$\mathrm{\nabla }{f}_{\left(1,1\right)}=<3,-1>$
Step 2
Therefore,
$‖\mathrm{\nabla }{f}_{p}‖=\sqrt{{\left(3\right)}^{2}+{\left(-1\right)}^{2}}=\sqrt{10}$
Hence, the value of $‖\mathrm{\nabla }fp‖$ is $\sqrt{10}$.

Ronnie Schechter

(b).
$v=<1,1>$
now,
From formula,
$<>{D}_{u}f\left(p\right)=\frac{1}{‖v‖}\mathrm{\nabla }{f}_{p}\cdot v$
Step 3
Then,
${D}_{u}f\left(p\right)=\frac{1}{‖v‖}\mathrm{\nabla }{f}_{p}\cdot v$
${D}_{u}\left(1,1\right)=\frac{1}{‖\begin{array}{cc}1& 1\end{array}‖}<3,-1>\cdot <1,1>$
${D}_{u}\left(1,1\right)=\frac{2}{\sqrt{2}}$

karton

(c)
now,
apply ${D}_{u}f\left(p\right)=\left(\mathrm{\nabla }{f}_{p}\right)\mathrm{cos}\theta$, so
${D}_{u}f\left(1,1\right)=\sqrt{10}\mathrm{cos}{45}^{\circ }$
${D}_{u}f\left(1,1\right)=\frac{\sqrt{20}}{2}$
Hence, the rate of change of x is $\frac{\sqrt{20}}{2}$

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