Chris Cruz

2021-12-27

Let

(a) Calculate

(b) Find the rate of change of f in the direction

(c) Find the rate of change of f in the direction of a vector making an angle of

Stuart Rountree

Beginner2021-12-28Added 29 answers

Step 1

As per given by the question,

There are given that the$f(x,y)=x{e}^{{x}^{2}-y},P(1,1)$

Now,

(a). Calculate$\Vert \mathrm{\nabla}fp\Vert$ .

$\mathrm{\nabla}f=<{f}_{x}(x,y),{f}_{y}(x,y)>$

$\mathrm{\nabla}f=<x\left(2x{e}^{{x}^{2}-y}\right)+{e}^{{x}^{2}-y},-x{e}^{{x}^{2}-y}>$

$\mathrm{\nabla}f=<\left(2{x}^{2}{e}^{{x}^{2}-y}\right)+{e}^{{x}^{2}-y},-x{e}^{{x}^{2}-y}>$

Now,

Evaluate$\mathrm{\nabla}f$ at the point $P(1,1)$ .

$\mathrm{\nabla}{f}_{(1,1)}=<2\left(1\right){e}^{\left(1\right)-1}+{e}^{\left(1\right)-1},-\left(1\right){e}^{1-1}>$

$\mathrm{\nabla}{f}_{(1,1)}=<3,-1>$

Step 2

Therefore,

$\Vert \mathrm{\nabla}{f}_{p}\Vert =\sqrt{{\left(3\right)}^{2}+{(-1)}^{2}}=\sqrt{10}$

Hence, the value of$\Vert \mathrm{\nabla}fp\Vert$ is $\sqrt{10}$ .

As per given by the question,

There are given that the

Now,

(a). Calculate

Now,

Evaluate

Step 2

Therefore,

Hence, the value of

Ronnie Schechter

Beginner2021-12-29Added 27 answers

(b).

$v=<1,1>$

now,

From formula,

$<>{D}_{u}f\left(p\right)=\frac{1}{\Vert v\Vert}\mathrm{\nabla}{f}_{p}\cdot v$

Step 3

Then,

${D}_{u}f\left(p\right)=\frac{1}{\Vert v\Vert}\mathrm{\nabla}{f}_{p}\cdot v$

${D}_{u}(1,1)=\frac{1}{\Vert \begin{array}{cc}1& 1\end{array}\Vert}<3,-1>\cdot <1,1>$

$D}_{u}(1,1)=\frac{2}{\sqrt{2}$

now,

From formula,

Step 3

Then,

karton

Expert2022-01-09Added 613 answers

(c)

now,

apply

Hence, the rate of change of x is

22+64

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