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I have been trying to do this question but I'm completely lost.

Be $f(x,y)$ a differentiable function, the maximum value of $Duf$(0,2) is equal to 2 and occurs when $u = (\frac{\sqrt 2}{2},\frac{\sqrt 2}{2})$.
If $x(s,t) = s²t$, $y(s,t) = 2se^t$ and $F(s,t) = f(x(s,t),y(s,t))$ then $\frac{\partial F}{\partial t}$(1,0) is:

a)2$\sqrt 2$
b)3$\sqrt 2$
c)0
d)$\sqrt 2$
e)3$\frac{\sqrt 2}{2}$

I did the chain rule: $\frac{\partial F}{\partial t} = \frac{\partial F}{\partial x}*\frac{\partial x}{\partial t} + \frac{\partial F}{\partial y}*\frac{\partial y}{\partial t}$

And also the derivatives:
$\frac{\partial x}{\partial t} = s² = 1$
$\frac{\partial y}{\partial t} = 2se^t = 2$

Then I put everything together: $\frac{\partial F}{\partial t}(1,0) = \frac{\partial F}{\partial x}*1 + \frac{\partial F}{\partial y}*2$

After this I don't know how to proceed.

If there are any english mistakes then I am sorry. It isn't my first language.

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Your english was good! (It would've been better to say "Let $f(x,y)$ be a ..." instead, but this was the only mistake)

The directional derivative of $f$ at $(0,2)$ in the direction of $\underline{u}$ means

$$\frac{\sqrt{2}}{2}\frac{\partial f}{\partial x}\Bigg{\vert}_{(0,2)}+\frac{\sqrt{2}}{2}\frac{\partial f}{\partial y}\Bigg{\vert}_{(0,2)}=2\implies\frac{\partial f}{\partial x}\Bigg{\vert}_{(0,2)}+\frac{\partial f}{\partial y}\Bigg{\vert}_{(0,2)}=2\sqrt{2}$$

We know that $\underline{u}$ maximizes the directional derivative at $(0,2)$, which means that the angle between $\nabla f(0,2)$ and $\underline{u}$ is $0$. Then there exists a vector $\underline{v}$ such that $\underline{u}\cdot\underline{v}=0$ which means

$$\nabla_\underline{v}f(0,2)=0$$

We can find $\underline{v}$ easily, the only condition is that $\underline{u}\cdot\underline{v}=0$

$$\underline{v}=\left<\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right>$$

So

$$\frac{\sqrt{2}}{2}\frac{\partial f}{\partial x}\Bigg{\vert}_{(0,2)}-\frac{\sqrt{2}}{2}\frac{\partial f}{\partial y}\Bigg{\vert}_{(0,2)}=0\implies\frac{\partial f}{\partial x}\Bigg{\vert}_{(0,2)}=\frac{\partial f}{\partial y}\Bigg{\vert}_{(0,2)}$$

Then we can conclude that

$$\frac{\partial f}{\partial x}\Bigg{\vert}_{(0,2)}=\sqrt{2}$$

So

$$\frac{\partial F}{\partial t}\Bigg{\vert}_{(1,0)}=\frac{\partial f}{\partial x}\Bigg{\vert}_{(0,2)}\frac{\partial x}{\partial t}\Bigg{\vert}_{(1,0)}+\frac{\partial f}{\partial y}\Bigg{\vert}_{(0,2)}\frac{\partial y}{\partial t}\Bigg{\vert}_{(1,0)}$$

Where the partials with respect to $x$ and $y$ are evaluated at $(0,2)$ because $x(1,0)=0$ and $y(1,0)=2$

$$\boxed{\therefore\frac{\partial F}{\partial t}\Bigg{\vert}_{(1,0)}=3\sqrt{2}}$$


Using the chain rule means we have $z=f(x,y)$ with $x=x(s,t)$ and $y=y(s,t)$ so then

$$\frac{\partial z}{\partial t}=\frac{\partial f}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial t}$$

The way you tried to use the chain rule would not work because you defined

$$F(s,t)=f(x(s,t),y(s,t))\equiv g(s,t)$$

where $g$ is just some function of $s$ and $t$. So it wouldn't make sense to take partials of $F$ with respect to $x$ and $y$ because they've been already substituted and you would just get $0$.

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You are given that the maximum value of $D_uf(0,2)$ is $2$, occurring at $u=(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$. This means that $$\begin{aligned}2&=\nabla f(0,2)\cdot u\\&=(\frac{\partial f(0,2)}{\partial x},\frac{\partial f(0,2)}{\partial y})\cdot(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\end{aligned}\tag{1}$$ and as $D_uf(0,2)$ has a maximum value of $2$, the direction of the derivative vector $(\frac{\partial f(0,2)}{\partial x},\frac{\partial f(0,2)}{\partial{y}})$ will be the same as that of $u$, so that $$\frac{\partial f(0,2)}{\partial x}=\frac{\partial f(0,2)}{\partial y}$$ thus from equation $(1)$ we have $$\frac{\partial f(0,2)}{\partial x}=\frac{\partial f(0,2)}{\partial y}=\sqrt{2}$$ Going back to the chain rule you have written, this can be written in terms of the derivative of function $f(x,y)$:- $$\begin{align}\frac{\partial F(0,1)}{\partial t}&=\frac{\partial f(0,2)}{\partial t}\\&=\frac{\partial f(0,2)}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial f(0,2)}{\partial y}\frac{\partial y}{\partial t}\\&=\frac{\partial f(0,2)}{\partial x}\times1+\frac{\partial f(0,2)}{\partial y}\times2\\&=3\sqrt{2}\end{align}$$

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