If $y=-2x+1$ is an equation of the line tangent to the the curve $f(x,y)=2$ at point $(1,3)$ and $f_x(1,3)=4$.
Then find the maximum directional derivative of $f(x,y)$ in the direction of $(1,3)$.
Since the funntion $f(x,y)$ is not given how do I proceed with this?
Best Answer
I assume you mean directional derivative at the point $$(1,3)$$. For that, you can use the fact that if $f$ is differentiable, then: $$\frac{\partial f}{\partial\hat{u}}(1,3)=f_x(1,3)u_1+f_y(1,3)u_2$$ Here $\hat{u}=\left(u_1,u_2\right)$ is our direction. We also know from properties of the inner product: $$ f_x(1,3)u_1+f_y(1,3)u_2=\vec{\nabla}f(1,3)\cdot\left(u_1,u_2\right)=\Vert{\vec{\nabla}f(1,3)}\Vert\overset{=1}{\overbrace{\Vert{\hat{u}}\Vert}}\cos{(\theta)} $$ Meaning the maximum occurs when $\theta=0$ or in other words, when $\hat{u}=\frac{\vec{\nabla}f(1,3)}{\Vert\vec{\nabla}f(1,3)\Vert}$. So, all we have to do is calculate $f_y(1,3)$.
We know that $y=2x+1$ (I assume the minus is a mistake since the point does not lie on this line) implies $f(x,y(x))=2$. Define $g(x)=f(x,2x+1)\equiv 2$, and use the chain rule to get: $$ g'(1)=0=f_x(1,3)+2f_y(1,3)\overset{f_x(1,3)=4}{\overbrace{\Longrightarrow}}f_y(1,3)=-2 $$ So the direction in which the directional derivative is maximal will be: $$ \hat{u}_{{\rm max}}=\frac{(4,-2)}{\sqrt{16+4}}=\left(\frac{4}{\sqrt{20}},-\frac{2}{\sqrt{20}}\right) $$