Problem about the derivative of an unknown function

[Granger]

Homework Statement

$$f:\mathbb{R^2}\to\mathbb{R}$$ a differentiable function in the origin so:

$$f(t,t) =t^3+t$$ and $$f(t,-2t)=2t$$

Calculate $$D_vf(0,0)$$

$$v=(1,3)$$

Homework Equations

3. The Attempt at a Solution [/B]

I have no idea on how to approach this problem.
I know that because f is differentiable we have

$$D_vf(0,0)= Df(0,0)v$$

So I should be able to determine the partial derivatives. But how can I do it?

My biggest obstacle is not knowing the value of $$f(0,0)$$ (and therefore not being able to take limit definition of partial derivative)

 

[LCKurtz]

Well, for what it's worth, $f(0,0)$ = 0 from either of your equations given about $f$, putting $t=0$.

 

[LCKurtz]

After looking a bit more carefully at your problem, here's another hint. If you have $f(x,y)$ and you want to know the rate of change along a path $\vec R(t) = \langle x(t), y(t)\rangle$, have you seen the formula$$
\frac{df}{dt} = \nabla f \cdot \frac{d \vec R}{dt}\text{ ?}$$

 

[Granger]

After looking a bit more carefully at your problem, here's another hint. If you have $f(x,y)$ and you want to know the rate of change along a path $\vec R(t) = \langle x(t), y(t)\rangle$, have you seen the formula$$
\frac{df}{dt} = \nabla f \cdot \frac{d \vec R}{dt}\text{ ?}$$

Is what you mean equivalent to the derivative of a composition of functions?

I think I got it. What I did was to differentiate both the equations given obtaining:

$$\frac{df}{dx}(t,t) + \frac{df}{dy}(t,t) = 3t^2+1$$

$$\frac{df}{dx}(t,-2t) -2 \frac{df}{dy}(t,-2t) = 2$$

Then making t=0:
$$\frac{df}{dx}(0,0) + \frac{df}{dy}(0,0) = 1$$

$$\frac{df}{dx}(0,0) -2 \frac{df}{dy}(0,0) = 2$$

Solving this system we obtain the partial derivatives of f in the point (0,0).

 

[LCKurtz]

Is what you mean equivalent to the derivative of a composition of functions?

I think I got it. What I did was to differentiate both the equations given obtaining:

$$\frac{df}{dx}(t,t) \color{red}{\bf +}\frac{df}{dy}(t,t) = 3t^2+1$$

$$\frac{df}{dx}(t,-2t) -2 \frac{df}{dy}(t,-2t) = 2$$

Aren't you missing a plus sign that I inserted above? Other than that it should work.

 

[Granger]

Aren't you missing a plus sign that I inserted above? Other than that it should work.

Yes, thanks!