At the point (1,2), the function f(x,y) has a derivative of 2 in the direction toward (2,2) and a derivative of -2 in the direction toward (1,1).
- Find f_x(1,2) and f_y(1,2).
- Find the derivative of f at (1,2) in the direction towards the point (4,6).
I'm not sure how to start, I know how to find f_x and f_y when given f(x,y), but how do I do this without being given f(x,y) to start with? Help is appreciated! Thanks in advance.
Since the directional derivative of $f$ in the direction of $\mathbf v$ is given by $$D_{\mathbf v}f=\nabla f\cdot \mathbf v,$$ here you have at $(1,2)$ \begin{align} \mathbf v=(2,2):\quad D_{\mathbf v}f&=\nabla f\cdot \mathbf v=(f_x,f_y)\cdot (2,2)=2f_x+2f_y=2,\\ \mathbf v=(1,1):\quad D_{\mathbf v}f&=\nabla f\cdot \mathbf v=(f_x,f_y)\cdot (1,1)=f_x+f_y=-2,\\ \end{align} that is, \begin{align} 2f_x(1,2)+2f_y(1,2)&=2,\\ f_x(1,2)+f_y(1,2)&=-2. \end{align} But this system to has no solution for $f_x(1,2)$ and $f_y(1,2)$. This makes me think the directional derivative values should have been something other than $\pm 2$ and/or the specified vectors should not have been $(1,1)$ and $(2,2)$.
More generally, suppose you were told that at the point $(a,b)$ that the directional derivative of $f$ in the direction of $\mathbf{v}_1=(v_1,v_2)$ is $d_1$, while the directional derivative of $f$ in the direction of $\mathbf{v}_2=(v_3,v_4)$ is $d_2$, and we want to know $f_x(a,b)$ and $f_y(a,b)$. Then, as above, we'd have \begin{align} D_{{\mathbf v}_1}f&=\nabla f\cdot {\mathbf v}_1=(f_x,f_y)\cdot (v_1,v_2)=\color{blue}{v_1f_x(a,b)+v_2f_y(a,b)=d_1},\\ D_{{\mathbf v}_2}f&=\nabla f\cdot {\mathbf v}_2=(f_x,f_y)\cdot (v_3,v_4)=\color{blue}{v_3f_x(a,b)+v_4f_y(a,b)=d_2}.\\ \end{align} The highlighted portion in blue is two (linear) equations in the two unknowns $f_x(a,b)$ and $f_y(a,b)$, which has a unique solution so long as the determinant of the associated coefficient matrix, $v_1v_4-v_2v_3$ is nonzero.
In your stated example, this determinant is zero.
Ok, back to your second question. If we had $f_x(a,b)$ and $f_y(a,b)$ above, we could answer a more general version of your second bullet point,
as follows: $$ \text{At }(a,b):\quad D_{{\mathbf w}}f=\nabla f\cdot {\mathbf w}=(f_x,f_y)\cdot (w_1,w_2)=\color{red}{w_1f_x(a,b)+w_2f_y(a,b)}, $$ the answer being the number in red since we'd know all four of those pieces.