# Directional derivative conceptual question

Suppose that $D_{\vec{u}}\;f(1,2)=-5$ and $D_{\vec{v}}\;f(1,2)=10$, where $\vec{u}=\langle\frac{3}{5}, \frac{-4}{5}\rangle$ and $\vec{v}=\rangle\frac{4}{5},\frac{3}{5}\rangle$
Part a) Find $f_x(1,2)$ and $f_y(1,2)$
Part b) Find directional derivative of $f$ at $(1,2)$ in the direction of the origin.

Attempt: I solved part a). It was easy. $f_x(1,2)=5$ and $f_y(1,2)=10$

So in part b) for the vector $\vec{t}$ in the direction from the origin to $(1,2)$:

$$D_{\vec{t}}\;f(1,2)=\nabla f(1,2)\cdot \vec{t}=\|\nabla f(1,2)\|\cos\theta$$

Since the vector $\vec{t}$ goes from the origin to $(1,2)$ I have $\cos\theta=\frac{1}{\sqrt{5}}$.

Then, I have:

$$D_{\vec{t}}\;f(1,2)=\|\nabla f(1,2)\|\cos\theta=\|\langle 5,10\rangle\|\frac{1}{\sqrt{5}}=5$$

So derivative of the vector that goes FROM $(1,2)$ to the origin ,i.e., $(-t)=\langle-1,-2\rangle$ is $(-5)$. However, in the answer key it is $(-5\sqrt{5})$. What mistake am I making? Any hints please.

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Your angle $\theta$ is between $(1,2)$ and the $x$-axis; for the dot product formula to be valid it needs to be the angle between $(1,2)$ and $\nabla f=(5,10)=5(1,2)$, which is of course zero and $\cos\,0=1$, not $\frac{1}{\sqrt{5}}$.