Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\cos\theta=\frac{1}{\sqrt{5}}$.

enter image description here

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.

share|cite|improve this question
up vote 1 down vote accepted

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 $\frac{1}{\sqrt{5}}$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.