I have the following question:

Calculate the directional derivative of the function at the point and in direction indicated.

$f(x, y) = \arctan(xy)$ at $(1, 2)$ along the line $y = 2x$ in the direction of increasing $x$.

When I looked at the solution, I was confused about the way they solved it:
enter image description here

I understand that we need a gradient vector at (1,2) and some unit vector to give the direction.

I also understand that since it is in the direction of increasing x, x will be positive.
However, how did they get that it is going to be along the line (1,2).
Did they set x = t and then got parametric equations where x = t, and y = 2t?
If so, are the coefficient before t our vector that gives us the direction?

Also, I'm so confused about the use of parametric equations with directional derivatives. Could someone explain relationship between parametric equations and directional derivatives?

  • $\begingroup$ $$\pmatrix{x \\ y} = \pmatrix{t \\ 2t} = t\pmatrix{1 \\ 2}$$ So the direction of the line $y=2x$ is $\langle 1,2\rangle$. $\endgroup$ – user137731 Oct 16 '16 at 2:15

If you are going along the line $y = 2x$ then you find it's directional vector by looking at $\vec{OP}$ where $O = (0,0)$ and $P$ is a point on the line. Take $P = (1,2)$ then the directional vector is $\vec{OP} = \langle1,2 \rangle$.

  • $\begingroup$ But could we try another point? Not (1,2), but something else? Would the answer be different? $\endgroup$ – Jack Oct 16 '16 at 2:16
  • $\begingroup$ @Jack Try a different point on the line. What happens if you choose, say, the point $(-3,-6)$ (verify that this is on the line $y=2x$)? $\endgroup$ – user137731 Oct 16 '16 at 2:26
  • $\begingroup$ @Bye_World Since in (-3, -6), x = -3 (it's negative). But we need the direction in increasing x. So, if I try something like (3,6), it would have the same results as (1,2) vector, right? $\endgroup$ – Jack Oct 16 '16 at 2:30
  • $\begingroup$ @Jack Ah. You're right. It asks for $D_vf$ for along $y=2x$ for increasing $x$. Then yes try $(3,6)$ instead. You will get the same answer (but try it yourself anyway to see why). $\endgroup$ – user137731 Oct 16 '16 at 2:33
  • $\begingroup$ @Bye_World Thank you! That makes so much sense right now. So, just to clarify, we can either construct parametric equations and then pick the coefficient before "t" to be our direction vector, or we can choose any point that lies on our line? $\endgroup$ – Jack Oct 16 '16 at 2:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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