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: 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?

• $$\pmatrix{x \\ y} = \pmatrix{t \\ 2t} = t\pmatrix{1 \\ 2}$$ So the direction of the line $y=2x$ is $\langle 1,2\rangle$. – 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 $\vec{OP} = \langle1,2 \rangle$.
• @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 $y=2x$)? – user137731 Oct 16 '16 at 2:26
• @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). – user137731 Oct 16 '16 at 2:33