Finding the Direction Given Rate of Change

Question

I'm working on the problem:

In what direction at the point (2,0) does the function $f(x,y) = xy$ have rate of change of -1? Are there directions in which the rate is -3?

So, we know that the gradient of f(x,y) is = < y, x > (because of the partial derivatives)

Thus, when we plug in our points, we get <0,2>.

So, I then set up the equation of $-1=<0,2>*u$ (since u is the direction vector)

Since we're looking for u, we must determine a situation where 2 can be multiplied by the y-component to equal -1 as desired, which is -1/2.

Thus, I set my answer to be <0,-1/2> and said that a rate of -3 could be achieved similarly with a direction vector of <0,-6/2>.

Now, my issue comes into play here: In the x-component of the direction vector, since the x-component of our gradient is 0, couldn't our x-component for the direction vector be anything from -infinity to +infinity? Me setting it to 0 was just an arbitrary number, but the truth is that we can't actually determine it since we're multiplying by 0. Am I missing something important here?

Any help in the right direction would be greatly appreciated!

Answer 1

So the directional derivative in any direction would be given by

$$\langle 0, 2\rangle\cdot\langle a,b\rangle=2b$$

Thus we would have a derivative of $-1$ when $b=-\frac{1}{2}$. If $b$ is $-\frac{1}{2}$, then we need $a$ to be $±\frac{\sqrt 3}{2}$ so that the length of the vector is a unit.

That may be the key part you are missing: the direction must be a unit vector. Now can you see why $2b$ can never be $3$?