I am working on a question that asks me to find a direction $v$ so that the directional derivative of $f(x,y) = xe^{xy}+y$, at the point $P=(2,0)$, is seven.
I've computed that the gradient of $f$ at $P$ is $(1,1)$, so the directional derivative is the dot product of $(1,1)$ and a vector $v$, but I am getting stumped on trying to find its components.
Vector $v$ has to be a unit vector so it would be $(x,y)$ divided by its norm:
$$v = \left(\frac{x}{\sqrt{x^2+y^2}} , \frac{y}{\sqrt{x^2+y^2}}\right).$$
Then I set the dot product as:
$$(1, 1) \cdot \left(\frac{x}{\sqrt{x^2+y^2}} , \frac{y}{\sqrt{x^2+y^2}}\right) = 7,$$
and now to simplify everything:
$$\frac{\left(x+y\right)}{\sqrt{x^2+y^2}} = 7$$
$$x+y=7\sqrt{x^2+y^2}$$
$$\left(x+y\right)^2=49\left(x^2+y^2\right)$$
$$x^2+2xy+y^2=49x^2+49y^2$$
$$2xy=48x^2+48y^2$$
So you're left with: $xy=24x^2+24y^2$.
I don't know how to proceed since there is no system of equations and I have two variables here.
Ok, I see what you're doing. You already computed $\nabla f(P) = (1,1)$ and now are trying to find a unit vector $v$ with $$(1,1) \cdot v = 7.$$
There are several ways to proceed from your calculations.
1) Try to solve for $x$, as a function of $y$, by applying the quadratic formula. What can you conclude about the expression you get? Is there any possible real solution $(x,y)$?
2) Go back to the expression $(1,1)\cdot v$, where you know $\|v\|=1$. Is there any convenient bound you can place on the magnitude of the dot product?
3) Instead of $(x,y)$, write your vector as $(\cos \theta, \sin\theta)$. Try to solve for $\theta$.