Find the unit vector $v$ with the property that $D_{v}f(a)$ is as large as possible. In a multivariate environment Find the unit vector $\mathbf{v}$ with the property that $D_{\mathbf{v}}f(\mathbf{a})$ is as large as possible.
Note: I'm aware that this can be sloved by using the gradient, but at this point in my text I haven't "learned" what that is so I don't have that tool at my disposal.
Given the following values:
$f(x,y) = x^{2} + xy$
$\mathbf{a} = (2,1)$ and $\mathbf{v} = (v_{1},v_{2})$.
I attempted to approach this question in the same fashion as would be done in single variable calculus. In that situation I would first take the derivative of my function, but since we are asked for the value that would maximize my derivative function it means I would have to take the second derivative of the original function. From here if the math was nice I would have a variable I can isolate for and find its value.
So I thought perhaps a similar approach could be used in a multivariable setting. Using the definition of the directional derivative, the given information, and after some simplifying:
$$D_{\mathbf{v}}f(\mathbf{a}) = \lim_{t \to 0} \frac{f(\mathbf{a} + t\mathbf{v}) - f(\mathbf{a})}{t}\\ = \lim_{t \to 0} \frac{t(5v_{1} +v_{1}^{2}t + 2v_{2} + tv_{1}v_{2})}{t} \\ = \lim_{t \to 0} \ (5v_{1} +v_{1}^{2}t + 2v_{2} + tv_{1}v_{2}) \\ = 5v_{1} + 2v_{2}$$
So this would be the first derivative. Here is where I'm stuck. I'm not sure what to take the directional derivative of here. If I were to take the directional derivative of this final function, something along the lines of say
$$\lim_{t \to 0} \frac{f(\mathbf{a} + t\mathbf{w}) - f(\mathbf{a})}{t}$$ where $\mathbf{w}$ would be my directional derivative for this second function after doing the math:
$$\lim{t \to 0} \frac{5(2+tw_{1})+2(1+tw_{2}) - 12}{t} \\ = 5w_{1} + 2w_{2}$$
Doing some rearranging I arrive at:
$$\Rightarrow\ w_{2} = \frac{-5w_{1}}{2} $$
THis then to me means that the vector that maximizes my directional derivative is of the general from $$\mathbf{w} = (w_{1}, w_{2}) = (w_{1}, \frac{-5w_{1}}{2})$$.
Choosing $w_{1} = 1$, since it appears $w_{1}$ can be a free variable, I am left with a vector $(1, \frac{25}{4})$. From this I can get the norm of this vector $\|\mathbf{w}\| = \sqrt{\frac{29}{4}}$. Which would then allow me to turn it into a unit vector by expressing it as $\frac{\mathbf{w}}{\|\mathbf{w}\|}$.
Is this the way to approach this sort of situation?
It feels like this approach wouldn't work if I were faced with a more complicated function to begin with.
 A: In this case - you’re trying to choose a unit $v$ that maximizes $a \cdot v$. By Cauchy Schwartz, the maximum value of this is $|a||v|$ which occurs when $a$ and $v$ are the same direction. When $v$ is a unit vector, this gives the maximum. $a$ is $(5,2)$, so after normalization, that’s your answer.
In general, the gradient of a function is the direction of highest increase.
There is a way of taking a derivative over a constrained set (Lagrange Multipliers), but it’s more complicated than you need here.
A: I believe the simplest way in general is to use:
$$\text{The maximum of } D_{u}f({\bf{a}}) \text{ occurs when } u \text{ points in the direction of }\nabla{f}({\bf{a}})$$
In order to prove the above we only need to remember
$$D_{u}f({\bf{a}})=\nabla f({\bf{a}}) \cdot u = \|\nabla f({\bf{a}})\|\|u\|\cos(\theta).$$
Because $u$ is of unit length we obtain
$$D_uf({\bf{a}}) = \|\nabla f({\bf{a}})\|\cos(\theta).$$
The maximum will then occur when $u$ points in the direction of $\nabla f({\bf{a}})$ because 1. $\|\nabla f({\bf{a}})\|$ is constant and 2. when $u$ points in the direction of $\nabla f({\bf{a}})$, $\theta = 0$.

Let us apply, for completeness and comparison the above to your problem.
We have $\nabla f(x,y) = (2x + y, x)$ and $\nabla f({\bf{a}})=(5,2)$ so our sought after vector $u$ is $u= \frac{1}{\sqrt{29}}(5,2)$
A: Assume $\lVert v \rVert = 1$. It is easy to verify that $D_vf(a) = \nabla f(a) \cdot v$. We have by Cauchy-Schwarz that $\nabla f(a) \cdot v \leq \lVert\nabla f(a)\rVert$. This bound is attained when $v = \frac{\nabla f(a)}{\lVert \nabla f(a) \rVert}$.
