1. ## Directional Derivatives 2

Compute the directional derivative of the following function along unit vectors at the indicated points in directions parallel to the given vector:

$f(x,y) = x^y, \ (x_0,y_0) = (e,e), \ \bold{d} = 5\bold{i} + 12\bold{j}$

2. What are the partials and their values at (e,e)?
What is the unit vector in the direction of d?

3. The partials are:

$\frac{\partial{f}}{\partial{x}} = yx^{y-1}$

and

$\frac{\partial{f}}{\partial{y}} = x^y \log{x}$

So, at (e, e) you get:

For the 1st one: $ee^{e-1} = e^e$

For the 2nd one: $e^e \log{e}$

And the unit vector of $\bold{d}$ is:

$\bold{v} = \frac{5}{13}\bold{i} + \frac{12}{13}\bold{j}$

The parallel part is what confuses me...

4. First $\ln (e) = 1$.
You are correct about $v$ it is a unit vector in the same direction as $d$.
Parallel means a scalar multiple.

Now what is the directional derivative?

5. So, I need to use $\nabla f(\bold{x}) \cdot \alpha\bold{v} = \alpha[\nabla f(\bold{x})] \cdot \bold{v}$

Right?

6. I give you a very cautious yes is may be correct.
But because $v$ is a unit vector then $\alpha = 1$.
You did this in your other posting. So why is this different?

7. If parallel means a scalar multiple, then $\alpha\bold{v}$, where $\alpha$ is some number. Would that not cover the parallel vectors? I think this is the difference between this problem and the one I posted before.

8. Originally Posted by Aryth
If parallel means a scalar multiple, then $\alpha\bold{v}$, where $\alpha$ is some number. Would that not cover the parallel vectors? I think this is the difference between this problem and the one I posted before.
You are overthinking this problem!
If we want a directional derivative parallel to $i + 2j - 2k$ we change that vector to the following: $\frac{1}{3}\left( {i + 2j - 2k} \right)$.
You make the vector into a unit vector.
Again I say that you are over thinking this problem.

9. I got it. Thanks. I figured it out, Haha.