Compute the directional derivative of the following function along unit vectors at the indicated points in directions parallel to the given vector:
$\displaystyle f(x,y) = x^y, \ (x_0,y_0) = (e,e), \ \bold{d} = 5\bold{i} + 12\bold{j}$
Compute the directional derivative of the following function along unit vectors at the indicated points in directions parallel to the given vector:
$\displaystyle f(x,y) = x^y, \ (x_0,y_0) = (e,e), \ \bold{d} = 5\bold{i} + 12\bold{j}$
The partials are:
$\displaystyle \frac{\partial{f}}{\partial{x}} = yx^{y-1}$
and
$\displaystyle \frac{\partial{f}}{\partial{y}} = x^y \log{x}$
So, at (e, e) you get:
For the 1st one: $\displaystyle ee^{e-1} = e^e$
For the 2nd one: $\displaystyle e^e \log{e}$
And the unit vector of $\displaystyle \bold{d}$ is:
$\displaystyle \bold{v} = \frac{5}{13}\bold{i} + \frac{12}{13}\bold{j}$
The parallel part is what confuses me...
You are overthinking this problem!
If we want a directional derivative parallel to $\displaystyle i + 2j - 2k$ we change that vector to the following: $\displaystyle \frac{1}{3}\left( {i + 2j - 2k} \right)$.
You make the vector into a unit vector.
Again I say that you are over thinking this problem.