Question

Find the directional derivative of the function f at the point p in the direction of the vector v, where f(x,y)= e^(xy) * (cosx+siny); p=(pi/2,0); v=5i-j.

Answer 1

I know that to find the directional derivative, you find the gradient of f. So you compute the partial derivatives of f(x,y) wrt x and y. Then you take the dot product of the gradient of f with the vector v.

Answer 2

I'm confused about the vector v. My homework won't accept the result of the dot product in terms of i,j, or k.

Answer 3

they probably just mean
\[\nabla f(x,y) \cdot \vec{v} =\frac{\partial f}{\partial x}\hat{i} \cdot 5\hat{i}-\frac{\partial f}{\partial y}\hat{j} \cdot \hat{j} + 0\hat{k} \cdot \hat{k} = etc\]

and then just jour answer in the end

Answer 4

not wth v, unitvector v

Answer 5

or else u are getting a project of your gradient on that vector * the magnitude of that vector

Answer 6

projection*

Answer 7

that depends,
have a look at
http://en.wikipedia.org/wiki/Directional_derivative#Generally_applicable_definition
I don't know the context of that homework, so I went for the most general definition, that tiffany_rhodes also mentioned in the second part here

Anyway, that was not the point, the point was, where do the terms i, j and k come in. You only see them if you write it out like in components
It can also be done in a way that you do not write i j and ka anywhaere:

Answer 8

like this
\[\nabla f(x,y) \cdot \vec{v} = \begin{pmatrix} \frac{\partial f}{\partial x}\\ \frac{\partial f}{\partial y}\\ 0 \end{pmatrix} \cdot \begin{pmatrix} 5\\ -1\\ 0 \end{pmatrix}\]

Answer 9

Then you forget about the i j and k

Answer 10

and it is also true that they drop out of the final answer