# I've looked everywhere... can somebody help.

• Dec 4th 2009, 03:43 AM
billym
I've looked everywhere... can somebody help.
I've searched and searched for this type of question but I can't find an example.

Can somebody please show me how to do this type of question:

Let $\displaystyle f(x,y)=x^2+3xy+xy^3$ and let $\displaystyle {\bf v}=(1,2)$

Calculate $\displaystyle \frac{\partial^2f}{\partial {\bf v}^2}$

My exams start this week and I'm out of time. If somebody could help me with this that would be great.
• Dec 4th 2009, 07:05 AM
tonio
Quote:

Originally Posted by billym
I've searched and searched for this type of question but I can't find an example.

Can somebody please show me how to do this type of question:

Let $\displaystyle f(x,y)=x^2+3xy+xy^3$ and let $\displaystyle {\bf v}=(1,2)$

Calculate $\displaystyle \frac{\partial^2f}{\partial {\bf v}^2}$

My exams start this week and I'm out of time. If somebody could help me with this that would be great.

Either the question is boringly trivial and uninteresting, or else you, or someone else, made a mistake copying it: $\displaystyle v$ is a constant vector and any function's derivative wrt to a constant is zero, even if you meant to derivate wrt a vector (what is that?) and not wrt some variable.
Perhaps you, or someone else, wanted to check what some partial second derivative wrt the constant vector is, for example $\displaystyle \frac{\partial^2 f}{\partial x^2}(1,2)$ or something like this. Who knows...

Tonio
• Dec 5th 2009, 12:26 AM
billym
I assure you it's not me. I think my lectures just have a fetish for unorthodox notation (which would explain why I can't find this anywhere).

I think $\displaystyle \frac{\partial f}{\partial {\bf v}}$ is just meant to be the directional derivative.

Would that make this a second order directional derivative?!
• Dec 5th 2009, 10:10 AM
shawsend
Quote:

Originally Posted by billym
I've searched and searched for this type of question but I can't find an example.

Can somebody please show me how to do this type of question:

Let $\displaystyle f(x,y)=x^2+3xy+xy^3$ and let $\displaystyle {\bf v}=(1,2)$

Calculate $\displaystyle \frac{\partial^2f}{\partial {\bf v}^2}$

My exams start this week and I'm out of time. If somebody could help me with this that would be great.

You want the second derivative at the point $\displaystyle (x,y)$ in the direction of the unit vector $\displaystyle \bold{v}=\{v_x i, v_y j \}$. Then from my calculations I get:

$\displaystyle \frac{\partial^2 f}{\partial \bold{v}^2}(x,y)=f_{xx}(x,y)(v_x^2)+f_{yy}(x,y)(v_ y^2)+2f_{xy}(x,y) (v_x v_y)$

Look, I'm not sure, but come Monday and nobody corrects me in here, this is going on my test paper.