Question

Hi
I have this function of two variables:
f(x,y)=x^2-y^2+xy

And I need to find the gradient at the point (1,2).
I tried an get the gradient at the point to (0,-11).
It is true?
Thanks in advance!

Answer 1

I'm getting something different, show your work and I can help

Answer 2

\[f(x,y)=x^2-y^2+xy\]
P=(1,2)

To find the gradient I first finde the derivate with respect to x and after with respect to y:
\[gradient f(x,y)=f_x(x,y),f_y(x,y)\] <=>
\[gradient f(x,y)=f_x(2x-2y+2), f_y(-3y^2+x)\]

Now i insert the point P(1,2) :
\[gradient f(2*1-2*2+2), (-3*2^2+1)\] = (0,-11)

That's how I understand it :/

Answer 3

I see, so I think I see the issue here, when you compute \(f_x\) and \(f_y\) these are partial derivatives of \(f\) with respect to x and y, so when you write your gradient:

\[\nabla f = \langle f_x, f_y \rangle = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right\rangle\]

I think you're having some issues with calculating your partial derivatives, so I'll help walk you through it :D

I'll do the partial with respect to x, then you try to do it with respect to y on your own and I'll help.

\[\frac{\partial}{\partial x} ( x^2-y^2+xy) = \frac{\partial}{\partial x} ( x^2) +\frac{\partial}{\partial x} ( -y^2) +\frac{\partial}{\partial x} (xy) \]
First step, I just look at each individual term sincethe derivative is a linear operator, you probably already know this, but it just lets us look at each term individually so we're not distracted.
First term:
\[\frac{\partial}{\partial x} ( x^2)=2x \]
Second term:
\[\frac{\partial}{\partial x} ( -y^2) =0\]
Since y is independent of x, it's a constant with respect to a change in x and it's 0 just like any other constant would be.
Last term:
\[\frac{\partial}{\partial x} (xy)=y \]
Similar situation here, y is a constant, but x isn't, so just like the derivative of \(\frac{\partial}{\partial x} (cx) = c\) that derivative is y.

So:

\[\frac{\partial f}{\partial x} = 2x+y\]

So now try to calculate \(f_y\)

Answer 4

Then: \[\frac{ ∂f }{ ∂y }=x-2y\]

Answer 5

Perfect. Now calculate the gradient vector at the point and you're done

Answer 6

Do you get (4,-3) ? :/

Answer 7

\[2*1+2=4\]
\[1-2*2=-3\]
right?

Answer 8

Yup I got (4,-3) too

Answer 9

Thank you very much!! :)

Answer 10

Yeah you're welcome :)

Answer 11

I also have to find the \[D_uf\] I don't know the name for it in english but think you also call it \[D_uf\] ?

Answer 12

I get \[D_uf=\frac{ 11*\sqrt{5} }{5 }\]
It is right?

Answer 13

Sorry you need some informations:

Answer 14

I have the direction R=(4,-2)

Answer 15

And the point (0,0) is a saddle piont, right?

Answer 16

you should compute \[\Large f_{xx}f_{yy}-f_{xy}f_{yx}\]

\[\Large = f_{xx}f_{yy}-f^2_{xy}\]

and switch x's and y's around...

If it is \(<0\), it is a saddle point.