# Differentiation of f(x,y)

• Mar 30th 2006, 10:26 AM
paloma
Differentiation of f(x,y)
f(x,y) = x^2 + y^2 - 3 abs(xy)

What does the vertical cross section of the surface along a diagonal through the origin (ie. the plane y = x = t) look like?

I graphed this in Scientific Workplace and got a paraboloid, but I don't see a paraboliod in the graph of the function, can someone please help me to see how to get this?
Thanks
• Mar 30th 2006, 10:30 AM
paloma
Directional direvitives
I also have a question on the directional direvatives at the origin and whether or not they exist. For the partial direvative of x at the origin I got 0 and same for partial of y.
So I was using Duf(0,0) =fx cos() + fy sin() to find, but since both my fx and fy are 0 all I get is 0.
Is it possible that I got the right answer in that Duf(0,0) = 0?
Thanks
• Mar 31st 2006, 04:27 AM
topsquark
Quote:

Originally Posted by paloma
f(x,y) = x^2 + y^2 - 3 abs(xy)

What does the vertical cross section of the surface along a diagonal through the origin (ie. the plane y = x = t) look like?

I graphed this in Scientific Workplace and got a paraboloid, but I don't see a paraboliod in the graph of the function, can someone please help me to see how to get this?
Thanks

Actually, it's a parabola, not a paraboloid (which is a 3-D object).

I think you're thinking too hard about it. If you are trying to project the function onto the x=y=t plane, just set x = t and y = t:
$\displaystyle f(t,t)=t^2+t^2-3t^2=-t^2$
which is clearly a parabola.

-Dan
• Mar 31st 2006, 04:33 AM
topsquark
Quote:

Originally Posted by paloma
I also have a question on the directional direvatives at the origin and whether or not they exist. For the partial direvative of x at the origin I got 0 and same for partial of y.
So I was using Duf(0,0) =fx cos() + fy sin() to find, but since both my fx and fy are 0 all I get is 0.
Is it possible that I got the right answer in that Duf(0,0) = 0?
Thanks

Remember that the directional derivative is a vector, so
$\displaystyle D_uf(x,y) |_{(0,0)} = f_x(0,0) \hat{i} + f_y(0,0) \hat{j}$

Since both the partial derivatives are zero, then, the directional derivative is the zero vector (which, of course, has a magnitude of zero.) Zero is certainly an allowed value (being a real number) so we CAN say the directional derivative exists at the origin, but that it simply has no magnitude.

-Dan
• Mar 31st 2006, 05:08 AM
topsquark
Quote:

Originally Posted by topsquark
Remember that the directional derivative is a vector, so
$\displaystyle D_uf(x,y) |_{(0,0)} = f_x(0,0) \hat{i} + f_y(0,0) \hat{j}$

Since both the partial derivatives are zero, then, the directional derivative is the zero vector (which, of course, has a magnitude of zero.) Zero is certainly an allowed value (being a real number) so we CAN say the directional derivative exists at the origin, but that it simply has no magnitude.

-Dan

Whoops! I made a boo boo! :o

The directional derivative is, of course, not a vector. It is defined as:
$\displaystyle D_uf(x,y)=(f_x \hat i + f_y \hat j ) \cdot \hat u$ where $\displaystyle \hat u$ is a unit vector in the direction of u.

So your original expression for the directional derivative was correct. My apologies.

The rest of my discussion is almost unchanged. The only difference is the comment about the zero vector, which no longer has meaning with the correction.

-Dan
• Mar 31st 2006, 09:02 AM
ThePerfectHacker
Quote:

Originally Posted by paloma
I also have a question on the directional direvatives at the origin and whether or not they exist. For the partial direvative of x at the origin I got 0 and same for partial of y.
So I was using Duf(0,0) =fx cos() + fy sin() to find, but since both my fx and fy are 0 all I get is 0.
Is it possible that I got the right answer in that Duf(0,0) = 0?
Thanks

No the directional derivative do not exist. Because the function $\displaystyle f(x)=|x|$ is no differencialbe at the origin. Thus, when you are given the function,
$\displaystyle f(x,y)=x^2+y^2-3|xy|$ it is equivalen to,
$\displaystyle f(x,y)=x^2+y^2-3|x||y|$ no you cannot take the partial derivatives at zero because by the previous explanation the functions $\displaystyle |x|,|y|$ have no derivatives thus, no partial derivatives.
• Mar 31st 2006, 12:45 PM
topsquark
Ahem! Forgot about the absolute value part of the function!

-Dan
• Apr 1st 2006, 09:53 AM
paloma
Thank you!
Thank you both very much for helping me out with these two problems! I really appreciate your help.
I just have one last question about the abs(x) not being differentiable, why is that? I haven't taken math courses in over three years so I don't remember all the rules. Please explain this part to me.
Thank you so much. :)
• Apr 1st 2006, 11:47 AM
topsquark
Quote:

Originally Posted by paloma
Thank you both very much for helping me out with these two problems! I really appreciate your help.
I just have one last question about the abs(x) not being differentiable, why is that? I haven't taken math courses in over three years so I don't remember all the rules. Please explain this part to me.
Thank you so much. :)

Consider the graph of y = |x|. What is the first derivative for x < 0? It is -1. For x > 0 the derivative is +1. What is the derivative at x = 0? Obviously the first derivative has a discontinuity at x = 0 since it abruptly goes from -1 to +1. So we can't define a first derivative there.

-Dan
• Apr 1st 2006, 04:20 PM
ThePerfectHacker
Quote:

Originally Posted by paloma
Thank you both very much for helping me out with these two problems! I really appreciate your help.
I just have one last question about the abs(x) not being differentiable, why is that? I haven't taken math courses in over three years so I don't remember all the rules. Please explain this part to me.
Thank you so much. :)

First, let me remind you what absolute value means.
For all real numbers,
$\displaystyle |x|=\left\{ \begin{array}{ccc}x& \mbox{for}& x\geq 0 \\-x& \mbox{for}& x<0\end{array}\right$
--------
By Definition of Derivative we have,
$\displaystyle \lim_{\Delta x\to 0}\frac{|x+\Delta x|-|x|}{\Delta x}$

If $\displaystyle x>0$ we have, $\displaystyle |x|=x$ and $\displaystyle |x+\Delta x|=x+\Delta x$. Thus, the limit becomes,
$\displaystyle \lim_{\Delta x\to 0}\frac{x+\Delta x-x}{\Delta x}=\frac{\Delta x}{\Delta x}=1$

If $\displaystyle x<0$ we have, $\displaystyle |x|=-x$ and $\displaystyle |x+\Delta x|=-x-\Delta x$. Thus, the limit beomes,
$\displaystyle \lim_{\Delta x\to 0}\frac{-x-\Delta x+x}{\Delta x}=\frac{-\Delta x}{\Delta x}=-1$

If $\displaystyle x=0$ we have, $\displaystyle |x|=0$ and $\displaystyle |x+\Delta x|=|\Delta x|$. Thus, the limit beomes,
$\displaystyle \lim_{\Delta x\to 0}\frac{|\Delta x|}{\Delta x}$ but the limit does not exits. Because,
$\displaystyle \lim_{\Delta x\to 0^+}\frac{|\Delta x|}{\Delta x}=\frac{\Delta x}{\Delta x}=1$ and,
$\displaystyle \lim_{\Delta x\to 0^-}\frac{|\Delta x|}{\Delta x}=\frac{-\Delta x}{\Delta x}=-1$. Thus, the left sided limit does not coincide with the right handed limit thus, the function is not differenciable.

Thus, we have,
$\displaystyle (|x|)'=\left\{ \begin{array}{ccc}1 &\mbox{for}&x>0\\ \mbox{Does not exist}&\mbox{for}&x=0\\-1&\mbox{for}&x<0\end{array}\right$.

Below is a graph demonstrating the derivative, notice the point $\displaystyle x=0$ is not in its domain.
• Apr 3rd 2006, 10:18 AM
paloma
Thank you so much for that explanation. It really did help me to figure out what was going on.
Thanks again!
• Apr 3rd 2006, 02:09 PM
ThePerfectHacker
Welcome