View Full Version : Math problem: Partial differention

09-17-2008, 01:32 PM
I have the answer but don't know how to get to it.


I really don't understand the last step. Please try to explain it to me. You don't need to know anymore than the above to solve the problem, but I can post the full exercise if you want.

I know it's probably smarter to just send an e-mail to my docent, but this way would be more fun.

If the image doesn't appear as it should, then paste the following code into any Wikipedia page and press the button to preview it.
:<math>\frac{\partial^2 z}{\partial s^2} = \frac{\partial}{\partial s} \left( 2 \frac{\partial z}{\partial x} + 3 \frac{\partial z}{\partial y} \right) = 4 \frac{\partial z}{\partial x} + 12 \frac{\partial z}{\partial x \partial y} + 9 \frac{\partial z}{\partial y}</math>

09-17-2008, 01:39 PM
well, if I was to solve that problem, I would go about my usual method:

Stare blankly at it for a few minutes, and then either take a completely wild guess or say "fuck it", throw the paper out the window, and move out to the forest where I will live out my days surviving on berries and talking to trees.

Yeah. I got a "D" in high school trig.

09-17-2008, 03:48 PM
The coolest part about differentiation is the name.

09-17-2008, 05:52 PM
its been years since i studied maths and even then i dont remember touching on anything like that!

09-17-2008, 10:47 PM
you haven't posted a problem. post the whole exercise and i can help.

09-18-2008, 10:44 AM
you haven't posted a problem. post the whole exercise and i can help.

I don't need the solution to the problem, since that's what I posted. I just want to know how to solve this step:


I know how to get left part, but not to the right part.

09-18-2008, 10:49 AM
and i can't tell you unless i know what z or s are, or at least what they're a function of.

d/ds of (2 dz/dx + 3 dz/dy) is equal to what you'd think: 2 d^2z/dsdx + 3 d^2z/dsdy. beyond that i'd have to know what the functions are.

09-18-2008, 11:20 PM
z = f(x,y), and I think x and y is; x = 2s-t, y = 3+2t.

09-18-2008, 11:28 PM
edit: i think you mean y = 3x + 2t.

thus: d^2/ds^2(z) = d/ds ( dz/ds) = d/ds( dz/dx*dx/ds + dz/dy*dy/ds) (via the chain rule) = d/ds( dz/dx * 2 + dz/dy * 3)

now for the second part: d/ds(dz/dx) = d^2z/dxds = d^2z/dx^2*dx/ds + d^2z/dxdy*dy/ds and apply what dx/ds and dy/ds are again, and do the other term as well, and you end up with....

4 d^2z/dx^2 + 12 d^2z/dxdy + 9 d^2z/dy^2.

and yes, those squares are important, you shouldn't have dropped them all in what you've written out.