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Partial Derivatives:

  • #1

Homework Statement



Given:
[tex]\varphi(t)[/tex] – differentiable function.
[tex]z=z(x,y)[/tex] – differentiable function.

And there is the following equation:

[tex]x^2 + y^2 + z^2 = \varphi (ax+by+cz)[/tex]

where [tex]a,b,c[/tex] are constants,

Prove that:

[tex](cy - bz)\cdot \frac {\partial z}{\partial x} + (az-cx)\cdot \frac{\partial z}{\partial y} = bx - ay [/tex]

The Attempt at a Solution



I tried to take partial derivatives of both sides with respect to x and then with respect to y. But I don't know how to differentiate the right-hand side of the equation.
Also if I did, what should I had done next?
 

Answers and Replies

  • #2
diazona
Homework Helper
2,175
6
Well, first things first: do you know the chain rule?

In your case, it says that
[tex]\frac{\partial\varphi}{\partial x} = \frac{\partial\varphi}{\partial t}\frac{\partial t}{\partial x}[/tex]
and similarly for y. Use that on the right side and see what you get. (I'm not sure the solution will pop out of that but it is probably worth trying, at least)
 
  • #3
Yep, I know that chain rule. We get:

[tex] 2x + 2z \frac{ \partial z }{ \partial x } = \frac{ \partial \varphi } { \partial t } \left( a + c \cdot \frac{ \partial z }{ \partial x } \right) [/tex]

and:

[tex] 2y + 2z \frac{ \partial z }{ \partial y } = \frac{ \partial \varphi } { \partial t } \left( b + c \cdot \frac{ \partial z }{ \partial y } \right) [/tex]

From here I have no idea how to continue. What should I do?
 
  • #4
392
0
Hint: You did the hard part already! :)

Spoiler below:




Solve each equation for [tex]\frac{ \partial \varphi } { \partial t } [/tex] and ....
 
  • #5
What do you mean to solve the equation "for" something?
I suppose I could isolate the [tex] \frac{\partial \varphi}{\partial t}[/tex] if I knew it's multiplier wasn't zero...

Edit:
I did that. It leads to nothing.
It's obvious that I miss something obvious but after a week on this problem and a dead-line of 2 days ahead, I am in doubt I could solve it.
 
Last edited:

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