Partial Derivative

  • #1
198
0
Whoops got it now, didn't carry out my substitutions far enough.

Homework Statement


[tex]
z = x^2 + 2y^2
[/tex]
[tex]
x = rcos(\theta)
[/tex]
[tex]
y = rsin(\theta)
[/tex]

Homework Equations





The Attempt at a Solution


Find [tex](\partial z/\partial x)[/tex] (theta is constant)

dz = 2xdx + 4ydy
dx = cos[tex](\theta)[/tex]dr - rsin[tex](\theta)[/tex]d[tex]\theta[/tex]
dy = sin[tex](\theta)[/tex]dr + rcos[tex](\theta)[/tex]d[tex]\theta[/tex]

Unfortunately I'm not really quite sure where to go from here, I know that
[tex](\frac{ \partial z } { \partial x} )[/tex] is 2x when y is constant. But how to factor in theta being constant?
I suppose I could reduce
dx to dx = cos[tex](\theta)[/tex]dr
and dy = sin[tex](\theta)[/tex]dr
 
Last edited:

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
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Whoops got it now, didn't carry out my substitutions far enough.

Homework Statement


[tex]
z = x^2 + 2y^2
[/tex]
[tex]
x = rcos(\theta)
[/tex]
[tex]
y = rsin(\theta)
[/tex]

Homework Equations





The Attempt at a Solution


Find [tex](\partial z/\partial x)[/tex] (theta is constant)

dz = 2xdx + 4ydy
dx = cos[tex](\theta)[/tex]dr - rsin[tex](\theta)[/tex]d[tex]\theta[/tex]
dy = sin[tex](\theta)[/tex]dr + rcos[tex](\theta)[/tex]d[tex]\theta[/tex]

Unfortunately I'm not really quite sure where to go from here, I know that
[tex](\frac{ \partial z } { \partial x} )[/tex] is 2x when y is constant. But how to factor in theta being constant?
If [itex]\theta[/itex] is a constant, then [itex]d\theta= 0[/itex]

I suppose I could reduce
dx to dx = cos[tex](\theta)[/tex]dr
and dy = sin[tex](\theta)[/tex]dr
Yes, that is exactly correct. Then dz= dx+ dy= ?
 

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