Chain rule

  • Thread starter sinbad30
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I am given Z = f (x, y), where x= r cosθ and y=r sinθ

I found

∂z/∂r = ∂z/∂x ∂x/∂r + ∂z/∂y ∂y/∂r = (cos θ) ∂z/∂x + (sin θ) ∂z/∂y and

∂z/∂θ = ∂z/∂x ∂x/∂θ + ∂z/∂y ∂y/∂θ= (-r sin θ) ∂z/∂x + (r cos θ) ∂z/∂y

I need to show that

∂z/∂x = cos θ ∂z/∂r - 1/r * sin θ ∂z/∂θ and

∂z/∂y = sin θ ∂z/∂r + 1/r * cos θ ∂z/∂θ

Any ideas?

Thanks
 

Answers and Replies

  • #2
ehild
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I am given Z = f (x, y), where x= r cosθ and y=r sinθ

I found

∂z/∂r = ∂z/∂x ∂x/∂r + ∂z/∂y ∂y/∂r = (cos θ) ∂z/∂x + (sin θ) ∂z/∂y and

∂z/∂θ = ∂z/∂x ∂x/∂θ + ∂z/∂y ∂y/∂θ= (-r sin θ) ∂z/∂x + (r cos θ) ∂z/∂y

I need to show that

∂z/∂x = cos θ ∂z/∂r - 1/r * sin θ ∂z/∂θ and

∂z/∂y = sin θ ∂z/∂r + 1/r * cos θ ∂z/∂θ

Any ideas?

Thanks

∂z/∂r = (cos θ) ∂z/∂x + (sin θ) ∂z/∂y

∂z/∂θ = (-r sin θ) ∂z/∂x + (r cos θ) ∂z/∂y

It is two equations for the "unknowns" ∂z/∂x and ∂z/∂y. Multiply the first equation by r(cosθ), the second one with (sinθ) and subtract them. What do you get?

ehild
 
Last edited:

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