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Partial Derivates using Chain Rule

  • Thread starter Sheldinoh
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  • #1
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Homework Statement



Find: ∂f/∂x
f(r,θ)=rsin^2(θ), x=rcosθ, y=rsinθ

The Attempt at a Solution


∂f/∂r=sin^2θ ∂r/∂x=-cosθ/x
∂f/∂θ=2*r*cosθ*sinθ ∂θ/∂x=-1/sqrt(1-(x^2)/(r^2)

∂f/∂x = -sin^2θcosθ/x^2 + -2*r*cosθ*sinθ/sqrt(1-(x^2)/(r^2)
∂f/∂x = -y-sqrt(x^2+y^2) / (x^2+y^2)^3/2

THE ANSWER IN THE BACK OF THE BOOK IS :
∂f/∂x = -xy^2 / (x^2+y^2)^3/2
 

Answers and Replies

  • #2
33,171
4,858

Homework Statement



Find: ∂f/∂x
f(r,θ)=rsin^2(θ), x=rcosθ, y=rsinθ

The Attempt at a Solution


∂f/∂r=sin^2θ ∂r/∂x=-cosθ/x
∂f/∂θ=2*r*cosθ*sinθ ∂θ/∂x=-1/sqrt(1-(x^2)/(r^2)
To find ∂r/∂x and ∂θ/∂x, it's helpful to have r as a function of x and y, and θ as a function of x and y.

r = sqrt(x2 + y2), θ = tan-1(y/x)

From the above, ∂r/∂x = x/sqrt(x2 + y2), and
∂θ/∂x = (-y/x2)/(1 + (y/x)2).

Your ∂θ/∂x looks wrong. Using the above I got the same answer as in the book.

∂f/∂x = -sin^2θcosθ/x^2 + -2*r*cosθ*sinθ/sqrt(1-(x^2)/(r^2)
∂f/∂x = -y-sqrt(x^2+y^2) / (x^2+y^2)^3/2

THE ANSWER IN THE BACK OF THE BOOK IS :
∂f/∂x = -xy^2 / (x^2+y^2)^3/2
To find ∂r/∂x and ∂θ/∂x, you need r as a function of x and y, and θ as a function of x and y.

r = sqrt(x2 + y2), θ = tan-1(y/x)
 

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