Partial Derivates using Chain Rule

[Sheldinoh]

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

 

[Mark44]

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)