Solve 4x-y^3=32xy

Solve the equation

x = \frac{y ^{3}}{4 - 32 y}

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Testing for symmetry about the origin

Testing for symmetry about the x-axis

Testing for symmetry about the y-axis

Not symmetry with respect to the origin

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r = 0 r = - \frac{8 sin ( 2 θ ) + 2 16 sin ^{2} ( 2 θ ) + sin ^{3} ( θ ) cos ( θ )}{sin ^{3} ( θ )} r = \frac{- 8 sin ( 2 θ ) + 2 16 sin ^{2} ( 2 θ ) + sin ^{3} ( θ ) cos ( θ )}{sin ^{3} ( θ )}

Evaluate

4 x - y^{3} = 32 x y

Move the expression to the left side

4 x - y^{3} - 32 x y = 0

To convert the equation to polar coordinates,substitute x for r cos (θ) and y for r sin (θ)

4 cos (θ) \times r - (sin (θ) \times r)^{3} - 32 cos (θ) \times r sin (θ) \times r = 0

Factor the expression

- sin^{3} (θ) \times r^{3} - 32 cos (θ) sin (θ) \times r^{2} + 4 cos (θ) \times r = 0

Simplify the expression

- sin^{3} (θ) \times r^{3} - 16 sin (2 θ) \times r^{2} + 4 cos (θ) \times r = 0

Factor the expression

r (- sin^{3} (θ) \times r^{2} - 16 sin (2 θ) \times r + 4 cos (θ)) = 0

When the product of factors equals 0,at least one factor is 0

r = 0 - sin^{3} (θ) \times r^{2} - 16 sin (2 θ) \times r + 4 cos (θ) = 0

Solution

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r = 0 r = - \frac{8 sin ( 2 θ ) + 2 16 sin ^{2} ( 2 θ ) + sin ^{3} ( θ ) cos ( θ )}{sin ^{3} ( θ )} r = \frac{- 8 sin ( 2 θ ) + 2 16 sin ^{2} ( 2 θ ) + sin ^{3} ( θ ) cos ( θ )}{sin ^{3} ( θ )}

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Find the derivative with respect to x

Find the derivative with respect to y

\frac{d y}{d x} = \frac{- 32 y + 4}{3 y ^{2} + 32 x}

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