# Calculus problem?

What equation must be satisfied at any point where the tangent line to 4x^2y^2 + 3y =2x is vertical?

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• 4x^2y^2 + 3y -2x = 0. Implicit differentiation

gives 4(2xy^2 + 2x^2yy') + 3y' = 2, ie.,

8xy^2 -2 + y'(3 + 8x^2y) = 0, ie., (3 +8x^2y)y' =

2(1- 4xy^2). Clearly, for 3+8x^2y = 0, y' is

undefined and the tangent line is vertical.

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• dx/dy must be 0, so let's calculate dx/dy:

8x y^2 dx/dy + 8x^2 y + 3 = 2 dx/dy =>

dx/dy = (8x^2 y + 3)/(2 - 8x y^2).

For this fraction to be zero, you must have x^2 y = -3/8.

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• dx/dy = 0

4 * x^2 * y^2 + 3y = 2x

4 * (x^2 * 2y * dy + 2x * dx * y^2) + 3 * dy = 2 * dx

4 * 2 * (x^2 * y * dy + x * y^2 * dx) + 3 * dy = 2 * dx

8x^2 * y * dy + 8xy^2 * dx + 3 * dy = 2 * dx

(8x^2 * y + 3) * dy = (2 - 8xy^2) * dx

(8x^2 * y + 3) / (2 - 8xy^2) = dx/dy

dx/dy = 0

8x^2 * y + 3 = 0

8x^2 * y = -3

x^2 * y = -3/8

y = -3 / (8 * x^2)

4x^2 * y^2 + 3y = 2x

4 * x^2 * (9 / (64 * x^4)) + 3 * (-3 / (8 * x^2)) = 2x

36 * x^2 / (64 * x^4)  -  27 / (8 * x^2)  =  2x

9 / (16x^2) - 54 / (16x^2) = 2x

(9 - 54) / (16x^2) = 2x

-45 = 32x^3

-45/32 = x^3

x = (-45/32)^(1/3)

y = -3 / (8 * x^2)

y = -3 / (8 * (-45/32)^(2/3))

y = (-3/8) * (-32/45)^(2/3)

y = (-3/8) * (1024/2025)^(1/3)

y = (-3/8) * (2^10 / (3^4 * 5^2))^(1/3)

y = (-3/8) * (2^9 * 2 * 5 * 3^2 / (3^6 * 5^3))^(1/3)

y = (-3/8) * (2^9 / (3^6 * 5^3))^(1/3) * (2 * 5 * 3^2)^(1/3)

y = (-3/8) * (2^3 / (3^2 * 5)) * 90^(1/3)

y = (-3/8) * (8/45) * 90^(1/3)

y = (-3/45) * (8/8) * 90^(1/3)

y = (-1/15) * 90^(1/3)

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• Hint:

dx/dy = 0

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