What is the slope of the curve y³ + xy² = 4 at the point where y = 2?

To find the slope of the curve at the point where y = 2, we first need to find the corresponding x-value by substituting y into the equation of the curve.

The curve is given by:

y³ + xy² = 4

Substituting y = 2:

(2)³ + x(2)² = 4

8 + 4x = 4

4x = 4 – 8

4x = -4

x = -1

Now, we have the point (-1, 2). To find the slope at this point, we can use implicit differentiation.

Differentiating both sides of y³ + xy² = 4 with respect to x:

3y²(dy/dx) + (y²)(1) + (x)(2y)(dy/dx) = 0

Rearranging the equation:

3y²(dy/dx) + y² + 2xy(dy/dx) = 0

(3y² + 2xy)(dy/dx) + y² = 0

Solving for dy/dx:

(dy/dx) = -y² / (3y² + 2xy)

Substituting x = -1 and y = 2 into the equation:

(dy/dx) = -(2)² / (3(2)² + 2(-1)(2))

=( -4 ) / ( 12 – 4 )

= -4 / 8 = -1/2

Therefore, the slope of the curve at the point where y = 2 is -1/2.

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