How do you find the slope of the curve $y^{3} - x y^{2} = 4$ $y^3-xy^2=4$ at the point where y=2?

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Answer 1 · 2018-07-02T06:59:39

The slope of the curve is $\frac{1}{2}$ $1/2$

We need to start by finding the x-coordinate.

$2^{3} - x {(2)}^{2} = 4$ $2^3 -x(2)^2= 4$

$- 4 x = - 4$ $-4x = -4$

$x = 1$ $x=1$

Now we differentiate.

$3 y^{2} (\frac{d y}{d x}) - y^{2} - 2 x y (\frac{d y}{d x}) = 0$ $3y^2(dy/dx) -y^2 - 2xy(dy/dx) =0$

$\frac{d y}{d x} (3 y^{2} - 2 x y) = y^{2}$ $dy/dx(3y^2 -2xy) = y^2$

$\frac{d y}{d x} = \frac{y}{3 y - 2 x}$ $dy/dx = y/(3y - 2x)$

The slope is simply the value of $\frac{d y}{d x}$ $dy/dx$ at $(1, 2)$ $(1,2)$

$\frac{d y}{d x} = \frac{2}{6 - 2} = \frac{2}{4} = \frac{1}{2}$ $dy/dx = 2/(6- 2) = 2/4=1/2$

Hopefully this helps!

How do you find the slope of the curve y^3-xy^2=4y3−xy2=4y^3-xy^2=4 at the point where y=2?