How do you find the slope of the curve #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 #1/2#

We need to start by finding the x-coordinate.

#2^3 -x(2)^2= 4#

#-4x = -4#

#x=1#

Now we differentiate.

#3y^2(dy/dx) -y^2 - 2xy(dy/dx) =0#

#dy/dx(3y^2 -2xy) = y^2#

#dy/dx = y/(3y - 2x)#

The slope is simply the value of #dy/dx# at #(1,2)#

#dy/dx = 2/(6- 2) = 2/4=1/2#

Hopefully this helps!