How do you Use implicit differentiation to find the equation of the tangent line to the curve x^3+y^3=9 at the point where x=-1 ?

1 Answer
AJ Speller
Sep 26, 2014

We begin this problem by finding the point of tangency.

Substitute in the value of 1 for x.

x^3+y^3=9
(1)^3+y^3=9
1+y^3=9
y^3=8

Not sure how to show a cubed root using our math notation here on Socratic but remember that raising a quantity to the 1/3 power is equivalent.

Raise both sides to the 1/3 power

(y^3)^(1/3)=8^(1/3)

y^(3*1/3)=8^(1/3)

y^(3/3)=8^(1/3)

y^(1)=8^(1/3)

y=(2^3)^(1/3)

y=2^(3*1/3)

y=2^(3/3)

y=2^(1)

y=2

We just found that when x=1, y=2

Complete the Implicit Differentiation

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

Substitute in those x and y values from above =>(1,2)

3(1)^2+3(2)^2(dy/dx)=0

3+3*4(dy/dx)=0

3+12(dy/dx)=0

12(dy/dx)=-3

(12(dy/dx))/12=(-3)/12

(dy)/dx=(-1)/4=-0.25 => Slope = m

Now use the slope intercept formula, y=mx+b

We have (x,y) => (1,2)

We have m = -0.25

Make the substitutions

y=mx+b

2 = -0.25(1)+b

2 = -0.25+b

0.25 + 2=b

2.25=b

Equation of the tangent line ...

y=-0.25x+2.25

To get a visual with the calculator solve the original equation for y.

y=(9-x^3)^(1/3)

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