Question

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` ?

Answer 1

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)`