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
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)#