How do you find the equation of the tangent and normal line to the curve $y=x^3$ at x=2?

Question

8550 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-22T17:08:35

Equation of tangent is $12x-y-16=0$ and that of normal is $x+12y-98=0$

The slope of the tangent is given by value of first derivative at that point i.e. here at $x=2$

As $y=x^3$ , we have $(dy)/(dx)=3x^2$

and at $x=2$ , its value is $3*2^2=12$

When we seek a tangent at $x=2$ , it means tangent at $(2,2^3)$ i.e. at $(2,8)$

As tangent passes through $(2,8)$ and has a slope $12$ , its equation is $y-8=12(x-2)$ or $12x-y-16=0$

As normal is perpendicular to tangent, its slope is $(-1)/12$

and hence equation of normal is $y-8=-1/12(x-2)$

or $x+12y-98=0$

How do you find the equation of the tangent and normal line to the curve y=x^3y=x^3 at x=2?