Question

How do you find the equation of the tangent and normal line to the curve `y=1/x^3` at `(2,1/8)`?

Answer 1

The slope of the tangent line is the first derivative evaluated at x = 2.
The slope of the normal line will be perpendicular to the tangent.
Use the point-slope form to find the equation.

Explanation:

To find the slope of the tangent line, `m_t`, we compute `dy/dx` and then evaluate it at `x = 2`

`dy/dx = -3/x^4`

`m_t= -3/2^4= -3/16`

The slope of the normal line, `m_n`, can be found using the following equation:

`m_n= -1/m_t`

`m_n= -1/(-3/16)`

`m_n = 16/3`

Using the point-slope form of the equation of a line, we find that the equation of any line passing through the point `(2,1/8)` is:

`y = m(x-2)+1/8" [1]"`

Substitute the value of `m_t` into equation [1], to obtain the equation of the tangent line:

`y = -3/16(x-2)+1/8`

Substitute the value of `m_n` into equation [1], to obtain the equation of the normal line:

`y = 16/3(x-2)+1/8`