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

Apr 28, 2017

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 $\frac{\mathrm{dy}}{\mathrm{dx}}$ and then evaluate it at $x = 2$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3}{x} ^ 4$

${m}_{t} = - \frac{3}{2} ^ 4 = - \frac{3}{16}$

The slope of the normal line, ${m}_{n}$, can be found using the following equation:

${m}_{n} = - \frac{1}{m} _ t$

${m}_{n} = - \frac{1}{- \frac{3}{16}}$

${m}_{n} = \frac{16}{3}$

Using the point-slope form of the equation of a line, we find that the equation of any line passing through the point $\left(2 , \frac{1}{8}\right)$ is:

$y = m \left(x - 2\right) + \frac{1}{8} \text{ [1]}$

Substitute the value of ${m}_{t}$ into equation [1], to obtain the equation of the tangent line:

$y = - \frac{3}{16} \left(x - 2\right) + \frac{1}{8}$

Substitute the value of ${m}_{n}$ into equation [1], to obtain the equation of the normal line:

$y = \frac{16}{3} \left(x - 2\right) + \frac{1}{8}$