# What is the equation of the line normal to f(x)=x^2 + sin(x)  at x=pi?

Jan 28, 2016

$y - {\pi}^{2} = \frac{x - \pi}{1 - 2 \pi}$

#### Explanation:

Find the point the normal line will intercept.

$f \left(\pi\right) = {\pi}^{2} + \sin \left(\pi\right) = {\pi}^{2}$

The normal line will pass through the point $\left(\pi , {\pi}^{2}\right)$.

To find the slope of the normal line, first find the slope of the tangent line. Since the tangent line and normal line are perpendicular, their slopes will be opposite reciprocals of one another.

To find the slope of the tangent line, first find the derivative of the function.

$f \left(x\right) = {x}^{2} + \sin \left(x\right)$

The derivative of ${x}^{2}$ can be found through the power rule and the derivative of $\sin \left(x\right)$ is $\cos \left(x\right)$.

$f ' \left(x\right) = 2 x + \cos \left(x\right)$

The slope of the tangent line is

$f ' \left(\pi\right) = 2 \pi + \cos \left(\pi\right) = 2 \pi - 1$

Thus, the slope of the normal line will be the opposite reciprocal of $2 \pi - 1$, which is $- \frac{1}{2 \pi - 1} = \frac{1}{1 - 2 \pi}$.

The normal line passes through the point $\left(\pi , {\pi}^{2}\right)$ and has a slope of $\frac{1}{1 - 2 \pi}$. This can be expressed as

$y - {\pi}^{2} = \frac{x - \pi}{1 - 2 \pi}$

Graphed are the function and the normal line:

graph{(x^2+sinx-y)(y-pi^2-1/(1-2pi)(x-pi))=0 [-15.51, 20.53, -0.57, 17.46]}