# What is the equation of the line normal to  f(x)=xsin(3x+pi/4) at  x=pi/2?

Feb 4, 2017

$0.3809 x + y + 0.5124 = 0$. See the normal-inclusive Socratic graphs.

#### Explanation:

graph{(xsin(3x+.7854)-y)(0.3809x+y+0.5124)=0 [-40, 40, -20, 20]}

Uniform scale graphs.

graph{(xsin(3x+.7854)-y)(0.38x+y0.51)((x-1.5804)^2+(y+1.1107)^2-.01)=0 [0, 2, --100, 100]}

$f = x \sin \left(3 x + \frac{\pi}{4}\right)$ gives amplitude-increasing oscillations,

with zeros at x = 1/3(k-1/4)pi. k = 0, +-1, +-2, ... that have common

spacing $\frac{\pi}{3}$

The foot of the normal is $P \left(\frac{\pi}{2} , f \left(\frac{\pi}{2}\right)\right) = P \left(1.5708 , - 1.1107\right)$

$f ' = 3 x \cos \left(3 x + \frac{\pi}{4}\right) + \sin \left(3 x + \frac{\pi}{4}\right) = 2.6254$, at P.

Slope of the normal = $- \frac{1}{f '} = - 0.3809$, nearly.

So, the equation to the normal at P is

$y + 1.1107 = - 0.3809 \left(x - 1.5708\right)$ giving

$0.3809 x + y + 0.5124 = 0$