Question

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

Answer 1

`0.3809x+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=xsin(3x+pi/4)` gives amplitude-increasing oscillations,

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

spacing `pi/3`

The foot of the normal is `P(pi/2, f(pi/2))=P(1.5708, -1.1107)`

`f'=3xcos(3x+pi/4)+sin(3x+pi/4)=2.6254`, at P.

Slope of the normal = `-1/(f')=-0.3809`, nearly.

So, the equation to the normal at P is

`y+1.1107=-0.3809(x-1.5708)` giving

`0.3809x+y+0.5124=0`