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

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Answer 1 · 2017-02-04T05:24:22

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

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#