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

1 Answer
Noah G
Mar 5, 2017

The equation is #y = -sqrt(3)/4x + (sqrt(3)pi - 6)/12#

Explanation:

Start by finding the y-coordinate of tangency.

#f(pi/3) = sin((pi/3)/2 + pi)#

#f(pi/3) = sin(pi/6 + pi)#

#f(pi/3) = sin((7pi)/6)#

#f(pi/3) = -1/2#

Now we find the derivative of #f(x)#, using the chain rule. We let #y = sinu# and #u = x/2 + pi#. Then #dy/(du) = cosu# and #(du)/dx = 1/2# (since #π# is only a constant).

The derivative therefore is

#dy/dx = dy/(du) * (du)/dx#

#dy/dx = cosu * 1/2#

#dy/dx = 1/2cos(x/2 + pi)#

We now find the slope of the tangent.

#dy/dx|_(x = pi/3) = 1/2cos((pi/3)/2 + pi)#

#dy/dx|_(x = pi/3) = 1/2cos((7pi)/6)#

#dy/dx|_(x = pi/3) = 1/2(-sqrt(3)/2)#

#dy/dx|_(x= pi/3) = -sqrt(3)/4#

We now have all the information we need to find the equation of the line.

#y - y_1 = m(x- x_1)#

#y - (-1/2) = -sqrt(3)/4(x - pi/3)#

#y + 1/2 = -sqrt(3)/4x + (sqrt(3)pi)/12#

#y = -sqrt(3)/4x + (sqrt(3)pi - 6)/12#

Hopefully this helps!

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