What is the equation of the line that is normal to the polar curve f(theta)=2sin(3theta+pi/3) - 2thetacostheta at theta = pi/4?

1 Answer
Jul 27, 2018

y = (-8-8sqrt3)/(-12-4sqrt3+2pi)(x-((2-2sqrt3-pi)/4)) +((2-2sqrt3-pi)/4)

Explanation:

To find the equation of the line, we need a color(red)(poi nt), and we need a color(blue)(slope).

First, let's find the color(red)(poi nt).

Plug in pi/4 into f(theta).

f(pi/4)=2sin(3(pi/4)+pi/3)-2(pi/4)(cos(pi/4))
f(pi/4)=(sqrt2(2-2sqrt3-pi))/4

We have the polar coordinate (pi/4, (sqrt2(2-2sqrt3-pi))/4).
We have to convert this to Cartesian form. Using the formulas:
x = rcostheta
y = rcostheta

We find that the Cartesian coordinate is color(red)(((2-2sqrt3-pi)/4,(2-2sqrt3-pi)/4).

Good! We have found the color(red)(poi nt).

Now, we need to find the color(blue)(slope).

The formula for the tangent line of a polar function is:

dy/dx = ((dr)/(d theta)sintheta+rcostheta)/((dr)/(d theta)costheta-rsintheta)

If you notice, sintheta and costheta will be the same in this case, because sin(pi/4) and cos(pi/4) are the same, so for this case, our dy/dx is simply:

dy/dx = ((dr)/(d theta)+r)/((dr)/(d theta)-r)

We solved r already previously, which was (sqrt2(2-2sqrt3-pi))/4.

Now, we need to find (dr)/(d theta).

f'(theta)=(dr)/(d theta)=6cos(3theta+pi/3)+2thetasintheta-2costheta

Now, we plug in pi/4 and get:

f'(pi/4)=(dr)/(d theta)=(sqrt2(-10-6sqrt3+pi))/4

Going back to the tangent line formula:

dy/dx = ((sqrt2(-10-6sqrt3+pi))/4+(sqrt2(2-2sqrt3-pi))/4)/((sqrt2(-10-6sqrt3+pi))/4-(sqrt2(2-2sqrt3-pi))/4)

After some tedious simplification, we find this reduces to color(blue)(dy/dx = (-8-8sqrt3)/(-12-4sqrt3+2pi)

We have our color(red)(poi nt):
color(red)(((2-2sqrt3-pi)/4,(2-2sqrt3-pi)/4)

We have our color(blue)(slope):
color(blue)(dy/dx = (-8-8sqrt3)/(-12-4sqrt3+2pi)

All that is left is to plug in these values into the point slope formula:
(y - color(red)(y_1)) = color(blue)(m)(x - color(red)(x_1))

We get:

y - ((2-2sqrt3-pi)/4) = (-8-8sqrt3)/(-12-4sqrt3+2pi)(x-((2-2sqrt3-pi)/4))

So the equation is:

y = (-8-8sqrt3)/(-12-4sqrt3+2pi)(x-((2-2sqrt3-pi)/4)) +((2-2sqrt3-pi)/4)