What is the equation of the line that is normal to the polar curve f(theta)=-5theta- sin((3theta)/2-pi/3)+tan((theta)/2-pi/3) at theta = pi?

1 Answer
Jul 13, 2016

The line is y = (6 - 60pi + 4sqrt(3))/(9sqrt(3) -52) x + ((sqrt(3)(1 - 10pi) +2)^2)/(9sqrt(3) - 52)

Explanation:

This behemoth of an equation is derived through a somewhat lengthy process. I will first outline the steps by which the derivation will proceed and then perform those steps.

We are given a function in polar coordinates, f(theta). We can take the derivative, f'(theta), but in order to actually find a line in cartesian coordinates, we will need dy/dx.

We can find dy/dx by using the following equation:
dy/dx = (f'(theta)sin(theta) + f(theta)cos(theta))/(f'(theta)cos(theta) - f(theta)sin(theta))

Then we'll plug that slope into the standard cartesian line form:
y = mx + b
And insert the cartesian converted polar coordinates of our point of interest:
x = f(theta)cos(theta)
y = f(theta)sin(theta)

A few things that should be immediately obvious and will save us time down the line. We are taking a line tangent to the point theta = pi. This means that sin(theta) = 0 so...
1) Our equation for dy/dx will actually be:
dy/dx = f(pi)/(f'(pi))

2) Our equations for the cartesian coordinates of our point will become:
x = -f(theta)
y = 0

Starting to actually solve the problem, then, our first order of business is finding f'(theta). It isn't hard, just three easy derivatives with chain rule applied to two:
f'(theta) = -5 - 3/2 cos((3pi)/2 - pi/3) + 1/2 sec^2 (theta/2 - pi/3)

Now we want to know f(pi):
f(pi) = -5pi - sin((7pi)/6) + tan(pi/6)
= -5pi - 1/2 + 1/sqrt3
= (sqrt3(1 - 10pi) + 2)/(2sqrt3)

And f'(pi)...
f'(pi) = -5 - 3/2 cos((7pi)/6) + 1/2 sec^2 (pi/6)
= -5 + (3sqrt3)/4 + 2/3
= (9sqrt3 - 52)/12

With these in hand, we are ready to determine our slope:
dy/dx = f(pi)/(f'(pi))
= (sqrt3(1 - 10pi) + 2)/(2sqrt3) * 12/(9sqrt3 - 52)
=(6(1-10pi) + 4sqrt3)/(9sqrt3 - 52)

We can plug this in as m in y = mx +b. Recall that we previously determined that y=0 and x = -f(theta):
0 =-((6(1-10pi) + 4sqrt3)/(9sqrt3 - 52))((sqrt3(1 - 10pi) + 2)/(2sqrt3)) + b
0 =-((3(1-10pi) + 2sqrt3)/(9sqrt3 - 52))((sqrt3(1 - 10pi) + 2)/(sqrt3)) + b
0 =-((sqrt3(1-10pi) + 2)/(9sqrt3 - 52))(sqrt3(1 - 10pi) + 2) + b
b = ((sqrt3(1 - 10pi) + 2)^2)/(9sqrt3 - 52)

We can combine our previously determined m with our newly determined b to give the equation for the line:
y = (6 - 60pi + 4sqrt(3))/(9sqrt(3) -52) x + ((sqrt(3)(1 - 10pi) +2)^2)/(9sqrt(3) - 52)