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Polar functions

# Worked example: differentiating polar functions

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.G (LO)
,
FUN‑3.G.1 (EK)
,
FUN‑3.G.2 (EK)
An AP Calculus sample item where we find the rate of change of 𝘹 with respect to θ.

## Want to join the conversation?

• At , Sal says the function 'flips' it over into positive. Why is this? Why can't the function r be negative?
• I had the same question. This helped:

https://youtu.be/ii31Gy4quqo?list=PLX2gX-ftPVXXWNn8FQ8DfZ0N0YVJCxe-p

https://youtu.be/DqfG8jLyfzQ?t=446

Lets's evaluate a simpler function at a few points.

r(a) = 4 sin (a)
--------------
r(pi/2) = 4sin(pi/2) = 4*1 = 4
r(pi) = 4sin(pi) = 4*0 = 0
r(1.5pi) = 4sin(1.5pi) = 4*-1 = -4

The insight is that you look at the sign of the radius. Here it is negative. So you prepare to put the point where you expect it to, but you can't put it there because that would mean the radius was positive after all, that would be the point (4, 1.5pi). To plot (-4, 1.5pi), make the arrow of the radius point in the opposite direction, up instead of down.
• Why is zeta pi/2 at ?
• Theta is pi/2 at point P on the graph, which sits on the y axis. As you go around the unit circle, the axes represent certain divisions on the unit circle, either in radians (in this case) or in degrees. Here, we've gone one quarter of the way around the unit circle, which is made up of 2pi radians. A quarter of 2pi is pi/2. This is also the angle of theta at point P.