Steve had done the above steps for you, see the first of the Related Questions for you.
First, it might help if you looked at the graph
http://www.wolframalpha.com/input/?i=polar+plot+r%3D8cos(3%C3%98)
Steve told you:
|cos 3Ø| ? 1
so |8cos 3Ø| ? 8
| r | ? 8 , so that's done!
for the zeros's of r, Steve said:
8cos 3Ø = 0
cos 3Ø = 0
3Ø = ?/2 or 3Ø = 3?/2 <---- you should know this
Ø = ?/6 or Ø = ?/2
but the period of cos 3Ø is 2?/3 ,
so adding/subtracting 2?/3 to any of the answers already found will give more solutions.
starting with Ø= ?/2
e.g. ?/2 + 2?/3 = 7?/6
7?/6 + 2?/3 = 11?/6 <--- last one before 2?
Starting with Ø = ?/6
?/6 + 2?/3 = 5?/6
5?/6 + 2?/3 = 3?/2
3?/2 + 2?/3 = 13?/6 , which is > 2? by ?/6 , so we are repeating.
So for just one rotations, we have
Ø = ?/6, 5?/6 , 3?/2, ?/2, 7?/6, and 11?/6
If you find it easier to think in degrees
Ø = 30°, 150° ,270°, 90°, 210°, and 330°
checking one of these, e.g. Ø = 210°
r = 8cos(3(210°))
= 8 cos 630°
= 8(0) = 0
Notice that when you sketch tangents to the curve, you can do this at 30° (same as 210° line), 150° (same as 330° line) and at 90° (same as the 270°).
Find the polar equation r=8cos3theta, find the maximum value of lrl and any zeros of r. Verify your answers numerically.
|cos3θ| <= 1
cos3θ=0 when 3θ is an odd multiple of π/2
what is the next step to answering the question?
1 answer
1 answer