# Polar curves Maximum and minimum.

• Apr 5th 2011, 04:13 PM
IDontunderstand
Polar curves Maximum and minimum.
I do not understand how to find the maximum and minimum.

Find the maximum and minimum point form the pole.

a) r=6cos(x)

b)r=3-6sin(x)

Thank You.
• Apr 5th 2011, 07:06 PM
surjective
Hey,

First realize that the functions you are dealing with are the cos/sin functions. Do the following (i haven't specified an interval):

1) Differentiate the function, i.e. find r'(x)

2) Solve the equation r'(x)=0. You may get several x-values, depending on the interval you are considering.

3) Find r'(x) for values greater and smaller than the x value found in 2). From the found r'(x) in this part you can establish where the function increases and where it decreases.

4) Hence if the function increases up until x=k and then starts decreasing you'll know that at x=k you have the maximum function-value.
• Apr 6th 2011, 12:54 PM
IDontunderstand
I haven't learned how to differentiate yet? Is there another way?
• Apr 6th 2011, 04:39 PM
Soroban
Hello, IDontunderstand!

By the way, polar expressions are always written in terms of $\,r$ and $\,\theta.$

Quote:

Find the maximum and minimum points from the pole.

. . $a)\;r\:=\:6\cos\theta$

. . $b)\;r\:=\:3-6\sin\theta$

Without Calculus? .It can be done, but it takes a bit of Thinking.
Can you handle that?

$(a)\;r \:=\:6\cos\theta$

We know that $\cos\theta$ ranges between -1 and +1.

The maximum value of $\,r$ occurs when $\cos\theta = +1$
. . That is, when $\theta = 0.$
Hence, a maximum occurs at $(6,\,0)$

The minimum value of $\,r$ occurs when $\cos\theta = \text{-}1.$
. . That is, when $\theta = \pi.$
Hence, a minimum occurs at $(-6,\,\pi)$

$(b)\;r \:=\:3 - 6\sin\theta$

We know that $\sin\theta$ ranges between -1 and +1.

The maximum value of $\,r$ occurs when $\sin\theta = \text{-}1$
. . That is, when $\theta = \frac{3\pi}{2}$
Hence, a maximum occurs at $\left(9,\:\frac{3\pi}{2}\right)$

The minimum value of $\,r$ occurs when $\sin\theta = +1.$
. . That is, when $\theta = \frac{\pi}{2}$
Hence, a minimum occurs at $\left(-3,\:\frac{\pi}{2}\right)$