# Parametric and Polar Questions

• Apr 19th 2009, 04:34 PM
coolguy99
Parametric and Polar Questions
Reviewing for a test tomorrow, and most of the homework is pretty simple, but I just wanted to ask if you guys would check a few that I'm unsure on.

Turn this equation from Cartesian to polar.

x + y = 9

What I did..

square both sides,
x^2 + 2xy + y^2 = 81
x^2 + y^2 is r^2, so
r^2 + (2(r)cos(theta) times (r)sin(theta)) = 81
r^2 (1+ 2costheta sintheta) = 81
r = 9/sq.root (1+2cos(theta)sin(theta))

Next one..

find the area inside of the larger loop and outside of the smaller loop of the limacon r = 1/2 + cos(theta)

I solved this one through, and got pi/2 + (3root3)/4

but the answer in the back of the book has pi/4 instead of pi/2.. Could someone just show where I may have missed an extra 1/2 that would change my 1/2 into 1/4?

this was my integral
2 integral from 0 to 4pi/6 of [1/2 (1/2 + cos(theta))^2 dtheta]
• Apr 19th 2009, 04:43 PM
Calculus26
In the first why not simply rcos(t) +r sin(t) =9

then r= 9/[cos(t)+sin(t)] ? By the way there are pblms with your algebra --no need to square both sides

In the second your integral Calculates the area inside the outer loop

you have to subtract the area of the inner loop

2integral 1/2( 2pi/3 to pi) f(theta)^2(dtheta)
• Apr 19th 2009, 05:05 PM
coolguy99
alright, I fixed up those two, thanks a lot :)

I have one more question, a little easier this time

find all the points of intersection,

r=cos(3theta), r = sin(3theta)

cos3theta = sin3theta
1 = tan(theta)

so, 3theta = pi/4, 5pi/4
so theta = pi/12, 5pi/12

but when I graph it on my calculator (now allowed for the test, just doing it to check), there is one more. What did I miss?
• Apr 19th 2009, 05:13 PM
Calculus26
I assume you are looking for all solutions from 0 to 2pi. you only missed one?

This compresses the graph and so you have to extend beyond the usual

2 solutions . for 3theta go out six

3theta = pi/4, 5pi/4,9pi/4,13pi/4, 17pi/4, 21pi/4, etc

t = pi/12,5pi/12,3pi/4,13pi/12,17pi/12,7pi/4 the next one would be 25pi/12

which is outside the 2pi range.
• Apr 19th 2009, 05:20 PM
coolguy99
alright, that makes sense.

I lied though, it looks like I have one more question..

find the area between a large loop and the enclosed small lopp of the curve

r = 1 + 2cos(3theta)

I graphed it, it's three clovers, and inside each clover is a smaller clover. But how do you determine which bounds to use to get the outside loop, and then just the inside loop?
• Apr 19th 2009, 06:02 PM
Calculus26
For half of a big loop find the zero of 1+ 2 cos(3t)

Find the next zero --- between these 2 zeroes an entire small loop is generated

subtract the small loop from the big loop