Need Help with math

[Mirinee]

Lines with equal slopes are parallel to each other; lines with negative reciprocals are perpendicular.

Not to each other. You mean a line with gradient m is perpendicular to a line with gradient -1/m.

Eep. Converting from polar and rectangular. Um...
I hope the ones you get given are no more difficult than the example you gave.

R=cos(theta) on the polar graph (plot it ) is a circle of radius 1, its left side touching the origin, *draws*, so just look at that and say, oh, use the cartesian "equation of a circle", which is of the form
((X-A)^2)+((Y-B)^2) = C^2 , like whiteninja said.
The centre would be (0.5, 0), and its radius is 1, hence

(X-0.5)^2 + Y^2 = 1.

This might be a clumsy way of doing it, but I can't think of another way off the top of my head right now ;;^_^ .

I hope you won't be given anything other than
-equation of a straight line
-equation of a parabola
-equation of a circle

[zero132132]

Oh but I will. They decided to give me stuff such as some four-leaf clovers and some good old hyperbolas. Thanks for the help that actually helps a little. Guess I'll just have to use the chart method.

Edit: Go ahead and shut this one down mods, I have the test in about 2 hours and I won't have time to check back here.

[Mirinee]

They expect you to convert those into cartesian? I'd really like to know the proper way to do those problems. Would you tell me?

Anyway, good luck.

[zero132132]

Well... I just took it and well... I knew about half of the stuff on the test. I was a bit confused about what he was expecting us to do. All he expected us to do was to take something like y=(-1/squareroot(3))x and convert it to polar (that one turned out weird). With the graphing stuff, we had to graph cos^2(theta)=r on some weird graph dealie (bunch of circles with lines going out for all the parts of 2 pi). Oh well. Thanks for the help anyway. If I figure out how to convert the weird four-leaf clover to cartesian I'll get back to you. Don't expect it to be too soon though.