Question

Find a polar form of the equation x^2+y^2-8x=0.

a.) r= sin theta
b.) r=8
c.) r=8 sin theta
d.) r=8 cos theta

Answer 1

plugin :
x = rcos(theta)
y = rsin(theta)

in the given equaiton

Answer 2

so the x will be cosine and the y will be sin?

Answer 3

x will be r*cos(theta)
y will be r*sin(theta)

Answer 4

x^2+y^2-8x=0

plugin :
x = rcos(theta)
y = rsin(theta)


[r*cos(theta) ]^2 + [r*sin(theta)]^2 - 8*rcos(theta) = 0

Answer 5

simplify

Answer 6

okay, thanks for helping.

Answer 7

widout latex, its hard to type those sines/cosines :|

Answer 8

u sure u got how to simplify ? :)

Answer 9

im actually looking it up on google how to simplify this.

Answer 10

[r*cos(theta) ]^2 + [r*sin(theta)]^2 - 8*rcos(theta) = 0

r^2 [cos^2(theta) + sin^2(theta) ] - 8*rcos(theta) = 0

r^2[1] - 8*rcos(theta) = 0

r^2 = 8*rcos(theta)

r = 8cos(theta)

Answer 11

see if that makes some sense

Answer 12

yeah, it does. I've never done this before so i needed some step-by-step instruction.

Answer 13

good :) these are easy. u just need to remember below :

x = rcos(theta)
y = rsin(theta)

Answer 14

they give u : x^2+y^2 = r^2

Answer 15

to change to polar, u just substitute them for x and y, and simplify watever u can. thats all.

Answer 16

okay, that explains a lot. thank you so much.

Answer 17

remember x =rcostheta y =rsintheta and r^2 = x^2 + y^2 you may need them again in calc iv

Answer 18

this is calc? this is suppose to be trig....

Answer 19

O___________________________________o for me I had it in calc 3 and 4... wow my school must be that behind D:

Answer 20

well this is an online class for me, because i failed the trig at school so thats probably why i dont understand anything.
but thank you for giving the advice.

Answer 21

trig and polar wont lev u ever.. it pops up in many places... so u better master these now when u are at it :) good luck !