Question

polar functions

Answer 1

@jim_thompson5910 @robtobey @Spacelimbus @TuringTest

Answer 2

to convert cartesian to polar,put x=r*cos(theta),y=r*sin(theta)....try it...

Answer 3

x= 4 sin(theta)?

Answer 4

try x= 4 cos(theta).....similarly.y= 2 sin(theta)..and simplify....

Answer 5

u realise that this is the equation of the circle?

Answer 6

an ellipse

Answer 7

oh,yes ellipse.....sorry

Answer 8

(4cos(theta))^2/16 + (2sin(theta))^2/ 4 = 1

Answer 9

cos^2(theta) + sin^2(theta) =1

Answer 10

is that it?

Answer 11

actually the general form in polar is
r=p/(1+e*cos(theta))
do u know how to find eccentricity?

Answer 12

actually the general form in polar is
r=p/(1+e*cos(theta))
do u know how to find eccentricity?

Answer 13

no, i haven't learnt that

Answer 14

what is that?

Answer 15

ok...then try this r^2=a^2*cos(theta)^2+b^2*sin()theta^2
u know a and b,right?

Answer 16

is a = 4 and b =2?

Answer 17

r^2=a^2*cos(theta)^2+b^2*sin(theta)^2

Answer 18

yes,correct,just put this in general equation...and u get the polar form....

Answer 19

r^2 = 4^2 cos(theta)^2 + 2^2*sin(theta)^2

Answer 20

u got it?

Answer 21

this is what that looks like http://i.imgur.com/dRslp.png

Answer 22

here is solution
just do the straight substitution...x=rcos, y=rsin , then solve for r

http://www.wolframalpha.com/input/?i=polar+plot+r+%3D+4%2Fsqrt%28cos%28t%29^2%2B4sin%28t%29^2%29

Answer 23

Oh nice dumbcow I just did the same by plugging it in.

Answer 24

HOw do you find the end points for this?

Answer 25

is it -pi to pi?

Answer 26

on the ellipse its from -4 to 4 .... so find when r=4
--> cos^2 +4sin^2 = 1
--> (1-sin^2) +4sin^2 = 1
--> 3sin^2 = 0
--> sin = 0

theta = +-pi

Answer 27

nice thanks