Question

conversion cartesian to polar. help!!

Answer 1

1.) y^2 = x^3 / 2a - x the answer is r= 2a sin theta tan theta..
2.) y^2 = x(x-a) ^2 / 2a - x the answer is r^2 - 2ar sec theta + a^2 = 0..

Answer 2

step by step process pls.. im lost again

Answer 3

@AllTehMaffs ?

Answer 4

be right back. ill have my lunch.. :

Answer 5

hai

Answer 6

need your help..

Answer 7

im sorry. ill be back tomorrow. but pls.. help me with these problems.. i have to study for our final exam, identifying rocks blabla..

Answer 8

y^2 = x^3 / (2a - x) ?

y^2 = x(x-a) ^2 / (2a - x) ?

Answer 9

yes thats it. been equate in y^2

Answer 10

I can set em up for ya :)

Answer 11

really? Oh God. Thank you so much!!

Answer 12

ow by the way, any idea about parametric equations?? THANK YOU SO MUCH! YOURE THE BEST!!! :D

Answer 13

Yeah, a bit. What about them?

Answer 14

For the first one, start by noting that the answer

\[r = 2a \sin \theta \tan \theta = \frac{2a \sin^{2} \theta}{\cos \theta}\]

\[(2a - r \cos \theta)r^{2}\sin^{2}\theta= r^{3}\cos^{3}\theta\]
\[(2a - r \cos \theta)\sin^{2}\theta= rcos^{3}\theta\]

multiply through on the left side, move the r terms to the right side, apply sin^2=1-cos^2 on the right hand side. reduce and simplify.

Answer 15

For the second, first note that the given answer

\[r^{2}\cos \theta - 2ar +a^{2}\cos \theta= 0\]

now

\[(2a - r \cos \theta) r^{2} \sin^{2} \theta = r \cos \theta (r^{2}\cos^{2}\theta - 2ar \cos \theta + a^{2})\]
\[(2a - r \cos \theta) r \sin^{2} \theta = \cos \theta (r^{2}\cos^{2}\theta - 2ar \cos \theta + a^{2})\]

multiply through on both sides

First trick is to notice the 2ar sin^2 theta on the right, and 2ar cos^2 theta on the left.

\[2ar \sin^{2} \theta + 2ar \cos^{2}\theta = 2ar (\sin^{2} \theta + \cos^{2} \theta) = 2ar\]

On the last sin^2theta term, apply sin^2 = 1 - cos^2. Simplify the expression to get the answer.

Answer 16

1st)

\[2a \sin^{2} \theta - r \cos \theta \sin^{2}\theta = r \cos^{3} \theta\]

2nd)

\[2ar \sin^{2} \theta - r^{2}\sin^{2} \theta \cos \theta = r^{2} \cos^{3} \theta - 2ar \cos^{2} \theta + a^{2}\cos \theta\]