Question

Identify the curve by finding a Cartesian equation for the curve.

Answer 1

\[r=\tan \theta \sec \theta\]

@ganeshie8

Answer 2

r = sin/cos * 1/sin

Answer 3

Oh yeah, think that'll work?

Answer 4

maybe lol I am experimenting

Answer 5

What if we multiply both sides by cos theta...idk

Answer 6

r/cos = 1 like this?

Answer 7

r*costheta = 1 not sure though

Answer 8

oh you said multiply LMAO yeah

Answer 9

lool yeah

Answer 10

wait a minute

Answer 11

trying to get some identities to work

Answer 12

Wait isn't sec theta = 1/cos theta haha

Answer 13

So then 1/ cos theta would get cancelled out and you're left with r cos theta = tan theta ye?

Answer 14

yeah what did I put?

Answer 15

You put 1/ sin theta and I went with it haha

Answer 16

flutter I put 1/sin

Answer 17

LMAOOOO

Answer 18

Damn it nin, ok but yeah this makes sense now, we'll get a parabola :)

Answer 19

yeahh!
r = sin/cos * 1/cos

Answer 20

i should have stuck with x's and y's lmao

Answer 21

\[x = y/x \implies x^2 = y\] since we multiply both sides by x.

Answer 22

yeah

Answer 23

Lol no worries, here's your medal.

Answer 24

Thanks for helping :)

Answer 25

helping with nothing LMAO

Answer 26

Haha, what ever, least you put in the effort and fked up :P

Answer 27

Wait @ganeshie8 while you're here, now I need help going the other way :P

Answer 28

Find polar equation for the curve represented by the given cartesian equation.
4y^2 = x

Answer 29

the other way is always easy
\(\large x = r\cos \theta\)
\(\large y = r \sin \theta\)

Answer 30

So then \[4*rsin^2 \theta = rcos \theta\]

Answer 31

yes exept for the missing square on r on left side :)

Answer 32

Oh hold on I think the r should be squared

Answer 33

Yeah!

Answer 34

cancel r and try to get some good looking equation

Answer 35

The end result doesn't seem to look nice but give me a sec :P

Answer 36

\[\large r = \dfrac{1}{4}\cot\theta \csc\theta \]

Answer 37

or something like that...
in polar we represent function as \(\large r(\theta)\) in general..

Answer 38

Yup, got the exact same thing as \[\cot \theta = \cos \theta/ \sin \theta ~~~ \csc \theta = 1/ \sin \theta \]

Answer 39

as \(\large \theta\) changes, the function \(\large r(\theta)\) tells you how the distance from pole changes

Answer 40

most teachers/profs are particular about seeing the explicit form r(θ), just eliminating x and y may not impress them