Question

What is the Cartesian equation equivalent to the polar equation r = sec Ө csc Ө?

x = r cos Ө y = r sin Ө
x/r =cos Ө y/r = sin Ө

r^2 = x^2 + y^2

and then i get lost.
halp
pls

Answer 1

\[\Large\rm r=\sec \theta \csc \theta\]Do you remember how secant and cosecant relate to sine and cosine? :)
That will help us quite a bit here.

Answer 2

Yeah, sec = 1/cos
csc = 1/sin

Answer 3

\[\Large\rm r=\color{royalblue}{\sec \theta}\cdot\csc \theta\]Mmm ok that's a good start.
\[\Large\rm r=\color{royalblue}{\frac{1}{\cos \theta}}\cdot\frac{1}{\sin \theta}\]

Answer 4

If I divide both sides by r, I end up with:\[\Large\rm 1=\color{royalblue}{\frac{1}{r\cos \theta}}\cdot\frac{1}{\sin \theta}\]I'll divide by r again and get,\[\Large\rm \frac{1}{r}=\color{royalblue}{\frac{1}{r\cos \theta}}\cdot\frac{1}{r\sin \theta}\]

Answer 5

Maybe that wasn't extremely necessary, I just like to try and work with those first two identities that you listed :)

Answer 6

Do you see where we can plug them in?

Answer 7

Not really, my brain is super-fried atm :(

Answer 8

\[\Large\rm \frac{1}{r}=\frac{1}{\color{orangered}{r\cos \theta}}\cdot\frac{1}{\color{blue}{r\sin \theta}},\qquad \color{orangered}{x=r \cos \theta},\qquad \color{blue}{y=r \sin \theta}\]Ya? :d

Answer 9

oh!

Answer 10

so then
r = r cos θ * r sin θ
r = x*y

Answer 11

r = xy
ya looks good :)
So then undo your r using the square identity thing, ya?

Answer 12

\[\Large\rm r^2=x^2+y^2\]Therefore,\[\Large\rm r=\sqrt{x^2+y^2}\]

Answer 13

We can plug that in :d

Answer 14

so
xy = sqrt(x^2 +y^2)

Answer 15

and therefore
y = sqrt(x^2+y^2)/x

Answer 16

Well, we'd like to get it in the form \(\Large\rm y=x~stuff\).
See how you left a y^2 on the right?

Answer 17

oh.

Answer 18

So let's back up a step,\[\Large\rm xy=\sqrt{x^2+y^2}\]and square each side.

Answer 19

x^2 * y^2 = x^2 + y^2

Answer 20

Ok good.
Do you understand how to get your y's grouped up and isolated?
It's a tricky couple of algebra steps I guess.

Answer 21

Yeah, i'm gonna need some guidance on that part as well.
x^2 * y^2 - y^2 = x^2
no clue what next lol

Answer 22

Each term contains a y^2, let's factor a y^2 out of each :)\[\Large\rm y^2(\qquad -\qquad)=x^2\]What's going to be left in the brackets when we do that?

Answer 23

ofc.
y^2(x^2 -1) = x^2
so
y^2 = (x^2-1)/x^2
y = sqrt( (x^2-1)/x^2))

Answer 24

well +/- the sqrt of that

Answer 25

Oh, i had it backwards
its x^2 over that

Answer 26

Ah yes good :)

The original polar representation, if you graph it, it gives you 4 "branches".
Since we now have a function, we need both the plus and minus root, each gives us 2 "branches".

Answer 27

\[\Large\rm y=\pm\sqrt{\frac{x^2}{x^2-1}}\]Yay good job! \c:/

Answer 28

thanks a lot!!! :)