Question

The letters x and y represent rectangular coordinates. Write the following equation using polar coordinates (r, θ).
x^2 + y^2 - 4x = 0

Answer 1

is it rsin^2 tangent = 4cos tangent ?

Answer 2

@theEric

Answer 3

I don't think so. I won't say I'm 100% sure, because I make mistakes. But let's work through it together.

Do you know the first step in converting these?

Answer 4

welll
x= rcos θ
y = rsin θ

Answer 5

Right! That's a big part. And so what we do then is substitution.
So, check my math, but I'll do just the substitution and get...
\(\left(r\cos\theta\right)^2+\left(r\sin\theta\right)^2-4\left(r\cos\theta\right)=0\)
How does that look?

Answer 6

yes thats right

Answer 7

I got like one step ahead of you but idk if i did it rihgt :/

Answer 8

Okay! Well I'm going to distribute that power as my next step. If we got different equations, we should look at them!

I'll type mine out now.

Answer 9

I also took away some parenthesis, where I put it in blue.

\(r^2\cos^2\theta+r^2\sin^2\theta-4\color{blue}{r\cos\theta}=0\)

So, how do those changes look to you?

Answer 10

yeah thats what i got too :))

Answer 11

Cool!

Answer 12

And that is as far as i got as well haha

Answer 13

Okay! You asked
"is it rsin^2 tangent = 4cos tangent ?"
earlier. You didn't try a next step?

Answer 14

Thats one of the answers and it looked the closest so i just thought it might be that one haha

Answer 15

That's better than guessing randomly, I suppose!

So, the next step! Of all the annoying (difficult to remember) trigonometric equations, there are two I frequently find useful

1. \(\quad\tan\theta=\dfrac{\sin\theta}{\cos\theta}\)
and
2. \(\quad\cos^2\theta+\sin^2\theta=1\)

We'll use that second one. Usually the \(\cos^2\) and \(\sin^2\) share a common facter that you can distribute out.

Is that the case here?

Answer 16

Kind of our equation equals 0 tho?

Answer 17

The expressions in our equation equal \(0\), yep!
"x^2 + y^2 - 4x = 0"
and
\(r^2\cos^2\theta+r^2\sin^2\theta-4r\cos\theta=0\)

Answer 18

oh okay so whats the next step??

Answer 19

But we want to simplify what we have.
Look at this hint:
\(\color{red}{r^2}\color{blue}{\cos^2\theta}+\color{red}{r^2}\color{blue}{\sin^2\theta}-4r\cos\theta=0\)
When we see \(\cos^2\theta+\sin^2\theta\) we always want to see if we can isolate it. That's because it will turn into \(1\), which is a lot easier on the eyes! :)

Answer 20

Does this r^2cos^2θ+r^2sin^2 cancel out because of cos^2θ+sin^2θ=1?

Answer 21

OK so it become 1 and then what happens to the r^2?

Answer 22

Well, it can't just disappear on its own. You rearrange the equation so that \(\cos^2\theta+\sin^2\theta\) is by itself, and then you'll see how the magic works.

So,
\(\color{red}{r^2}\color{blue}{\cos^2\theta}+\color{red}{r^2}\color{blue}{\sin^2\theta}-4r\cos\theta=0\)
\(\color{red}{r^2}\left(\color{blue}{\cos^2\theta}+\color{blue}{\sin^2\theta}\right)-4r\cos\theta=0\)
Right?

Answer 23

Then can you see what happens next?

Answer 24

ya

Answer 25

r^2- 4rcosθ=0

Answer 26

Right! And then it's up to you whether you want to pull an \(r\) out to change it more or not! Both forms are useful for different things, so I'd say you're done!

Answer 27

I guess just look at the choices and see if it matches now. If not, simplify it one step further, and then check.

Answer 28

r = 4cos θ is an answer is that one right??

Answer 29

What do you think?

Answer 30

yes

Answer 31

Yep! Looks good! You add \(4r\cos\theta\) to both sides, and then divide by \(r\). You can divide by \(r\) as long as \(r\neq0\). And if \(r=0\), then you can see that the equation is useless anyway.

Answer 32

Awesomeee!! :) thanks :D

Answer 33

You're welcome! Good job! :)

Answer 34

Thx :)

Answer 35

Np!