Question

Help please! :)
Transform each polar equation in rectangular coordinates and identify its shape.
1. theta= 1.34 radians

2. r=tan(theta)sec(theta)

Answer 1

I do know how to convert polar to rectangular but I dont know how to with those questions

Answer 2

We just gotta be able to remember all the conversions we have xD I'm sure you have them somewhere. Either way, for the first problem we need to use this one:
\[\tan \theta = \frac{ y }{ x }\]

Answer 3

With your problem, we can actually take the tangent of both sides:
\[\theta = 1.34\]
\[\tan \theta = \tan(1.34)\]
Now use the conversion I just posted :P

Answer 4

I got 4.25, so how do I apply that to the conversion?

Answer 5

Right. Well, just look up. You can perform a direct substution with the conversion I posted. Like literally replace :3

Answer 6

I'm sorry, I still don't understand

Answer 7

So right now we have:
\[\tan \theta = 4.25\]
But one of our conversions, the one I posted replaces the tan(theta) part.
\[\tan \theta = \frac{ y }{ x }\]

Answer 8

Kinda see how you can replace it now?

Answer 9

okay so instead of y/x, the answer would just be tan(theta)=4.25, is that what you mean?

Answer 10

Well I mean replace tan(theta) with y/x
\[\frac{ y }{ x } = 4.25\]
\[y = 4.25x \]
See what I did?

Answer 11

oh! So that would be my answer?

Answer 12

Yep xD

Answer 13

thank you! Okay so how would I figure out the next one? :)

Answer 14

The next one requires us to TRY to get the equation into three conversions:
\[x = rcos \theta \]
\[y = rsin \theta\]
\[x ^{2}+y ^{2}= r ^{2} \]
So we want to try and see how we can get those 3 or some combination of them. So the first thing I'd advise is turn tan and sec into sines and cosines.

Answer 15

wow that seems really complicated! So should I turn tan into sin(theta)/cos(theta) and sec into 1/cos(theta)?

Answer 16

Right. So you would have this then:
\[r = \frac{ \sin \theta }{ \cos ^{2}\theta }\]
So now we need to get r paired together with those sines and cosines somehow. SO what do you think iw ould do next?

Answer 17

my best guess is to divide out 1/cos(theta) and put it with the r

Answer 18

Yeah, I'd just multiply both sides by cos^2 xD This gives us:
\[rcos ^{2}\theta = rsin \theta\]
So we want r to be paired with cos and sin. Problem is we have cos^2, but not r^2 and the other side is missing an r as well. Now you may not know, but think you have an idea how we get 2 r's for cosine and an r for sin?

Answer 19

I have no idea!

Answer 20

Thought I'd see xD Multiply both sides by r.

Answer 21

If I multiply an r in to both sides I get:
\[r ^{2}\cos ^{2}\theta = rsin \theta \]Now that we can use. So using the conversions above and what we have now, think you can turn it into the x's and y's we need?

Answer 22

\[x=r ^{2}\cos ^{2}\theta \] and \[y=rsin \theta \]

Answer 23

Well x = rcos(theta), but you have r^2cos^2. So it'd actually be x^2 = y. You see why?

Answer 24

because they are equal to eachother

Answer 25

Yeah, you just need to realize it was squared, so it becomes x^2.
\[r ^{2}\cos ^{2}\theta = (rcos \theta)^{2} = x ^{2} \]

Answer 26

thank you! Also do you what shape it would be if it were graphed? I know the first one

Answer 27

It's actually a parabola. It's just y = x^2 xD