Question

Convert the parametric equation into polar form:

Answer 1

\[x = \frac{ 3t }{ 1+t^{3} }\]
\[y = \frac{ 3t^{2} }{ 1+t^{3} }\]

Answer 2

Do you know the formula for converting from parametric to polar and the other way around?

Answer 3

Wasn't even provided a formula. I'm used to eliminating the parameter normally or doing simple rectangular equations to polar, but this is kinda smack in your face brand new, so not even sure where to start.

Answer 4

@Loser66 Feel free to jump in bud.

Answer 5

Probably should of asked this earlier, but what class is this for?

Answer 6

Well, this is my own kind of personal review and skimming through stuff, but this is end calc 2 intro calc 3.

Answer 7

@Jonask can probably help you better then, I am not into calculus 3 or even 2 yet.

Answer 8

You were thinking regular rectangular to polar and vice versa @Luigi0210 ? :P

Answer 9

Yea, exactly xD

Answer 10

\[x=\frac{3t}{1+t^3},y=\frac{3t^2}{1+t^3}\]
\[y=t\frac{3t}{1+t^3}=tx\]
so \[t=\frac{y}{x}\]
then
\[x=\frac{3t}{1+t^3}=\frac{\frac{y}{x}}{1+(\frac{y}{x})^3}=x\]
\[\frac{yx^3}{(x^3+y^3)x}=x\\x^2(x^3+y^3)=yx^3\\x^3+y^3=yx\]

Answer 11

Forgot a 3 in there, but yeah, I see it. I guess it's just being more clever and better at recognition than what I am, lol.

Answer 12

i think we can use polar coordintes wen we eliminated t @Loser66

Answer 13

the 3 doesnt have to be there notice that its part of x @Concentrationalizing

Answer 14

Is it? O.o It just looks like it disappeared once you replaced t with y/x

Answer 15

@Loser66 has a good substitution,also you ight jus substitute in the last equation to get
\[\cos^3\theta+\sin^3\theta=\cos\theta\sin\theta\]

Answer 16

no its not about clever,we are jus trying to eliminate t by making the two equations agree,the long way will be solving for t then substituting but recognising the combination is better in this case

Answer 17

@Loser66 it makes lots of sense

Answer 18

Exactly, recognizing it as a combination. I justimmediately see that it'd be hardtotake one of the parameters and solve for t. And then some recent problems had it where you would go from polar to parametric, so I was trying to think of that but backwards, except that didn't work. So....I suppose recent problems and what I'm used to just limited my thinking. So yes, I still say recognition, lol.

Answer 19

Yeah, and I never was able to figure out my last problem with the series solution @Loser66 haha

Answer 20

Well, from where you had it before was just fine. Although I still see a 3 needed in there, lol.

\[x^{3} + y^{3} = 3xy\]
\[r^{3}\cos^{3}\theta + r^{3}\sin^{3}\theta = 3rcos \theta r \sin \theta\]
\[r*r^{2}(\cos^{3}\theta + \sin^{3} \theta) = 3r^{2}\cos \theta \sin \theta\]
\[r = \frac{ 3\cos \theta \sin \theta }{ \cos^{3} \theta + \sin^{3} \theta}\]

dun dun dun

Answer 21

yes you are on the spot but still i dont c the 3 why it has to be here haha

Answer 22

\[r=\frac{\sin \theta \cos\theta}{\sin^3 \theta +\cos^3\theta}\]

Answer 23

\[x = \frac{ 3t }{ 1+t^{3}} \]
\[t = y/x\]
\[x = \frac{ \frac{ 3y }{ x } }{ 1+(\frac{ y }{ x })^{3} }\]
3 still would be there?

Answer 24

i see it...thank you,you r right

Answer 25

Alrighty ^_^ Lol, and yeah, luigi didnt realize it was cal2 either @Loser66 lol. But yeah, thanks everyone : )

Answer 26

Im all over the place with the problems I post. But most of the time expect cal 2, 3, diff eq, linear algebra or something xD

Answer 27

Im pretty much done with all my calc2 review after one section, then Im jumping back to finish off the series solutions to DEs and then calc 3 we go.

Answer 28

The first DE course I just did was easy. But when I did calc 2, we never quite touched on this last section with parametrics and polar and such. A lot of it is trig review, so np, but some of this is new. So I'm basically finishing off what I shouldve done before and the last chapter in DE we didn't finish. Once I'm done with that it's full-board calc 3 and linear algebra.

Answer 29

lool i jus started DE n calc 2...u can crawl easier wen u watch ocw lectures,the material is great

Answer 30

I'm a math major, but I also do language study as a hobby and personal interest.
Oh, special circumstances. Normally you can't take it before multivariable, but the reason I can is it's more of a education war between the community college here and the main university, at least it seems that way. The community college offers some of the higher level courses as a way to kinda steal from the university I suppose. If I took the DE course at uni it'd be above linear, multivariable, all of that.

Answer 31

Of course taking it at the community college BEFORE multivariable and linear has its minor challenges, but the calc 3 and linear portions of the course I hadto do were np for me to catch up on.

Answer 32

probabaly jus different naming...cos we do have enough tools...so its probably same calc....my calc 2 myt be ur 3,bcos i have calc 1 then calc 2,then adv calc 1 then adv calc 2....wats @Loser66 Major

Answer 33

Yeah, I know some places even have a calc 4. If I follow the university math path, its calc 1, calc 2, discrete, calc 3, linear algebra, ODE

Answer 34

Not really, but we had to have calc 2 first. Was a horrible class. So poorly organized, I got a B in it and felt like I learned nothing.

Answer 35

Well, I was talkign about discrete, lol.

Answer 36

B in discrete because it just was a stupid class.

Answer 37

It is, but the class I was in was stupid. Poorly taught course.

Answer 38

Bingo. She was so out of it, unorganized, sped through lectures, explained things poorly, it was a waste. I just figure itll be better for discrete 2.

Answer 39

Eh, I just was disappointed with it is all. Now just waiting for the semester to start up again. Oh, well I go back and forth between the community college and the university here. Community college is College of Southern Nevada, uni is UNLV

Answer 40

But I need to head out, been keeping family waiting, lol. Thanks again and late Merry Christmas.