Question

INSTANT FAN//MEDAL
What is distance around the graph of r = 12 sin θ ?

Answer 1

It appears that you simply deleted your previous post of this question, along with the info and questions I added. Why would you do that? Certainly this does not motivate me to spend more time with you, nor would it motivate anyone else to do so.

Answer 2

@mathmale I didn't delete it, I closed it, because I couldn't bump it again for 23:00 and I had no information regarding finding distance around a graph anywhere on the internet.

Answer 3

@czesc Don't let what @mathmale says get to you. He is supposed to be a moderator, but he is very rude

Answer 4

I vote: rewirte in (x,y) coordinates, see it's a circle, and use the perimeter formula.

Answer 5

the rewrite isn't to bad: multi by r to get ---> \[r^2=12r\sin(\theta)\] and recall that \[x^2+y^2=r^2 \text{ and } y=\sin(\theta)\]

Answer 6

then complete the square to get standard form \[(x-x_0)+(y-y_0)=r^2\]

Answer 7

Thanks for the compliment, @Samslays.

@czesc: It may help you to define what you mean by "distance around the graph." This is decidedly non-standard mathematical language. Perhaps you mean "arc length?" or even "perimeter?"

Answer 8

arc length is my bet.

Answer 9

Also, I'd suggest you actually graph r= 12 sin theta. If, as another user seems to claim, this is a circle, then rewriting the given equation in rectangular coordinate form may make this problem seem quite easy.

Answer 10

@mathmale thank you for your response, I appreciate it as always, regarding distance around the graph, i'm just as confused as you are as I did not write the question.

Answer 11

Yes indeed it is a circle resting with it's bottom on the x-axis.

Answer 12

Then arc length is easy! \[2\pi r\] if you want \[0<\theta\le2\pi\]

Answer 13

@Redcan @mathmale
the options for response are this, but when I try to find arc length of polar curve r=12*sin(t) from t=0 to t=2pi i get 24 pi


and the options are
6π
144π
36π
12π

Answer 14

r= 6

Answer 15

\[2*6*\pi\]

Answer 16

\[x^2+(y-6)^2=6^2\] ----> circle of radius 6