anonymous
  • anonymous
find a polar representation for the curve: x^2 + y^2 = 9
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

myininaya
  • myininaya
do you remember that \[r^2=x^2+y^2\]
anonymous
  • anonymous
yeah so r = 3
myininaya
  • myininaya
satellite I having a moment here doesn't r=3 include the equation r=-3?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

myininaya
  • myininaya
you know when we are talking about polar equations?
myininaya
  • myininaya
i would include r=-3 just in case because i'm having a memory issue right now
anonymous
  • anonymous
i think it is just \[r=9\]
myininaya
  • myininaya
\[r^2=9 => r=\pm 3\]
anonymous
  • anonymous
r is the radius, always non negative. you want \[r=f(\theta) but here r is constant
anonymous
  • anonymous
\[r=f(\theta)\]
myininaya
  • myininaya
you only need r=3
anonymous
  • anonymous
right i lunched it is \[r=3\]
anonymous
  • anonymous
\[x=rcos \theta\] \[y=rsin \theta\] r=3\[r ^{2}\cos ^{2}\theta+r ^{2}\sin ^{2}\theta=9\]
anonymous
  • anonymous
yeah i got that far
anonymous
  • anonymous
yeah but this says \[r=3\]
anonymous
  • anonymous
how do i simplify that?
anonymous
  • anonymous
too much work. r is the radius. it is a constant since you have a circle of radius 3
myininaya
  • myininaya
factor out r^2
anonymous
  • anonymous
did that
myininaya
  • myininaya
cos^2(theta)+sin^2(theta)=1
anonymous
  • anonymous
\[\cos ^{2}\theta+\sin ^{2}\theta=1\]
anonymous
  • anonymous
okay
myininaya
  • myininaya
you don't have to do it that way the easiest is just recalling \[r^2=x^2+y^2\]
anonymous
  • anonymous
you shouldn't think that hard! it is true that \[\cos ^{2}\theta+\sin ^{2}\theta=1\] but that is way too much work
anonymous
  • anonymous
so it's just r = 3 as my answeR?
myininaya
  • myininaya
yes
anonymous
  • anonymous
yes a circle looks like \[r= number\]
myininaya
  • myininaya
r=3 will include all points from the center that have distance 3 from it
anonymous
  • anonymous
in polar coordinates a circle is just r = a number?
anonymous
  • anonymous
yes that is correct
anonymous
  • anonymous
r after all stands for "radius" and circle is a figure where the radius is constant
anonymous
  • anonymous
okay thank you!
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=r+%3D+3+
anonymous
  • anonymous
here is one where r is not constant http://www.wolframalpha.com/input/?i=r+%3D+1%2Bsin%28theta%29

Looking for something else?

Not the answer you are looking for? Search for more explanations.