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anonymous
 4 years ago
Yikes, need to find the area and the name for these parametric equations:
x(t) = a cos(t) (2 cos(2 t)+1)
y(t) = a sin(t) (2 cos(2 t)+1)
What is...? I don't even...? How? I'll draw the graph....
anonymous
 4 years ago
Yikes, need to find the area and the name for these parametric equations: x(t) = a cos(t) (2 cos(2 t)+1) y(t) = a sin(t) (2 cos(2 t)+1) What is...? I don't even...? How? I'll draw the graph....

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Cartesian? Check me please? x^6+3 x^4 (y^23)+3 x^2 y^2 (y^2+2)+y^6 = y^4 dw:1342133559037:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It looks like of like a Dorbital from chemistry, but it's got little lobes on the vertical, so that means it's probably more like some weird Forbital. :D Help?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[x^6+3 x^4 (y^23)+3 x^2 y^2 (y^2+2)+y^6 = y^4\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I just chose some arbitrary "a" constant to sub.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342133866479:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is that \(\tan t\) @mahmit2012 ? Ty in advance btw, you always seem to be the one who's here to help when I get stuck on these challenge problems :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342133990522:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342134107025:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342134151837:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(x^2 + y^2 = \text{radius}^2\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342134219662:dw

asnaseer
 4 years ago
Best ResponseYou've already chosen the best response.0you can see a graph of this using wolfram, e.g.: http://www.wolframalpha.com/input/?i=plot+%7Bx%3Da*cos%28t%29%282cos%282t%29%2B1%29%2C+y%3Da*sin%28t%29%282cos%282t%29%2B1%29%7D+when+a%3D3 your graph is correct :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(\text{Area} = 3 \pi a^2\), correct ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Whoa cool link there, @asnaseer I didn't know you could do that with Wolfram, that's some pretty fancy syntax :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Does this graph actually have a name though? It's not as if I can just Google the image lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342134548532:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342134699157:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342134862828:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342134931824:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0*reading & taking notes* :3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0origin build with 1+cos2t=0 _> x=0 , y=0 common term between x , y

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342135079515:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342135156538:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0for exactly drawing d/dt=0 for x , y give you some critical points

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Clear, you are finding the polar lines, basically, where r = 0.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342135317896:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We just touched on polar stuff today, this question was from the "challenge" section of the written homework. The other curves have names that we discussed in the lecture like "carotid", "ellipse", and "circle" which are kind of obvious which is which.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yep! That looks like the other questions where we were asked to find the vertical and horizontal tangents

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dy/dt = 0 is for horizontal I believe

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342135476935:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342135505524:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0interesting problem !

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Does this funky thing have a name though? :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No! it has a nice shape .

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It just asks me to type in the "area" in the one text field. Which was correct yay! And then it has a text box under it for "name of curve". The others were easy, but this one I haven't got a clue. Maybe it's in the text, somewhere amidst all its pages but it's not easy to find.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It kind of looks like... one of those things when you roll a circle within a circle

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Try to solve one by yourself then think it's so easy and sure after that enjoy with the figure.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well I'm not going to sweat it, the hard part is done. Who cares about the name, you're right :D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1342136059673:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0first thing attracted me to math was figuring out the shape of equation.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dx = 3 sin(3t) dt dy = 2 cos(t)3 cos(3t) dt

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Are you solving mine?