## 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....

1. anonymous

Cartesian? Check me please? x^6+3 x^4 (y^2-3)+3 x^2 y^2 (y^2+2)+y^6 = y^4 |dw:1342133559037:dw|

2. anonymous

It looks like of like a D-orbital from chemistry, but it's got little lobes on the vertical, so that means it's probably more like some weird F-orbital. :-D Help?

3. anonymous

$x^6+3 x^4 (y^2-3)+3 x^2 y^2 (y^2+2)+y^6 = y^4$

4. anonymous

I just chose some arbitrary "a" constant to sub.

5. anonymous

|dw:1342133866479:dw|

6. anonymous

Is 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

7. anonymous

|dw:1342133990522:dw|

8. anonymous

|dw:1342134107025:dw|

9. anonymous

|dw:1342134151837:dw|

10. anonymous

$$x^2 + y^2 = \text{radius}^2$$

11. anonymous

|dw:1342134219662:dw|

12. anonymous

so a=1

13. asnaseer

you 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 :)

14. anonymous

$$\text{Area} = 3 \pi a^2$$, correct ?

15. anonymous

Whoa cool link there, @asnaseer I didn't know you could do that with Wolfram, that's some pretty fancy syntax :D

16. anonymous

Does this graph actually have a name though? It's not as if I can just Google the image lol

17. anonymous

|dw:1342134548532:dw|

18. anonymous

|dw:1342134699157:dw|

19. anonymous

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20. anonymous

|dw:1342134931824:dw|

21. anonymous

22. anonymous

origin build with 1+cos2t=0 _> x=0 , y=0 common term between x , y

23. anonymous

|dw:1342135079515:dw|

24. anonymous

|dw:1342135156538:dw|

25. anonymous

is it clear?

26. anonymous

for exactly drawing d/dt=0 for x , y give you some critical points

27. anonymous

Clear, you are finding the polar lines, basically, where r = 0.

28. anonymous

|dw:1342135317896:dw|

29. anonymous

We 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.

30. anonymous

Yep! That looks like the other questions where we were asked to find the vertical and horizontal tangents

31. anonymous

dy/dt = 0 is for horizontal I believe

32. anonymous

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33. anonymous

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34. anonymous

interesting problem !

35. anonymous

Does this funky thing have a name though? :-D

36. anonymous

No! it has a nice shape .

37. anonymous

It 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.

38. anonymous

It kind of looks like... one of those things when you roll a circle within a circle

39. anonymous

Try to solve one by yourself then think it's so easy and sure after that enjoy with the figure.

40. anonymous

Well I'm not going to sweat it, the hard part is done. Who cares about the name, you're right :-D

41. anonymous

|dw:1342136059673:dw|

42. anonymous

first thing attracted me to math was figuring out the shape of equation.

43. anonymous

dx = 3 sin(3t) dt dy = 2 cos(t)-3 cos(3t) dt

44. anonymous

Are you solving mine?

45. anonymous

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