## anonymous 2 years ago write each polar equation in rectangular form: r=-3sec(Theta) r=costheta+sintheta r=5/3costheta+8sin

1. anonymous

^theta after the last sin

2. myininaya

You need to use these equations: $r^2=x^2+y^2 ; rcos(\theta)=x ; rsin(\theta)=y$

3. anonymous

i did, but like it ended up like x^2+y^2= 25+9x^2r^2-30xr/64y^2 for the last one

4. myininaya

We want the equations just in terms of x and y instead of r and theta.

5. anonymous

i don't know what i did wrong

6. myininaya

so the last one is: $r=\frac{5}{3}\cos(\theta)+8\sin(\theta)$?

7. anonymous

no uh (5)/(3costheta+8sintheta)

8. myininaya

oh so you meant 5/(3costheta)+8sin(theta)) and not 5/3cos(theta)+8sin(theta)

9. anonymous

mhm 5 / (3cos+8sin)

10. myininaya

$r= \frac{5}{3\cos(\theta)+8\sin(\theta)}$ Multiply both sides by that one fraction's denominator

11. myininaya

$r(3\cos(\theta)+8\sin(\theta))=5$ Distribute then substitute the equations I gave you

12. anonymous

3xr+8ry=5 but then you'll need to put r on one side cause you would need to square it. and the answer comes out all weird

13. myininaya

well rcos(theta) is x not just cos(theta same thing with the y deal rsin(theta) is y not just sin(theta)

14. anonymous

whoops how about the secant one

15. myininaya

put it in terms of cos by using an identity

16. myininaya

then put it in some form where you can use the equations I gave you

17. anonymous

sosoososososo it would be x^3+y^2x=-3 AND uh the other one x^2+y^2=x+y?

18. myininaya

for the first one? no...

19. myininaya

I haven't looked at the second one... You are making this way too hard it looks like.

20. myininaya

first one you did this as first step right?: $r=\frac{-3}{\cos(\theta)}$

21. anonymous

yep

22. myininaya

Notice this only has a cos in it... how do you get cos with the r over there so you can use the equation rcos(theta)=x

23. anonymous

multiply both sides with r

24. myininaya

no

25. myininaya

try again

26. myininaya

How do you undo multiplication?

27. myininaya

Or division?

28. anonymous

oh omg

29. myininaya

we have division by cos to undo that you ?

30. anonymous

x=-3

31. myininaya

yes

32. anonymous

thanks

33. myininaya

so for that second one...

34. myininaya

you actually got it right

35. myininaya

but how did you go about it you didn't choose a complicated way right?

36. anonymous

yep uh multiplied both side with r and then substituted the stuff in

37. myininaya

oh good

38. myininaya

do you have any other questions? do you think you got this better?

39. anonymous

uh yea i have a few more

40. anonymous

for standard forms of polar equation uh how would graph r=2cos3theta

41. myininaya

Plug in values of theta and see what the r output is

42. myininaya

then remember how you graph using polar coordinates r is the distance from the origin and direction is given by r to from the angle theta is the angle

43. anonymous

like how would you do this with a graphing calculator

44. myininaya

So for example if I plug in theta=0 $r=2 \cos (3 \cdot 0)=2\cos(0)=2(1)=2 => \text{ graph the point } (2,0^o)$ |dw:1388269865874:dw|

45. myininaya

find 0 degrees and then go to the circle with radius 2

46. anonymous

cause wksht says not to use a table

47. myininaya

|dw:1388269927627:dw| then plug in more points

48. anonymous

oh i know how to graph but not without plugging in points

49. myininaya

I haven't looked at a calculator in ages

50. myininaya

did you try looking up how to input polar equations into whatever calculator type you have

51. myininaya

some calculators require a difference procedure in order to do certain things

52. myininaya
53. myininaya

find mode and choose polar or pol or polargc

54. myininaya

then just type your equation in

55. anonymous

u what's the website

56. anonymous

*uh

57. myininaya

What ?

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