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anonymous

  • 2 years ago

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

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  1. anonymous
    • 2 years ago
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    ^theta after the last sin

  2. myininaya
    • 2 years ago
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    You need to use these equations: \[r^2=x^2+y^2 ; rcos(\theta)=x ; rsin(\theta)=y \]

  3. anonymous
    • 2 years ago
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    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
    • 2 years ago
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    We want the equations just in terms of x and y instead of r and theta.

  5. anonymous
    • 2 years ago
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    i don't know what i did wrong

  6. myininaya
    • 2 years ago
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    so the last one is: \[r=\frac{5}{3}\cos(\theta)+8\sin(\theta) \]?

  7. anonymous
    • 2 years ago
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    no uh (5)/(3costheta+8sintheta)

  8. myininaya
    • 2 years ago
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    oh so you meant 5/(3costheta)+8sin(theta)) and not 5/3cos(theta)+8sin(theta)

  9. anonymous
    • 2 years ago
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    mhm 5 / (3cos+8sin)

  10. myininaya
    • 2 years ago
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    \[r= \frac{5}{3\cos(\theta)+8\sin(\theta)}\] Multiply both sides by that one fraction's denominator

  11. myininaya
    • 2 years ago
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    \[r(3\cos(\theta)+8\sin(\theta))=5 \] Distribute then substitute the equations I gave you

  12. anonymous
    • 2 years ago
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    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
    • 2 years ago
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    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
    • 2 years ago
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    whoops how about the secant one

  15. myininaya
    • 2 years ago
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    put it in terms of cos by using an identity

  16. myininaya
    • 2 years ago
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    then put it in some form where you can use the equations I gave you

  17. anonymous
    • 2 years ago
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    sosoososososo it would be x^3+y^2x=-3 AND uh the other one x^2+y^2=x+y?

  18. myininaya
    • 2 years ago
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    for the first one? no...

  19. myininaya
    • 2 years ago
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    I haven't looked at the second one... You are making this way too hard it looks like.

  20. myininaya
    • 2 years ago
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    first one you did this as first step right?: \[r=\frac{-3}{\cos(\theta)}\]

  21. anonymous
    • 2 years ago
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    yep

  22. myininaya
    • 2 years ago
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    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
    • 2 years ago
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    multiply both sides with r

  24. myininaya
    • 2 years ago
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    no

  25. myininaya
    • 2 years ago
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    try again

  26. myininaya
    • 2 years ago
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    How do you undo multiplication?

  27. myininaya
    • 2 years ago
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    Or division?

  28. anonymous
    • 2 years ago
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    oh omg

  29. myininaya
    • 2 years ago
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    we have division by cos to undo that you ?

  30. anonymous
    • 2 years ago
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    x=-3

  31. myininaya
    • 2 years ago
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    yes

  32. anonymous
    • 2 years ago
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    thanks

  33. myininaya
    • 2 years ago
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    so for that second one...

  34. myininaya
    • 2 years ago
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    you actually got it right

  35. myininaya
    • 2 years ago
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    but how did you go about it you didn't choose a complicated way right?

  36. anonymous
    • 2 years ago
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    yep uh multiplied both side with r and then substituted the stuff in

  37. myininaya
    • 2 years ago
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    oh good

  38. myininaya
    • 2 years ago
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    do you have any other questions? do you think you got this better?

  39. anonymous
    • 2 years ago
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    uh yea i have a few more

  40. anonymous
    • 2 years ago
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    for standard forms of polar equation uh how would graph r=2cos3theta

  41. myininaya
    • 2 years ago
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    Plug in values of theta and see what the r output is

  42. myininaya
    • 2 years ago
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    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
    • 2 years ago
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    like how would you do this with a graphing calculator

  44. myininaya
    • 2 years ago
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    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
    • 2 years ago
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    find 0 degrees and then go to the circle with radius 2

  46. anonymous
    • 2 years ago
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    cause wksht says not to use a table

  47. myininaya
    • 2 years ago
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    |dw:1388269927627:dw| then plug in more points

  48. anonymous
    • 2 years ago
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    oh i know how to graph but not without plugging in points

  49. myininaya
    • 2 years ago
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    I haven't looked at a calculator in ages

  50. myininaya
    • 2 years ago
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    did you try looking up how to input polar equations into whatever calculator type you have

  51. myininaya
    • 2 years ago
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    some calculators require a difference procedure in order to do certain things

  52. myininaya
    • 2 years ago
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    http://mathbits.com/MathBits/TISection/PreCalculus/polargraphs.htm

  53. myininaya
    • 2 years ago
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    find mode and choose polar or pol or polargc

  54. myininaya
    • 2 years ago
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    then just type your equation in

  55. anonymous
    • 2 years ago
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    u what's the website

  56. anonymous
    • 2 years ago
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    *uh

  57. myininaya
    • 2 years ago
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    What ?

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