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music101

  • 2 years ago

Convert the polar equation to rectangular form and identify the graph. r= 7/ (2cosΘ+5sinΘ)

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  1. music101
    • 2 years ago
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    |dw:1393722700861:dw|

  2. sourwing
    • 2 years ago
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    y = r sin(theta) x = r cos(theta) x^2 + y^2 = r^2 plug and chug

  3. music101
    • 2 years ago
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    yea so how would i plug it in

  4. music101
    • 2 years ago
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    cuz i did this last week and i kinda forgot sorry

  5. myininaya
    • 2 years ago
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    Try getting your cos and sin with r over there by multiplying both sides by 2cos(theta)+5sin(theta)

  6. music101
    • 2 years ago
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    okay. i did that

  7. myininaya
    • 2 years ago
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    So you have: \[2rcos(\theta)+5rsin(\theta)=7 ?\]

  8. music101
    • 2 years ago
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    why would you put r w/ each one of the terms isnt there just 1?

  9. myininaya
    • 2 years ago
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    \[r=\frac{7}{2\cos(\theta)+5 \sin(\theta)} => r(2\cos(\theta)+5 \sin(\theta))=7\] Distributive property!

  10. music101
    • 2 years ago
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    oh ok i gotchu

  11. myininaya
    • 2 years ago
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    you should have it from here. Use the equations sour gave you.

  12. music101
    • 2 years ago
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    thannk you(:

  13. music101
    • 2 years ago
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    what if the question's like r= -3sinΘ ?

  14. myininaya
    • 2 years ago
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    Try multiplying r on both sides recall rsin(theta)=y and r^2=x^2+y^2

  15. music101
    • 2 years ago
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    hmm ok...

  16. music101
    • 2 years ago
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    so r^2= -3rsinΘ?

  17. myininaya
    • 2 years ago
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    right so now you are able to replace rsin(theta) with y and r^2 with x^2+y^2

  18. music101
    • 2 years ago
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    okay

  19. music101
    • 2 years ago
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    then youd have?

  20. music101
    • 2 years ago
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    the square root of that right?

  21. myininaya
    • 2 years ago
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    just do exactly what I said replace r sin(theta) with y replace r^2 with x^2+y^2

  22. music101
    • 2 years ago
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    i did tht

  23. myininaya
    • 2 years ago
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    you are done what else do you want?

  24. music101
    • 2 years ago
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    is tht my final answer

  25. music101
    • 2 years ago
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    oh ok

  26. myininaya
    • 2 years ago
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    putting an equation in terms of just x and y ( or just x or just y) is called a cartesian equation if that was the directions to just right as a cartesian equation then you are done

  27. myininaya
    • 2 years ago
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    write*

  28. music101
    • 2 years ago
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    right, sorry. i missed a problem on my homework it was r= 2sinΘ-4cosΘ

  29. myininaya
    • 2 years ago
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    multiply both sides by r

  30. music101
    • 2 years ago
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    do you square both sides 1st

  31. music101
    • 2 years ago
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    thts what i emant

  32. myininaya
    • 2 years ago
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    replace r^2 with x^2+y^2 replace r sin(theta) with y replace r cos(theta) with x

  33. music101
    • 2 years ago
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    ok i did that: x^2+y^2= 2y-4x

  34. myininaya
    • 2 years ago
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    yep. :) that is right you wrote the poloar equation as a cartesian equation if that is all the directions said to do then you are done.

  35. music101
    • 2 years ago
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    do you stop there

  36. music101
    • 2 years ago
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    ok

  37. music101
    • 2 years ago
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    aand i missed r= 3+3cosΘ

  38. music101
    • 2 years ago
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    so i multiplied both sides by r 1st

  39. music101
    • 2 years ago
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    and got x^2+y^2=3r +3x but dont i need to get rid of the r?

  40. myininaya
    • 2 years ago
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    you could subtract both sides by 3x giving you \[x^2+y^2-3x=3r \] then square both sides and replace the r^2 with x^2+y^2

  41. music101
    • 2 years ago
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    why square both sides?

  42. myininaya
    • 2 years ago
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    I just prefer my answer without radicals and me squaring both sides will allow me to write my answer without radicals

  43. music101
    • 2 years ago
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    so what would be final answer then

  44. myininaya
    • 2 years ago
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    what if you with what I was saying take my equation and then square both sides and then replace the r^2 with x^2+y^2 you will have your final answer

  45. music101
    • 2 years ago
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    yeah but when you square the whole left side you get x^4+y^4-9x^2 = 3x^2+y^2?

  46. myininaya
    • 2 years ago
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    \[(x^2+y^2-3x)^2=(3r)^2 \] the us what you get when you square both sides your equation is not equivalent to that you squared each term sorta but not really it is whatever you do to one side of equation you can do to the other and the equation still holds

  47. myininaya
    • 2 years ago
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    \[\text {recall law of exponents} (3r)^2=3^2r^2=9r^2=9(x^2+y^2) \]

  48. music101
    • 2 years ago
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    so..

  49. music101
    • 2 years ago
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    im not following are you saying keep left side just squared in parentheses?

  50. myininaya
    • 2 years ago
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    It is easier than multiplying it out, don't you think?

  51. music101
    • 2 years ago
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    yes. so only change right side. like substitue

  52. myininaya
    • 2 years ago
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    You can multiply it out if you want. Just do it correctly, but the if the directions just say write as a rectangular (aka cartesian equation) than you are done. If the left hand side was already in terms of x and y. r^2 was the only thing that needed to be converted.

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