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Babynini

  • one year ago

Graph and convert this equation to rectangular coordinates polar equation: r=8cos(theta)

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  1. dan815
    • one year ago
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    |dw:1434650063164:dw|

  2. dan815
    • one year ago
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    |dw:1434650153992:dw|

  3. dan815
    • one year ago
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    what do u think an expression for r is? in terms of x and y

  4. Babynini
    • one year ago
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    How did you graph that? o.0

  5. dan815
    • one year ago
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    u click on draw button

  6. Babynini
    • one year ago
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    em, well I tried multiplying both sides by r and got r^2=8rcos(theta) (x^2+y^2)=8x

  7. dan815
    • one year ago
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    yep that is right

  8. Babynini
    • one year ago
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    haha i know that. But i meant how did you come up with that graph.

  9. Babynini
    • one year ago
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    but i think it needs to be simplified further than that. right?

  10. dan815
    • one year ago
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    thats the relationship between polar and cartesian

  11. Babynini
    • one year ago
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    (x^2+y^2)-8x = 0 now what?

  12. dan815
    • one year ago
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    u can put it in circle form

  13. dan815
    • one year ago
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    x^2+y^2=8x rewrite it as this first x^2-8x + y^2=0 ^------------ complete the square and put the constant on the right side then

  14. Babynini
    • one year ago
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    I don't know how to complete squares :/ that is where I got stuck.

  15. dan815
    • one year ago
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    um

  16. dan815
    • one year ago
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    watch youtube video, its faster

  17. dan815
    • one year ago
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    its going to look like this

  18. dan815
    • one year ago
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    (x-4)^2-16+y^2=0 (x-4)^2+y^2=16 (x-4)^2+y^2=4^2 radius 4, shift in x by 4

  19. dan815
    • one year ago
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    |dw:1434650531548:dw|

  20. Babynini
    • one year ago
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    hrm ok. I tried looking for a good youtube video but just got all confused =.=

  21. imqwerty
    • one year ago
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    we have the basic formulas - x = rcos(theta) y = rsin(theta) If x^2 + y^2 = r^2 and r^2 = 8rcos(theta) = 8(x) then you have been given that: x^2 + y^2 = 8x x^2 + 8x + y^2 = 0 Convert to the center-vertex form for circles.

  22. dan815
    • one year ago
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    um completing square therse nothing to it

  23. Babynini
    • one year ago
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    haha sigh, i know it should be super easy. But not for me xD

  24. dan815
    • one year ago
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    Any quadratic is in the form ax^2+bx+c, where a b and c are anything

  25. dan815
    • one year ago
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    okay no lets change this... to something simpler first

  26. Babynini
    • one year ago
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    k

  27. dan815
    • one year ago
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    x^2+bx+c lets say u have an equation of this form

  28. dan815
    • one year ago
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    to put in square form, okay so u need a bracket with the variable + a constant squares

  29. dan815
    • one year ago
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    (x+d)^2 okay when u square this what happens (x+d)(x+d)=x*(x+d) +d*(x+d) = x^2+dx+dx+d^2 = x^2+2dx+d^2 therefore if you want to rewrite x^2+bx+c in (x+d)^2 form then you know that the 2dx has add up to bx 2dx=bx d=b/2 so we see that it has to be in this form (x+b/2)^2 now the constant part will be d^2 or (b/2)^2, this we subtract away as it is in cluded when u square it

  30. dan815
    • one year ago
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    x^2+bx+c ---> becomes (x+b/2)^2-(b/2)^2+c

  31. dan815
    • one year ago
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    in your case we have the equation x^2+bx+c x^2-8x+0 which means b=-8 and c=0 okay so (x+b/2)^2 -(b/2)^2+c (x+(-8/2))^2 - (-8/2)^2 +0

  32. Babynini
    • one year ago
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    k..i'm trying to follow hah

  33. Babynini
    • one year ago
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    and then we just put the y^2 back in there?

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