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Polar Coordinates Tangent line

  1. Sep 6, 2013 #1
    1. The problem statement, all variables and given/known data
    I don't know how to make theta so
    ∅ = theta.
    find the slope of the tangent line at
    r = sin(6∅) when ∅ = pi/12



    2. Relevant equations
    y=rsin(6∅)
    x=rcos(6∅)
    r=sin(6∅)
    tangent line equation
    y-y' = m(x-x')
    m = dy/dx

    3. The attempt at a solution
    when ∅ = pi/12 then sin(6∅) = 1
    so
    r=1
    x=rcos(6∅) = 0
    y = rsin(6∅) = 1

    so If I did the previous parts correctly, then I'm now attempting to find the slope (m).
    I do not know how to find it.

    m = dy/dx =EITHER= (dy/dr) / (dx/dr) or (dy/d∅) / (dx/d∅). Either way I wouldn't know how to take the derivatives if I did know which one to choose.

    Am I doing this correctly? Am I on the right track? Thanks :)
     
  2. jcsd
  3. Sep 6, 2013 #2

    CAF123

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    You do not need calculus for this question. As you observed, when ∅=pi/12, sin(6∅)=1. Now recall the sin graph and this should give you the answer.

    Alternatively, you know r=r(∅) and you want r'(∅=pi/12).

    theta should also be in quick symbols - look for θ.
     
  4. Sep 6, 2013 #3
    so when r=sin6θ=1 that is when the sin function is at a max. therefore the slope is zero?
     
  5. Sep 6, 2013 #4

    CAF123

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    Precisely.
     
  6. Sep 6, 2013 #5
    why when I type zero into my online homework for the slope of the tangent curve if says i'm wrong ;-(
     
  7. Sep 7, 2013 #6

    CAF123

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    Not sure, try '0' instead of typing zero. Have you tried it again using calculus?

    If you parametrize the curve r, writing expressions for x and y, then your method in the OP is also good.
    ##(dy/dx) = (dy/d\theta) (d\theta/dx)##, but it is probably simpler to just find r'.
     
    Last edited: Sep 7, 2013
  8. Sep 7, 2013 #7
    it's not accepting 0 as an answer either ;-( booo
     
  9. Sep 7, 2013 #8

    ehild

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    r is maximum as function of theta, which does not mean that the slope of the curve is zero. See picture.

    The equation of the curve in polar coordinates is r=sin(6θ). How would you write the Cartesian coordinates x,y as functions of θ?

    ehild
     

    Attached Files:

    Last edited: Sep 7, 2013
  10. Sep 7, 2013 #9

    CAF123

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    Wow, somehow I completely ignored that ∅ was the polar angle.
    What I suggested in my opening post is wrong because you are not plotting r versus ∅. r and ∅ are polar representation.

    Likewise, evaluating r'(∅=pi/15) here is incorrect because we are not plotting r versus theta.

    Do as I said in my last post and parametrise the curve to get eqns for x and y. The find dy/dx.

    (Thanks ehild for letting me know of my mistakes)
     
  11. Sep 9, 2013 #10
    slope = dy/dx = (dy/dθ) / (dx/dθ)
    I went in and saw my teacher at this part but I lost the sheet she wrote on because I'm a noob ;-(. But it went something like dy/dr(some sin's and cosines) / dx/dr(some sin's and cosines)
    but I have no idea what she did now or how she did it or why it made sense at all. Any help appreciated.


    y = rsinθ
    x = rcosθ
     
  12. Sep 9, 2013 #11

    ehild

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    It is given that r=sin(6θ). So the Cartesian coordinates are x=sin(6θ)cos(θ) and y=sin(6θ)sin(θ).
    Derive both of them with respect to θ. The slope is dy/dx=(dy/dθ)/(dx/dθ).

    ehild
     
  13. Sep 9, 2013 #12

    you are so beautiful!! I love this website so much. I'll learn LaTex so soon!! 18 credits (4 physics classes + calculus three) is quite demanding and you have no idea how grateful I am I discovered this website and all of you!! You're all so patient with me it's amazing.
     
  14. Sep 10, 2013 #13
    dy/dθ = 6sinθcos(6θ) + sin(6θ)cos(θ)
    dx/dθ = 1/2(5cos(5θ) + 7cos(7θ)

    (dy/dθ)/(dx/dθ) = (6sinθcos(6θ) + sin(6θ)cos(θ)) / (1/2(5cos(5θ) + 7cos(7θ))
    when θ = ∏/12...
    (6sin(∏/12)cos(6∏/12) + sin(6∏/12)cos(∏/12)) / (1/2(5cos(5∏/12) + 7cos(7∏/12))
    (0 + .9659258) / (.6470476128 + -1.8117333)
    = -.8293446126
    Which my online thing says is wrong ;-(
     
  15. Sep 10, 2013 #14

    ehild

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    Are you sure you did it correctly? Try again. Leave it in terms in of 6θ and θ. As 6θ=pi/2, both derivatives become very simple at θ=pi/12.

    ehild
     
    Last edited: Sep 10, 2013
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