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Homework Help: Parametric Equation Differentiation

  1. Jun 8, 2008 #1
    1. The problem statement, all variables and given/known data
    A curve has parametric equaions:

    x = 2cot t, y = 2sin²t, 0 < t <= pi/2

    Find an expression for dy/dx in terms of the parameter t

    3. The attempt at a solution

    Not sure where to go. Do i need to make a Cartesian equation first?

    Thanks :)
  2. jcsd
  3. Jun 8, 2008 #2


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    Hi thomas49th! :smile:

    Hint: what is dy/dx in terms of dy/dt and dx/dt? :smile:
  4. Jun 8, 2008 #3
    dy/dt = 2cos²t
    dx/dt = -2cosec²t

    is that right?
  5. Jun 8, 2008 #4
    dy/dy is incorrect don't forget chain rule!!!

    dx/dt is right.
  6. Jun 8, 2008 #5
    4cos²t ???

    Is that right now

  7. Jun 8, 2008 #6

    [tex]y=2\sin^2 t[/tex]

    [tex]\frac{dy}{dt}=4\sin t\cdot\frac{d}{dt}\sin t[/tex]
  8. Jun 8, 2008 #7
    [tex]4\sin t\cdot\frac{d}{dt}\sin t[/tex]

    = 2sinxcosx

    now where do i go?

  9. Jun 8, 2008 #8
    Think how to get [tex] \frac{dy}{dx} [/tex] on its own. Basically eliminating the [tex] dt [/tex].
    Any ideas?
  10. Jun 8, 2008 #9
    Your constant is wrong.
  11. Jun 8, 2008 #10


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    Now you answer tiny-tims original question:
    Remember that, even though "dy/dx" is not a fraction, it is defined as the limit of a fraction, and you can always treat it as if it were a fraction.
  12. Jun 8, 2008 #11
    dx/dt = -2cosec²t
    dy/dt = 4sin(t)cos(t)

    now i dont know how to lay it out but i remeber somthing about flipping dy/dx around and everyone going you cant do that in the class. Can you show me?

  13. Jun 8, 2008 #12


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    Hi thomas49th! :smile:

    Draw a diagram …

    x = 2cot t, y = 2sin²t, is a curve.

    Draw any old curve, and pretend it's the right one.

    Now makes some marks on the curve to indicate equal amounts of t.

    Draw a right-angled triangle connecting two of those marks … one side along the curve, and the other two parallel to the x and y axes.

    Then dy/dx is the gradient (slope) of the long side, isn't it?

    And dy/dt is the right-hand side divided by the change in t.

    And dx/dt is the bottom side divided by the change in t.

    So (dy/dt)/(dx/dt) is the right side divided by the bottom side … which is the tangent of the slope of the long side … which is dy/dx … isn't it? :smile:
  14. Jun 8, 2008 #13
    Try dividing one equation by the other, but which one?
  15. Jun 8, 2008 #14
    (dy/dt) / (dx/dt)
    = dy/dt * dt/dx
    = dy/dx

    yes yes. so now where. That was fun!

  16. Jun 8, 2008 #15


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    Maths is fun sometimes! :biggrin:
    So now apply your fun formula to:
  17. Jun 8, 2008 #16
    ahhh i see

    [tex]\frac{dy}{dx} = \frac{4\sin t \cos t}{-2cosec^{2}t}[/tex]

    which can be simplified to

    [tex]\frac{dy}{dx} = \frac{-2\cos t}{\sin t}[/tex]

    [tex]= -2 \cot t[/tex]

    Is that right?
  18. Jun 8, 2008 #17
    No problem.
  19. Jun 8, 2008 #18


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    erm … 1/cosec²t = sin²t. :rolleyes:

    So dy/dx = … ? :smile:
  20. Jun 8, 2008 #19
    o yea! I mean -2sin³t cost


    Ok. Now for the next bit i need to find the tangent where t = pi/4

    For this should i get all trig terms the same using trig identities i know or sub t= pi/4 in straight away.

    Thanks :)
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