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    How do you differentiate this, leaving it in terms of the parameter t ?

    x= 4sin3t y=3cos3t
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    (Original post by kirstieluu)
    How do you differentiate this, leaving it in terms of the parameter t ?

    x= 4sin3t y=3cos3t
    Are you trying to find \frac{dy}{dx}? Use the chain rule:

    \displaystyle \frac{dy}{dx} = \frac{dy}{dt} \times \frac{dt}{dx}
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    (Original post by tjf8)
    dy/dx = dy/dt / dx/dt
    dy/dx = d(3cos3t)/dt / d(4sin3t)/dt
    dy/dx = -9sin3t/12cos3t
    dy/dx = -3/4 tan3t

    I think...
    Posting full solutions is against the rules of this forum. Please try to guide towards an answer instead.
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    differentiation both
    dy/dt and dx/dt

    then write one over the other e.g. dy/dt over dx/dt (this is the same as dy/dx because the dt can cancel down)

    write your solution on here and ill let you know if I agree (because we're not allowed to post solutions, only help you and check you've got it right/understand how to do it)
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    (Original post by notnek)
    Posting full solutions is against the rules of this forum. Please try to guide towards an answer instead.
    I was unaware of that – thank you.
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    Sorry, i wrote completely the wrong question.
    I was meant to ask .. x=sect y=tant

    Would you still use the chain rule on this? I got up to.. ?dy/dx= sec^t/sect tant

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    (Original post by kirstieluu)
    Sorry, i wrote completely the wrong question.
    I was meant to ask .. x=sect y=tant

    Would you still use the chain rule on this? I got up to.. ?dy/dx= sec^t/sect tant

    Yes the method is exactly the same and your working is correct.

    You should be able to simplify from there.
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    (Original post by notnek)
    Yes the method is exactly the same and your working is correct.

    You should be able to simplify from there.
    When i simplified it.. i got sec/tant
    But the answer has got cosec in it and i dont know where that comes from?
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    (Original post by kirstieluu)
    When i simplified it.. i got sec/tant
    But the answer has got cosec in it and i dont know where that comes from?
    Write sec and tan in terms of sin and cos.
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    (Original post by kirstieluu)
    When i simplified it.. i got sec/tant
    But the answer has got cosec in it and i dont know where that comes from?
    I assume that you know that \tan t = \frac{\sin t}{\cos t}

    And that you know how to divide by a fraction
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    (Original post by kirstieluu)
    When i simplified it.. i got sec/tant
    But the answer has got cosec in it and i dont know where that comes from?
    In addition to TenOfThem's post, note that we define secant as follows

    \displaystyle \sec t = \frac{1}{\cos t}.

    There's nothing "official" about this (it's just a man-made definition of 1 over cos t).
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    I've got sect/tant = sec/1 x cost/sint

    Now I'm stuck
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    (Original post by kirstieluu)
    I've got sect/tant = sec/1 x cost/sint

    Now I'm stuck
    What's

    \displaystyle \frac{1}{\frac{1}{3}}

    ?
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    I get it now. Thanks for the help guys!
 
 
 
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