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    I need help to solve the following:

    A curve is given parametrically by
    x=t^2
    y=t^3

    Find dy/dx in terms of t

    Optain the equation of the tangent at the point P with parameter t and verify that it cuts the curve again at Q with parameter -t/2
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    (Original post by shawrie777)
    I need help to solve the following:

    A curve is given parametrically by
    x=t^2
    y=t^3

    Find dy/dx in terms of t

    Optain the equation of the tangent at the point P with parameter t and verify that it cuts the curve again at Q with parameter -t/2
    Actually Dom, the OP was expected to use parametric differentiation.

    x=t^2

    y=t^3

    \frac{dx}{dt}=2t

    \frac{dy}{dt}=3t^2

    \frac{dy}{dx}=\frac{dy}{dt} \times \frac{dt}{dx}= \frac{3t}{2}
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    To differentiate parametric equations note that:

     \frac{dy}{dx}= \frac{dy}{dt} \times \frac{dt}{dx}= \frac{dy}{dt} \times \frac{1}{\frac{dx}{dt}}

    You can easily work out \frac{dy}{dt}and\frac{dx}{dt} so the differentiation should not be too bad.

    The next bit is nothing to do with parametrics but rather just normal geometry which you should be able to do. If not just post and Ill give some more hints.
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    To Dom and Mr M: Can I direct your attention to the posting guide which is a sticky at the top of the maths forum, specifically to the bit about posting full solutions. This is an argument which has rages over many threads but it is generally considered frowned upon to write out full solutions. It is more beneficial to the op if you give out hints because this ensures that they have a good understanding of the problem rather than just a copy of the answers.
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    Actually I do find the next bit difficult. Surely I need to have a value to p to be able to find the equation of the tangent.
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    (Original post by The Muon)
    To Dom and Mr M: Can I direct your attention to the posting guide which is a sticky at the top of the maths forum, specifically to the bit about posting full solutions. This is an argument which has rages over many threads but it is generally considered frowned upon to write out full solutions. It is more beneficial to the op if you give out hints because this ensures that they have a good understanding of the problem rather than just a copy of the answers.
    I would dispute that I gave a full solution as:

    a) Dom had already posted dy/dx before I referred to it.

    b) I left the meat of the question for the OP to finish.
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    (Original post by Dom K. Jordan)
    Where does it say that? You get the same result anyway solving the cartesian equation then subbing t back in, I find it easier that way.
    What would you do if you cannot find the Cartesian equation?
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    (Original post by shawrie777)
    Actually I do find the next bit difficult. Surely I need to have a value to p to be able to find the equation of the tangent.
    I think the question is asking for a general point P. You know what the gradient at P will be - you have just worked it out (dy/dx). You also know a general point; the vales of x and y given in the question (note that these are in terms of t). You just have to put these into the equation of a straight line (I think this is in C1) and you will have the tangent to the graph at point P
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    (Original post by Dom K. Jordan)
    Um t= sqrt x
    y = x sqrt x
    = x^3/2

    thus dy/dx = 3/2x^1/2 = 3/2t
    Dom,

    Two points:

    x = t^2 does not imply t=\sqrt{x}

    dy/dx = 3/2t looks awfully like \frac{dy}{dx}=\frac{3}{2t}

    Learn LaTeX soon if you can find time.
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    LaTeX{} is something that will solve all your display confusions:

    http://www.thestudentroom.co.uk/wiki/LaTex
 
 
 
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