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Math Help - Parametric Equation Question

  1. #1
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    Parametric Equation Question

    Hello Everyone again! I'm trying to do this question and I'm rather stumped on the last part. Parametric Equation is turning into a nightmare for me .

    The full question is as follows:

    Show that the tangent at the point P, with parameter t, on the curve x=3t^2, y=3t^2 has equation y=tx-t^3. Prove that this tangent will cut the curve again at the point Q(\frac{3t^2}{4}, - \frac{t^3}{4}). Find the coordinates of the possible positions of P if the tangent to the curve at P is the normal to the curve at Q.

    To prove for the tangent, I solved (is there a better word) the parametric equation into a cartesian equation, which is: y=\frac{2x^{\frac{3}{2}}}{3\sqrt{3}} Implicitly differentiating, the gradient would be \frac{\sqrt{x}}{\sqrt{3}}=t (From solving the first parametric equation for t). Solve then for tangent.

    The gradient therefore is correct and shown as y=tx-t^3. Intersecting the gradient and the curve, y=tx-t^3=\frac{2x^{\frac{3}{2}}}{3\sqrt{3}}. Solving for x, I get x=3t^2 or \frac{3t^2}{4} Since the former option is already taken by P, Q will have to take the other option. Using then x=\frac{3t^2}{4}, and substituting it into y=\frac{2x^{\frac{3}{2}}}{3\sqrt{3}}, I get the value of y=\pm \frac{t^3}{4}. Since the positive solution does not satisfy, the negative must be the correct answer.

    Since the tangent to the curve at P is the normal to the curve at Q, and the tangent at P is t, we can assume that the tangent at Q is -\frac{1}{t}. Here I'm stuck, since I'm not sure how to proceed. Can someone help me over this?

    Thanks in advance!!
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  2. #2
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    Re: Parametric Equation Question

    are you sure you've posted this correctly? if x=3t2 and y=3t2 then y = x, dy/dx = 1
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  3. #3
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    Re: Parametric Equation Question

    Ooops sorry, it should be x=3t^2, y=2t^3. All other information is correct. Sorry!
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  4. #4
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    Re: Parametric Equation Question

    This is a mess. First off. Is Q your curve given by x(t) and y(t)? Is it a point? I can't tell.

    Next, the tangent to your curve is given by dy/dx and part of the whole idea of parametric formulation is that you can immediately say

    dy/dx = (dy/dt) / (dx/dt)

    Using your curve we get dy/dx = (6t2 / 6t) = t = sqrt(x/3), you found this, the hard way.

    As for the rest of it. I really don't understand what it is you are trying to show. Tangent lines normal to the curve at Q? What does that mean?
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  5. #5
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    Re: Parametric Equation Question

    I think the question is very clear on what is Q. I'm asked to prove that the tangent cuts the curve again at the point Q.

    I don't know about the last part, which is why I'm asking. The question is as the book puts it. I myself don't understand how the tangent at a certain point in a curve can be a normal at another point in the same curve, which is why I'm posting the question and asking for help.
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  6. #6
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    Re: Parametric Equation Question

    How about this?
    Parametric Equation Question-intersect.jpg

    -Dan
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  7. #7
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    Re: Parametric Equation Question

    Now I should be able to get my answer...I only plotted positive y values for the curve :P
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  8. #8
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    Re: Parametric Equation Question

    Hi,
    The first attachment solves your problem; fortunately, this was a text book problem so the main equation has a simple solution. The second attachment is a graph showing one of the two pairs of points P and Q which satisfy your problem.

    Parametric Equation Question-mhfcalc20a.png

    Parametric Equation Question-mhfcalc20b.png
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  9. #9
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    Re: Parametric Equation Question

    Hi again,
    Sorry, the derivation in my previous posting has a silly algebraic error. The conclusion is correct; I just mistyped. Here is a corrected version.

    Parametric Equation Question-mhfcalc20c.png
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