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 (Speechless).

The full question is as follows:

Show that the tangent at the point P, with parameter t, on the curve $\displaystyle x=3t^2$, $\displaystyle y=3t^2$ has equation $\displaystyle y=tx-t^3$. Prove that this tangent will cut the curve again at the point $\displaystyle Q(\frac{3t^2}{4}, - \frac{t^3}{4})$. 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: $\displaystyle y=\frac{2x^{\frac{3}{2}}}{3\sqrt{3}}$ Implicitly differentiating, the gradient would be $\displaystyle \frac{\sqrt{x}}{\sqrt{3}}=t$ (From solving the first parametric equation for t). Solve then for tangent.

The gradient therefore is correct and shown as $\displaystyle y=tx-t^3$. Intersecting the gradient and the curve, $\displaystyle y=tx-t^3=\frac{2x^{\frac{3}{2}}}{3\sqrt{3}}$. Solving for $\displaystyle x$, I get $\displaystyle x=3t^2$ or $\displaystyle \frac{3t^2}{4}$ Since the former option is already taken by P, Q will have to take the other option. Using then $\displaystyle x=\frac{3t^2}{4}$, and substituting it into $\displaystyle y=\frac{2x^{\frac{3}{2}}}{3\sqrt{3}}$, I get the value of $\displaystyle y=\pm \frac{t^3}{4}$. 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 $\displaystyle t$, we can assume that the tangent at Q is $\displaystyle -\frac{1}{t}$. Here I'm stuck, since I'm not sure how to proceed. Can someone help me over this?

Thanks in advance!!

Re: Parametric Equation Question

are you sure you've posted this correctly? if x=3t^{2} and y=3t^{2} then y = x, dy/dx = 1

Re: Parametric Equation Question

Ooops sorry, it should be $\displaystyle x=3t^2, y=2t^3$. All other information is correct. Sorry!

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 = (6t^{2} / 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?

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. :cool:

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Re: Parametric Equation Question

How about this?

Attachment 29840

-Dan

Re: Parametric Equation Question

Now I should be able to get my answer...I only plotted positive y values for the curve :P

2 Attachment(s)

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.

Attachment 29846

Attachment 29847

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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.

Attachment 29850