# parametric equations help

• May 13th 2010, 08:53 AM
Tweety
parametric equations help
The curve with parametric equations $\displaystyle x = 3t^{2} , y = 4t^{3}$ along with the tangent at the point with parameter t = 1

find the equation if this tangent and the coordinates of the point where it meets the curve again.

$\displaystyle \frac{dx}{dt} = 6t$

$\displaystyle \frac{dy}{dt} = 12t^{2}$

$\displaystyle \frac{dy}{dx} = \frac{12t^{2}}{6t} = 2t$

when t = 1

$\displaystyle x = 3 , y = 4 , \frac{dy}{dx} = 2$

equation of tangent = $\displaystyle y -4 = 2(x-3)$

$\displaystyle y = 2x-2$

however to find the coordinates of where the curve crosses the tangent again, would I not just be able to sub in the values of x and y into the tnagent equation?

$\displaystyle 4t^{3} = 2(3t^{2})-2$

$\displaystyle 4t^{3} = 6t^{2} - 2$ can't seem to solve from here,

any help appreicated.

thanks
• May 13th 2010, 10:52 AM
Hello Tweety
Quote:

Originally Posted by Tweety
The curve with parametric equations $\displaystyle x = 3t^{2} , y = 4t^{3}$ along with the tangent at the point with parameter t = 1

find the equation if this tangent and the coordinates of the point where it meets the curve again.

$\displaystyle \frac{dx}{dt} = 6t$

$\displaystyle \frac{dy}{dt} = 12t^{2}$

$\displaystyle \frac{dy}{dx} = \frac{12t^{2}}{6t} = 2t$

when t = 1

$\displaystyle x = 3 , y = 4 , \frac{dy}{dx} = 2$

equation of tangent = $\displaystyle y -4 = 2(x-3)$

$\displaystyle y = 2x-2$

however to find the coordinates of where the curve crosses the tangent again, would I not just be able to sub in the values of x and y into the tnagent equation?

$\displaystyle 4t^{3} = 2(3t^{2})-2$

$\displaystyle 4t^{3} = 6t^{2} - 2$ can't seem to solve from here,

any help appreicated.

thanks

Your working is perfect up to here. Good job!

What you need to realise now is that the tangent at $\displaystyle t=1$ effectively cuts the curve in two coincident points where $\displaystyle t = 1$.

So $\displaystyle t=1$ will be a repeated root of your final equation, which is more simply written as:
$\displaystyle 2t^3-3t^2+1=0$
In other words $\displaystyle (t-1)^2$, or $\displaystyle (t^2-2t+1)$, is a factor of the LHS of:
$\displaystyle 2t^3 - 3t^2 +1 =0$
So we get:
$\displaystyle (t^2-2t+1)(2t+1)=0$
(Check it out: multiply out the brackets and you get $\displaystyle 2t^3-3t^2+1$.)

So the other value of $\displaystyle t$ is ...?

I'm sure you can take it from here.