parametric equations help

Sep 2008
631
2
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
 

Grandad

MHF Hall of Honor
Dec 2008
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South Coast of England
Hello 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.

Grandad
 
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