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Thread: parametric equations help

  1. #1
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    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
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  2. #2
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    Hello Tweety
    Quote Originally Posted by Tweety View Post
    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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