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    The curve C has the equations:

    x=3t^2, y=2t^3

    a) find an equation of the tangent at the point where t=2.

    What's the procedure for a question like this? Finding dx/dt and dy/dt then subbing in t=2 and then rearranging for an equation in the form y=...x...?

    how would I start this thanks
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    - Find dy/dx by the chain rule as you said  \frac{dy}{dx} = \frac{dy}{dt}.\frac{dt}{dx} and sub in t = 2 to get the gradient at that point
    - Sub in t = 2 into the parametric equations to find coordinate values (x,y) at the tangent
    - Use the formula from c1  y - y_1 = m(x - x_1) , where (x_1, y_1) is your coordinates at the tangent, m is your gradient at the tangent
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    Someone asked me to change the title of this thread to "C4 Integration", but since it doesn't actually have anything to do with integration I think I'll let it be.
 
 
 
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Updated: April 11, 2011
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