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Math Help - Differentiation Help?

  1. #1
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    Differentiation Help?

    Find the equation of the tangent to the curve y=4x^2+c at the point x=t. If the tangent passes through the point (1,0), find c in terms of t. Help anyone?
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  2. #2
    Senior Member nikhil's Avatar
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    y=4x^2+c
    dy/dx=8x
    at x=t
    dy/dx=8t and y=4t^2+c
    Then equation of tangent is
    y-(4t^2+c)=8t(x-t)
    y=4t^2+c+8tx-8t^2 or
    y=8tx-4t^2+c
    but the tangent also passes through (1,0) so its equation can also be
    y-0=8t(x-1) or
    y=8tx-8t
    so two equations of same tangent are
    1)y=8tx-4t^2+c
    2)y=8tx-8t
    so by comparing both we get
    -8t=-4t^2+c
    or
    c=4t^2-8t
    so now we also expressed c in terms of t
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  3. #3
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    thx mate.
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