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Math Help - [SOLVED] Finding the equation of curve

  1. #1
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    Talking [SOLVED] Finding the equation of curve

    The gradient of the tangent to a curve at the point (x,y) where x>3, is inversely proportional to (x-3), and the curve passes through the points (4,0) and (6,ln9). Show that the equation of the curve is y=2ln(x-3).

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  2. #2
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    Quote Originally Posted by stpmmaths View Post
    The gradient of the tangent to a curve at the point (x,y) where x>3, is inversely proportional to (x-3), and the curve passes through the points (4,0) and (6,ln9). Show that the equation of the curve is y=2ln(x-3).

    Thanks
    If f'(x) = \dfrac k{x-3} then f(x)= k \cdot \ln(x-3)+C

    (4, 0) satisfies the equation:

    0 = k\ln(4-3)+C~\implies~C=0

    (6, \ln(9)) satisfies the equation:

    \ln(9)= k \cdot \ln(6-3)~\implies~k=\dfrac{\ln(9)}{\ln(3)}=2
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    It's that easy only. Thanks.
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