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
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 $\displaystyle f'(x) = \dfrac k{x-3}$ then $\displaystyle f(x)= k \cdot \ln(x-3)+C$
$\displaystyle (4, 0)$ satisfies the equation:
$\displaystyle 0 = k\ln(4-3)+C~\implies~C=0$
$\displaystyle (6, \ln(9))$ satisfies the equation:
$\displaystyle \ln(9)= k \cdot \ln(6-3)~\implies~k=\dfrac{\ln(9)}{\ln(3)}=2$