Moderate -0.3 This is a straightforward application of differentiation and normal line formula. Students find the y-coordinate at x=3, differentiate ln(x/3) to get 1/x, evaluate at x=3 to get gradient 1/3, then use perpendicular gradient -3 in point-slope form. All steps are routine C3 techniques with no conceptual challenges, making it slightly easier than average.
3. The point \(P\) lies on the curve with equation \(y = \ln \left( \frac { 1 } { 3 } x \right)\). The \(x\)-coordinate of \(P\) is 3 . Find an equation of the normal to the curve at the point \(P\) in the form \(y = a x + b\), where \(a\) and \(b\) are constants.
(5)
3. The point $P$ lies on the curve with equation $y = \ln \left( \frac { 1 } { 3 } x \right)$. The $x$-coordinate of $P$ is 3 . Find an equation of the normal to the curve at the point $P$ in the form $y = a x + b$, where $a$ and $b$ are constants.\\
(5)\\
\hfill \mbox{\textit{Edexcel C3 2006 Q3 [5]}}