A curve is given by the parametric equations $x = 2e^{2t} + 4e^{t}$, $y = 5te^{2t}$.
(i)[6]
Find $\frac{dy}{dx}$ as a function of $t$ and hence determine the coordinates of the stationary point, with each coordinate correct to 2 decimal places.
(ii)[3]
Find the gradient of the normal to the curve at the point of intersection with the $x$-axis.
Worked solution & mark scheme
This 9-mark question has a full step-by-step worked solution and mark scheme. One marking point: “Obtain the result $\dfrac{dx}{dt} = 4e^{2t} + 4e^t$” …