9709 P33 - Jun 2021 - Q3
1644
The parametric equations of a curve are
\(x = t + \ln(t + 2), \quad y = (t - 1)e^{-2t}\),
where \(t > -2\).
(a) Express \(\frac{dy}{dx}\) in terms of \(t\), simplifying your answer.
(b) Find the exact \(y\)-coordinate of the stationary point of the curve.
Solution
(a) To find \(\frac{dy}{dx}\), we first find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).
\(\frac{dx}{dt} = 1 + \frac{1}{t+2}\).
Using the product rule, \(\frac{dy}{dt} = e^{-2t} - 2(t-1)e^{-2t} = e^{-2t} - 2te^{-2t} + 2e^{-2t} = (3 - 2t)e^{-2t}\).
Now, \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{(3 - 2t)e^{-2t}}{1 + \frac{1}{t+2}} = \frac{(3 - 2t)(t + 2)}{t + 3} e^{-2t}\).
(b) A stationary point occurs when \(\frac{dy}{dx} = 0\).
Setting \((3 - 2t)(t + 2) = 0\), we solve for \(t\):
\(3 - 2t = 0 \Rightarrow t = \frac{3}{2}\).
Substitute \(t = \frac{3}{2}\) into \(y = (t - 1)e^{-2t}\):
\(y = \left(\frac{3}{2} - 1\right)e^{-3} = \frac{1}{2}e^{-3}\).
