Moderate -0.3 This is a straightforward parametric differentiation question requiring standard techniques: find dx/dt and dy/dt, compute dy/dx = (dy/dt)/(dx/dt), evaluate at t=0 to get the gradient, find the point on the curve at t=0, then write the tangent equation. While it involves the product rule and chain rule, these are routine A-level skills with no conceptual challenges or novel problem-solving required.
3 The parametric equations of a curve are
$$x = ( t + 1 ) \mathrm { e } ^ { t } , \quad y = 6 ( t + 4 ) ^ { \frac { 1 } { 2 } }$$
Find the equation of the tangent to the curve when \(t = 0\), giving the answer in the form \(a x + b y + c = 0\) where \(a , b\) and \(c\) are integers.
3 The parametric equations of a curve are
$$x = ( t + 1 ) \mathrm { e } ^ { t } , \quad y = 6 ( t + 4 ) ^ { \frac { 1 } { 2 } }$$
Find the equation of the tangent to the curve when $t = 0$, giving the answer in the form $a x + b y + c = 0$ where $a , b$ and $c$ are integers.
\hfill \mbox{\textit{CAIE P2 2015 Q3 [6]}}