Challenging +1.2 This is a standard Further Maths parametric calculus question requiring the chain rule formula for d²y/dx², followed by a mean value integral calculation. While it involves multiple steps and exponential functions, the techniques are routine for FP1 students: differentiate parametrically using dx/dt and dy/dt, apply the quotient rule, then integrate using substitution. The algebraic manipulation is moderately involved but follows established procedures without requiring novel insight.
6 The curve \(C\) is defined parametrically by
$$x = 4 t - t ^ { 2 } \quad \text { and } \quad y = 1 - \mathrm { e } ^ { - t }$$
where \(0 \leqslant t < 2\). Show that at all points of \(C\),
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { ( t - 1 ) \mathrm { e } ^ { - t } } { 4 ( 2 - t ) ^ { 3 } }$$
Show that the mean value of \(\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }\) with respect to \(x\) over the interval \(0 \leqslant x \leqslant \frac { 7 } { 4 }\) is
$$\frac { 4 e ^ { - \frac { 1 } { 2 } } - 3 } { 21 }$$
6 The curve $C$ is defined parametrically by
$$x = 4 t - t ^ { 2 } \quad \text { and } \quad y = 1 - \mathrm { e } ^ { - t }$$
where $0 \leqslant t < 2$. Show that at all points of $C$,
$$\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } } = \frac { ( t - 1 ) \mathrm { e } ^ { - t } } { 4 ( 2 - t ) ^ { 3 } }$$
Show that the mean value of $\frac { \mathrm { d } ^ { 2 } y } { \mathrm {~d} x ^ { 2 } }$ with respect to $x$ over the interval $0 \leqslant x \leqslant \frac { 7 } { 4 }$ is
$$\frac { 4 e ^ { - \frac { 1 } { 2 } } - 3 } { 21 }$$
\hfill \mbox{\textit{CAIE FP1 2008 Q6 [8]}}