| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2002 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Curve motion: find dy/dt |
| Difficulty | Standard +0.3 This is a straightforward multi-part question testing standard techniques: finding a normal line equation, integration to find a curve equation from its derivative, and basic connected rates of change using the chain rule. Part (iii) simply requires dy/dt = (dy/dx) × (dx/dt) with given values—a direct application with no problem-solving insight needed. Slightly above average due to the three-part structure, but all components are routine A-level procedures. |
| Spec | 1.07m Tangents and normals: gradient and equations1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08d Evaluate definite integrals: between limits |
| Answer | Marks | Guidance |
|---|---|---|
| Puts \(y = 0\), \(x = \frac{23}{3}\) | B1, M1, M1, A1 | Correct only. Use of \(m_1m_2 = -1\). Correct form – though may put \(y=0\) at start. Correct only |
| (ii) \(y = 12(2x+1)^{-1} \div -1 + 2\) | M1, A1, M1, A1 | For \(12(2x+1)^{-k}\) - no other "\(x\)" anywhere. For \(k = -1\) and \(\div 2\). Needs an attempt at integration, plus use of \(C\) |
| \(y = -\frac{6}{(2x+1)} + c\), \(c = 7\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} = \frac{4}{3} \times 0.3 = 0.4\) | B1, M1, A1 | Fact only. Correct relation between rates of change used. Correct only (condone use of \(\delta x\), \(\delta y\)). Nb could get M1 A1 for (ii) if in (i) |
**(i)** At $P(1,5)$, $x = 1$, $m = \frac{4}{3}$
Gradient of normal $= -\frac{3}{4}$
Eqn of normal: $y - 5 = -\frac{3}{4}(x-1)$
Puts $y = 0$, $x = \frac{23}{3}$ | B1, M1, M1, A1 | Correct only. Use of $m_1m_2 = -1$. Correct form – though may put $y=0$ at start. Correct only
**(ii)** $y = 12(2x+1)^{-1} \div -1 + 2$ | M1, A1, M1, A1 | For $12(2x+1)^{-k}$ - no other "$x$" anywhere. For $k = -1$ and $\div 2$. Needs an attempt at integration, plus use of $C$
$y = -\frac{6}{(2x+1)} + c$, $c = 7$ | |
**(iii)** $\frac{dx}{dt} = 0.3$
$\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} = \frac{4}{3} \times 0.3 = 0.4$ | B1, M1, A1 | Fact only. Correct relation between rates of change used. Correct only (condone use of $\delta x$, $\delta y$). Nb could get M1 A1 for (ii) if in (i)9 A curve is such that $\frac { \mathrm { d } y } { \mathrm {~d} x } = \frac { 12 } { ( 2 x + 1 ) ^ { 2 } }$ and $P ( 1,5 )$ is a point on the curve.\\
(i) The normal to the curve at $P$ crosses the $x$-axis at $Q$. Find the coordinates of $Q$.\\
(ii) Find the equation of the curve.\\
(iii) A point is moving along the curve in such a way that the $x$-coordinate is increasing at a constant rate of 0.3 units per second. Find the rate of increase of the $y$-coordinate when $x = 1$.
\hfill \mbox{\textit{CAIE P1 2002 Q9 [11]}}