| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/2 (Pre-U Mathematics Paper 2) |
| Year | 2011 |
| Session | June |
| Marks | 15 |
| Topic | Connected Rates of Change |
| Type | Chain rule with three variables |
| Difficulty | Challenging +1.2 This is a multi-part integration and related rates problem requiring area calculation, algebraic manipulation, and implicit differentiation. While it involves several steps and techniques (integration, solving for intersection points, implicit differentiation with respect to time), each individual component is standard A-level fare. The related rates portion (parts iii-iv) adds moderate complexity but follows predictable patterns once the constraint equation is established. More challenging than a routine question but doesn't require exceptional insight. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08e Area between curve and x-axis: using definite integrals1.08f Area between two curves: using integration |
| Answer | Marks | Guidance |
|---|---|---|
| \(P\) has \(x\)-coordinate \(k\) | B1 | |
| Region \(R\) has area \(\frac{1}{2}k \times k^2 - \int_0^{(k)} x^2 dx\) or \(\int_0^{(k)} kx - x^2 dx\) | M1 | |
| \(= \frac{1}{2}k^3 - \frac{1}{3}k^3\) | ||
| \(= \frac{1}{6}k^3\) | AG | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\int_0^a (kx - x^2)dx = \frac{1}{12}k^3\) or equivalent | M1 | |
| \(= \left[\frac{1}{2}kx^2 - \frac{1}{3}x^3\right]_0^a\) | A1 | |
| \(\Rightarrow k^3 - 6ka^2 + 4a^3 = 0\) | AG | A1 |
| Answer | Marks |
|---|---|
| Differentiate the implicit equation wrt \(t\) | M1 |
| \(3k^2 \frac{dk}{dt} - 12a \frac{da}{dt} k - 6a^2 \frac{dk}{dt} + 12a^2 \frac{da}{dt} = 0\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Make substitutions and obtain \(\frac{da}{dt} = 1\) | A1 | [4] |
| Answer | Marks |
|---|---|
| Differentiate the implicit equation wrt \(\sigma\) or \(k\) | M1 |
| \(3k^2 \frac{dk}{da} - 12ak - 6a^2 \frac{dk}{da} + 12a^2 = 0\) or \(3k^2 - 12a \frac{dk}{d\sigma} k - 6a^2 \frac{dk}{d\sigma} + 12a^2 \frac{da}{dk} = 0\) | A1 |
| Relate connected rates of change | M1 |
| Make substitutions and obtain \(\frac{da}{dt} = 1\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| (The formula \(\frac{da}{dt} = \frac{k^2 - 2}{2(k-1)}\) may appear) | ||
| Attempt to factorise \(k^3 - 6k + 4\) with linear factor \((k - 2)\) | M1 | |
| Obtain \((k - 2)(k^2 + 2k - 2)\) | A1 | |
| Solve quadratic factor and obtain either or both of \(k = \pm\sqrt{3} - 1\) | A1 | |
| Correctly substitute into derivative formula and attempt to simplify | M1 | |
| Obtain either or both of \(\frac{da}{dt} = 1 \pm \sqrt{3}\) | A1 | [5] |
**Part (i)**
$P$ has $x$-coordinate $k$ | B1
Region $R$ has area $\frac{1}{2}k \times k^2 - \int_0^{(k)} x^2 dx$ or $\int_0^{(k)} kx - x^2 dx$ | M1
$= \frac{1}{2}k^3 - \frac{1}{3}k^3$ |
$= \frac{1}{6}k^3$ | AG | A1 | [3]
**Part (ii)**
$\int_0^a (kx - x^2)dx = \frac{1}{12}k^3$ or equivalent | M1
$= \left[\frac{1}{2}kx^2 - \frac{1}{3}x^3\right]_0^a$ | A1
$\Rightarrow k^3 - 6ka^2 + 4a^3 = 0$ | AG | A1 | [3]
**Part (iii)**
Differentiate the implicit equation wrt $t$ | M1
$3k^2 \frac{dk}{dt} - 12a \frac{da}{dt} k - 6a^2 \frac{dk}{dt} + 12a^2 \frac{da}{dt} = 0$ | A1
(<3 errors) A1 CAO
Make substitutions and obtain $\frac{da}{dt} = 1$ | A1 | [4]
**OR:**
Differentiate the implicit equation wrt $\sigma$ or $k$ | M1
$3k^2 \frac{dk}{da} - 12ak - 6a^2 \frac{dk}{da} + 12a^2 = 0$ or $3k^2 - 12a \frac{dk}{d\sigma} k - 6a^2 \frac{dk}{d\sigma} + 12a^2 \frac{da}{dk} = 0$ | A1
Relate connected rates of change | M1
Make substitutions and obtain $\frac{da}{dt} = 1$ | A1
**Part (iv)**
(The formula $\frac{da}{dt} = \frac{k^2 - 2}{2(k-1)}$ may appear) |
Attempt to factorise $k^3 - 6k + 4$ with linear factor $(k - 2)$ | M1
Obtain $(k - 2)(k^2 + 2k - 2)$ | A1
Solve quadratic factor and obtain either or both of $k = \pm\sqrt{3} - 1$ | A1
Correctly substitute into derivative formula and attempt to simplify | M1
Obtain either or both of $\frac{da}{dt} = 1 \pm \sqrt{3}$ | A1 | [5]The curve $y = x^3$ intersects the line $y = kx$, $k > 0$, at the origin and the point $P$. The region bounded by the curve and the line, between the origin and $P$, is denoted by $R$.
\begin{enumerate}[label=(\roman*)]
\item Show that the area of the region $R$ is $\frac{1}{6}k^3$. [3]
\end{enumerate}
The line $x = a$ cuts the region $R$ into two parts of equal area.
\begin{enumerate}[label=(\roman*)]
\setcounter{enumii}{1}
\item Show that $k^3 - 6a^2k + 4a^3 = 0$. [3]
\end{enumerate}
The gradient of the line $y = kx$ increases at a constant rate with respect to time $t$. Given that $\frac{dk}{dt} = 2$,
\begin{enumerate}[label=(\roman*)]
\setcounter{enumii}{2}
\item determine the value of $\frac{da}{dt}$ when $a = 1$ and $k = 2$, [4]
\item determine the value of $\frac{da}{dt}$ when $a = 1$ and $k = 2$, expressing your answer in the form $p + q\sqrt{3}$, where $p$ and $q$ are integers. [5]
\end{enumerate}
\hfill \mbox{\textit{Pre-U Pre-U 9794/2 2011 Q9 [15]}}