| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2016 |
| Session | November |
| Marks | 9 |
| 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 connected rates of change question requiring basic differentiation of a rational function, chain rule application (dy/dt = dy/dx × dx/dt), and substitution of given values. The multi-part structure guides students through each step, making it slightly easier than average but still requiring competent calculus technique. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07m Tangents and normals: gradient and equations1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dy}{dx} = \frac{-3}{(2x-1)^2} \times 2\) | B1 | B1 for a single correct term (unsimplified) without \(\times 2\) |
| B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| e.g. Solve \(\frac{dy}{dx} = 0\) is impossible | B1\(\checkmark\) | Satisfactory explanation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| If \(x = 2\), \(\frac{dy}{dx} = \frac{-6}{9}\) and \(y = 3\) | M1* | Attempt at both needed |
| Perpendicular has \(m = \frac{9}{6}\) | M1* | Use of \(m_1 m_2 = -1\) numerically |
| \(\rightarrow y - 3 = \frac{3}{2}(x - 2)\) | DM1 | Line equation using \((2\), their \(3)\) and their \(m\) |
| Shows when \(x=0\) then \(y=0\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{dx}{dt} = -0.06\) | ||
| \(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} \rightarrow -\frac{2}{3} \times -0.06 = 0.04\) | M1 A1 |
## Question 7(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dy}{dx} = \frac{-3}{(2x-1)^2} \times 2$ | B1 | B1 for a single correct term (unsimplified) without $\times 2$ |
| | B1 | |
## Question 7(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| e.g. Solve $\frac{dy}{dx} = 0$ is impossible | B1$\checkmark$ | Satisfactory explanation |
## Question 7(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| If $x = 2$, $\frac{dy}{dx} = \frac{-6}{9}$ and $y = 3$ | M1* | Attempt at both needed |
| Perpendicular has $m = \frac{9}{6}$ | M1* | Use of $m_1 m_2 = -1$ numerically |
| $\rightarrow y - 3 = \frac{3}{2}(x - 2)$ | DM1 | Line equation using $(2$, their $3)$ and their $m$ |
| Shows when $x=0$ then $y=0$ | A1 | AG |
## Question 7(iv):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{dx}{dt} = -0.06$ | | |
| $\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} \rightarrow -\frac{2}{3} \times -0.06 = 0.04$ | M1 A1 | |7 The equation of a curve is $y = 2 + \frac { 3 } { 2 x - 1 }$.\\
(i) Obtain an expression for $\frac { \mathrm { d } y } { \mathrm {~d} x }$.\\
(ii) Explain why the curve has no stationary points.
At the point $P$ on the curve, $x = 2$.\\
(iii) Show that the normal to the curve at $P$ passes through the origin.\\
(iv) A point moves along the curve in such a way that its $x$-coordinate is decreasing at a constant rate of 0.06 units per second. Find the rate of change of the $y$-coordinate as the point passes through $P$.
\hfill \mbox{\textit{CAIE P1 2016 Q7 [9]}}