| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2017 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Curve motion: find dy/dt |
| Difficulty | Standard +0.3 Part (i) is a standard tangent line question requiring finding where y=0, then using differentiation. Part (ii) is a straightforward connected rates of change problem using the chain rule dy/dt = (dy/dx)(dx/dt) with given values - this is a textbook application of the technique with no conceptual challenges beyond the basic method. |
| Spec | 1.07l Derivative of ln(x): and related functions1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Crosses \(x\)-axis at \((6, 0)\) | B1 | \(x = 6\) is sufficient. |
| \(\frac{dy}{dx} = (0+) - 12(2-x)^{-2} \times (-1)\) | B2,1,0 | \(-1\) for each incorrect term of the three or addition of \(+ C\). |
| Tangent \(y = \frac{3}{4}(x-6)\) or \(4y = 3x - 18\) | M1 A1 | Must use \(dy/dx\), \(x =\) their 6 but not \(x = 0\) (which gives \(m = 3\)), and correct form of line equation. Using \(y = mx + c\) gets A1 as soon as \(c\) is evaluated. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| If \(x = 4\), \(dy/dx = 3\) | ||
| \(\frac{dy}{dt} = 3 \times 0.04 = 0.12\) | M1 A1FT | M1 for ("their m" from \(\frac{dy}{dx}\) and \(x = 4\)) \(\times 0.04\). Be aware: use of \(x = 0\) gives the correct answer but gets M0. |
## Question 5:
**Part (i):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Crosses $x$-axis at $(6, 0)$ | **B1** | $x = 6$ is sufficient. |
| $\frac{dy}{dx} = (0+) - 12(2-x)^{-2} \times (-1)$ | **B2,1,0** | $-1$ for each incorrect term of the three or addition of $+ C$. |
| Tangent $y = \frac{3}{4}(x-6)$ or $4y = 3x - 18$ | **M1 A1** | Must use $dy/dx$, $x =$ their 6 but not $x = 0$ (which gives $m = 3$), and correct form of line equation. Using $y = mx + c$ gets **A1** as soon as $c$ is evaluated. |
**Part (ii):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| If $x = 4$, $dy/dx = 3$ | | |
| $\frac{dy}{dt} = 3 \times 0.04 = 0.12$ | **M1 A1FT** | **M1** for ("their m" from $\frac{dy}{dx}$ and $x = 4$) $\times 0.04$. Be aware: use of $x = 0$ gives the correct answer but gets **M0**. |5 A curve has equation $y = 3 + \frac { 12 } { 2 - x }$.\\
(i) Find the equation of the tangent to the curve at the point where the curve crosses the $x$-axis.\\
(ii) A point moves along the curve in such a way that the $x$-coordinate is increasing at a constant rate of 0.04 units per second. Find the rate of change of the $y$-coordinate when $x = 4$.\\
\hfill \mbox{\textit{CAIE P1 2017 Q5 [7]}}