| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Connected Rates of Change |
| Type | Curve motion: find dy/dt |
| Difficulty | Moderate -0.3 This is a straightforward connected rates of change question requiring the chain rule dy/dt = (dy/dx)(dx/dt). Part (a) involves routine differentiation of a power function and substitution, while part (b) requires solving a simple equation. The topic is standard A-level content with no conceptual challenges, making it slightly easier than average. |
| Spec | 1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates |
| Answer | Marks | Guidance |
|---|---|---|
| 6(a): \(\frac{dy}{dx} = \left[\frac{1}{2}(5x-1)^{-1/2}\right] \times [5]\) | B1 B1 | |
| Use \(\frac{dy}{dt} = 2 \times \left(their \frac{dy}{dx}\ \text{when}\ x=1\right)\) | M1 | |
| \(\frac{5}{2}\) | A1 | |
| 6(b): \(2 \times their\ \frac{5}{2}(5x-1)^{-1/2} = \frac{5}{8}\) | M1 | oe |
| \((5x-1)^{1/2} = 8\) | A1 | |
| \(x = 13\) | A1 |
## Question 6:
**6(a):** $\frac{dy}{dx} = \left[\frac{1}{2}(5x-1)^{-1/2}\right] \times [5]$ | B1 B1 |
Use $\frac{dy}{dt} = 2 \times \left(their \frac{dy}{dx}\ \text{when}\ x=1\right)$ | M1 |
$\frac{5}{2}$ | A1 |
**6(b):** $2 \times their\ \frac{5}{2}(5x-1)^{-1/2} = \frac{5}{8}$ | M1 | oe
$(5x-1)^{1/2} = 8$ | A1 |
$x = 13$ | A1 |
---6 A point $P$ is moving along a curve in such a way that the $x$-coordinate of $P$ is increasing at a constant rate of 2 units per minute. The equation of the curve is $y = ( 5 x - 1 ) ^ { \frac { 1 } { 2 } }$.
\begin{enumerate}[label=(\alph*)]
\item Find the rate at which the $y$-coordinate is increasing when $x = 1$.
\item Find the value of $x$ when the $y$-coordinate is increasing at $\frac { 5 } { 8 }$ units per minute.
\end{enumerate}
\hfill \mbox{\textit{CAIE P1 2020 Q6 [7]}}