| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2003 |
| Session | June |
| Marks | 10 |
| 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 basic differentiation of a square root function and application of the chain rule dy/dt = (dy/dx)(dx/dt). Part (i) is routine differentiation, part (ii) is a standard textbook application of related rates with given values, and part (iii) is basic integration. All parts use well-practiced techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.08b Integrate x^n: where n != -1 and sums |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{dy}{dx} = 1/2(5x + 4)^{-1/2} \times 5\); \(x = 1, \frac{dy}{dx} = 5/6\) | B1B1, B1 [3] | \(1/2(5x + 4)^{-1/2} \times 5\) [B1 for each part]. Co |
| (ii) \(\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} = 5/6 \times 0.03\) \(\rightarrow 0.025\) | M1, A1∨ [2] | Chain rule correctly used. For (i) × 0.03 |
| (iii) realises that area \(\rightarrow\) integration | M1 | Realisation + attempt – must be \((5x + 4)^x\) |
| \(\int = (5x + 4)^{3/2} \div 3/2 \div 5\) | A1A1 [5] | For \((5x + 4)^{3/2} \div 3/2\). For \(\div 5\) |
| Use of limits \(\rightarrow 54/15 - 16/15 = 38/15 = 2.53\) | DM1, A1 [5] | Must use "0" to "1" |
**(i)** $\frac{dy}{dx} = 1/2(5x + 4)^{-1/2} \times 5$; $x = 1, \frac{dy}{dx} = 5/6$ | B1B1, B1 [3] | $1/2(5x + 4)^{-1/2} \times 5$ [B1 for each part]. Co
**(ii)** $\frac{dy}{dt} = \frac{dy}{dx} \times \frac{dx}{dt} = 5/6 \times 0.03$ $\rightarrow 0.025$ | M1, A1∨ [2] | Chain rule correctly used. For (i) × 0.03
**(iii)** realises that area $\rightarrow$ integration | M1 | Realisation + attempt – must be $(5x + 4)^x$
$\int = (5x + 4)^{3/2} \div 3/2 \div 5$ | A1A1 [5] | For $(5x + 4)^{3/2} \div 3/2$. For $\div 5$
Use of limits $\rightarrow 54/15 - 16/15 = 38/15 = 2.53$ | DM1, A1 [5] | Must use "0" to "1"10 The equation of a curve is $y = \sqrt { } ( 5 x + 4 )$.\\
(i) Calculate the gradient of the curve at the point where $x = 1$.\\
(ii) A point with coordinates $( x , y )$ moves along the curve in such a way that the rate of increase of $x$ has the constant value 0.03 units per second. Find the rate of increase of $y$ at the instant when $x = 1$.\\
(iii) Find the area enclosed by the curve, the $x$-axis, the $y$-axis and the line $x = 1$.
\hfill \mbox{\textit{CAIE P1 2003 Q10 [10]}}