CAIE P1 2020 March — Question 4 4 marks

Exam BoardCAIE
ModuleP1 (Pure Mathematics 1)
Year2020
SessionMarch
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicApplied differentiation
TypeRate of change on a curve
DifficultyStandard +0.3 This is a straightforward related rates problem requiring implicit differentiation of a simple quadratic. Students apply the chain rule (dy/dt = dy/dx × dx/dt) with given rates and solve a linear equation. While it requires understanding the relationship between rates, the algebra is minimal and the method is standard textbook material, making it slightly easier than average.
Spec1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates
4 A curve has equation \(y = x ^ { 2 } - 2 x - 3\). A point is moving along the curve in such a way that at \(P\) the \(y\)-coordinate is increasing at 4 units per second and the \(x\)-coordinate is increasing at 6 units per second. Find the \(x\)-coordinate of \(P\).
Question 4:
AnswerMarks Guidance
\(\frac{dy}{dx} = 2x-2\)B1
\(\frac{dy}{dx} = \frac{4}{6}\)B1 OE, SOI
\(\text{their}(2x-2) = \text{their}\,\frac{4}{6}\)M1 LHS and RHS must be *their* \(\frac{dy}{dx}\) expression and value
\(x = \frac{4}{3}\) oeA1
Total: 4 
## Question 4:

| $\frac{dy}{dx} = 2x-2$ | B1 | |
|---|---|---|
| $\frac{dy}{dx} = \frac{4}{6}$ | B1 | OE, SOI |
| $\text{their}(2x-2) = \text{their}\,\frac{4}{6}$ | M1 | LHS and RHS must be *their* $\frac{dy}{dx}$ expression and value |
| $x = \frac{4}{3}$ oe | A1 | |
| **Total: 4** | | |

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4 A curve has equation $y = x ^ { 2 } - 2 x - 3$. A point is moving along the curve in such a way that at $P$ the $y$-coordinate is increasing at 4 units per second and the $x$-coordinate is increasing at 6 units per second.

Find the $x$-coordinate of $P$.\\

\hfill \mbox{\textit{CAIE P1 2020 Q4 [4]}}
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