OCR PURE — Question 2 4 marks

Exam BoardOCR
ModulePURE
Marks4
PaperDownload PDF ↗
TopicChain Rule
TypeFind gradient at specific point
DifficultyEasy -1.2 This is a straightforward differentiation question requiring only the power rule (rewriting 1/t as t^(-1)) and substitution. It's simpler than average A-level questions as it involves a single-step differentiation of a basic function with no chain rule despite the topic label, followed by direct substitution.
Spec1.07b Gradient as rate of change: dy/dx notation
2 The number of people, \(n\), living in a small town is changing over time. In an attempt to predict the future growth of the town, a researcher uses the following model for \(n\) in terms of \(t\), where \(t\) is the time in years from the start of the research. \(n = 12500 + \frac { 5000 } { t }\), for \(t \geqslant 1\) Find the rate of change of \(n\) when \(t = 5\).
Question 2:
AnswerMarks Guidance
\(\frac{dn}{dt} = -5000t^{-2}\)M1, A1 Attempt differentiate
At \(t=5\), \(\frac{dn}{dt} = \frac{-5000}{25}\) ISWM1 Substitute \(t=5\) into their \(\frac{dn}{dt}\) which must include \(t^{-2}\)
(Rate of change of \(n\) is) \(-200\)A1
[4] 
# Question 2:
| $\frac{dn}{dt} = -5000t^{-2}$ | M1, A1 | Attempt differentiate |
|---|---|---|
| At $t=5$, $\frac{dn}{dt} = \frac{-5000}{25}$ ISW | M1 | Substitute $t=5$ into their $\frac{dn}{dt}$ which must include $t^{-2}$ |
| (Rate of change of $n$ is) $-200$ | A1 | |
| | [4] | |

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2 The number of people, $n$, living in a small town is changing over time. In an attempt to predict the future growth of the town, a researcher uses the following model for $n$ in terms of $t$, where $t$ is the time in years from the start of the research.\\
$n = 12500 + \frac { 5000 } { t }$, for $t \geqslant 1$\\
Find the rate of change of $n$ when $t = 5$.

\hfill \mbox{\textit{OCR PURE  Q2 [4]}}
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