OCR C3 — Question 9 11 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Functions
TypeExponential growth/decay model setup
DifficultyModerate -0.8 This is a straightforward exponential model question requiring standard techniques: substituting values to find k using logarithms, solving for doubling time, and differentiating to find rate of change. All parts follow routine procedures with no conceptual challenges beyond basic C3 exponential manipulation.
Spec1.06g Equations with exponentials: solve a^x = b1.06i Exponential growth/decay: in modelling context1.07j Differentiate exponentials: e^(kx) and a^(kx)

9. The number of bacteria present in a culture at time \(t\) hours is modelled by the continuous variable \(N\) and the relationship $$N = 2000 \mathrm { e } ^ { k t }$$ where \(k\) is a constant.
Given that when \(t = 3 , N = 18000\), find
  1. the value of \(k\) to 3 significant figures,
  2. how long it takes for the number of bacteria present to double, giving your answer to the nearest minute,
  3. the rate at which the number of bacteria is increasing when \(t = 3\).

AnswerMarks Guidance
(i) \(t = 3\), \(N = 18000 \Rightarrow 18000 = 2000e^{3k}\)M1
\(e^{3k} = 9\)M1
\(k = \frac{1}{3}\ln 9 = 0.732\) (3sf)M1, A1
(ii) \(4000 = 2000e^{0.7324t}\)B1
\(t = \frac{1}{0.7324}\ln 2 = 0.9464\) hoursM2
\(\therefore\) doubles in 57 minutes (nearest minute)A1
(iii) \(N = 2000e^{0.7324t}\), \(\frac{dN}{dt} = 0.7324 \times 2000e^{0.7324t} = 1465e^{0.7324t}\)M1, A1
when \(t = 3\), \(\frac{dN}{dt} = 13200 \therefore\) increasing at rate 13200 per hour (3sf)M1, A1 (11 marks)
Total (72 marks)
**(i)** $t = 3$, $N = 18000 \Rightarrow 18000 = 2000e^{3k}$ | M1 |
$e^{3k} = 9$ | M1 |
$k = \frac{1}{3}\ln 9 = 0.732$ (3sf) | M1, A1 |

**(ii)** $4000 = 2000e^{0.7324t}$ | B1 |
$t = \frac{1}{0.7324}\ln 2 = 0.9464$ hours | M2 |
$\therefore$ doubles in 57 minutes (nearest minute) | A1 |

**(iii)** $N = 2000e^{0.7324t}$, $\frac{dN}{dt} = 0.7324 \times 2000e^{0.7324t} = 1465e^{0.7324t}$ | M1, A1 |
when $t = 3$, $\frac{dN}{dt} = 13200 \therefore$ increasing at rate 13200 per hour (3sf) | M1, A1 | **(11 marks)** |

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**Total (72 marks)**
9. The number of bacteria present in a culture at time $t$ hours is modelled by the continuous variable $N$ and the relationship

$$N = 2000 \mathrm { e } ^ { k t }$$

where $k$ is a constant.\\
Given that when $t = 3 , N = 18000$, find\\
(i) the value of $k$ to 3 significant figures,\\
(ii) how long it takes for the number of bacteria present to double, giving your answer to the nearest minute,\\
(iii) the rate at which the number of bacteria is increasing when $t = 3$.

\hfill \mbox{\textit{OCR C3  Q9 [11]}}