OCR C3 2008 June — Question 7 9 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
Year2008
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Functions
TypeExponential growth/decay model setup
DifficultyModerate -0.3 This is a straightforward exponential growth question requiring standard techniques: substituting initial conditions to find constants, solving exponential equations using logarithms, and differentiating to find rate of change. All steps are routine C3 material with no novel problem-solving required, making it slightly easier than average.
Spec1.06a Exponential function: a^x and e^x graphs and properties1.06i Exponential growth/decay: in modelling context1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)

7 It is claimed that the number of plants of a certain species in a particular locality is doubling every 9 years. The number of plants now is 42 . The number of plants is treated as a continuous variable and is denoted by \(N\). The number of years from now is denoted by \(t\).
  1. Two equivalent expressions giving \(N\) in terms of \(t\) are $$N = A \times 2 ^ { k t } \quad \text { and } \quad N = A \mathrm { e } ^ { m t } .$$ Determine the value of each of the constants \(A , k\) and \(m\).
  2. Find the value of \(t\) for which \(N = 100\), giving your answer correct to 3 significant figures.
  3. Find the rate at which the number of plants will be increasing at a time 35 years from now.

AnswerMarks Guidance
(i)State \(A = 42\) B1
State \(k = \frac{1}{5}\)B1 or 0.11 or greater accuracy
Attempt correct process for finding \(m\)M1 involving logarithms or equiv
Obtain \(\frac{1}{5}\ln 2\) or 0.077A1 or 0.08 or greater accuracy
(ii)Attempt solution for \(t\) using either formula M1
Obtain 11.3A1 or greater accuracy; allow 11.3 ≥ 0.1
(iii)Differentiate to obtain form \(Be^{mt}\) M1
Obtain 3.235\(e^{0.077t}\)A1√ or equiv; following their \(A\) and \(m\)
Obtain 47.9A1 allow 48 or greater accuracy
(i) | State $A = 42$ | B1 | |
| State $k = \frac{1}{5}$ | B1 | or 0.11 or greater accuracy |
| Attempt correct process for finding $m$ | M1 | involving logarithms or equiv |
| Obtain $\frac{1}{5}\ln 2$ or 0.077 | A1 | or 0.08 or greater accuracy |

(ii) | Attempt solution for $t$ using either formula | M1 | using correct process (log'ms or T&I or …) |
| Obtain 11.3 | A1 | or greater accuracy; allow 11.3 ≥ 0.1 |

(iii) | Differentiate to obtain form $Be^{mt}$ | M1 | where $B$ is different from $A$ |
| Obtain 3.235$e^{0.077t}$ | A1√ | or equiv; following their $A$ and $m$ |
| Obtain 47.9 | A1 | allow 48 or greater accuracy |
7 It is claimed that the number of plants of a certain species in a particular locality is doubling every 9 years. The number of plants now is 42 . The number of plants is treated as a continuous variable and is denoted by $N$. The number of years from now is denoted by $t$.\\
(i) Two equivalent expressions giving $N$ in terms of $t$ are

$$N = A \times 2 ^ { k t } \quad \text { and } \quad N = A \mathrm { e } ^ { m t } .$$

Determine the value of each of the constants $A , k$ and $m$.\\
(ii) Find the value of $t$ for which $N = 100$, giving your answer correct to 3 significant figures.\\
(iii) Find the rate at which the number of plants will be increasing at a time 35 years from now.

\hfill \mbox{\textit{OCR C3 2008 Q7 [9]}}