OCR C3 Specimen — Question 5 8 marks

Exam BoardOCR
ModuleC3 (Core Mathematics 3)
SessionSpecimen
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicDifferentiating Transcendental Functions
TypeApplied rate of change
DifficultyModerate -0.3 This is a straightforward applied exponential question requiring standard techniques: evaluating a limit as t→∞, solving an exponential equation using logarithms, and differentiating to find a rate of change. All three parts are routine C3 exercises with no conceptual challenges beyond direct application of learned methods.
Spec1.06g Equations with exponentials: solve a^x = b1.06i Exponential growth/decay: in modelling context1.07k Differentiate trig: sin(kx), cos(kx), tan(kx)

5 At time \(t\) minutes after an oven is switched on, its temperature \(\theta ^ { \circ } \mathrm { C }\) is given by $$\theta = 200 - 180 \mathrm { e } ^ { - 0.1 t }$$
  1. State the value which the oven's temperature approaches after a long time.
  2. Find the time taken for the oven's temperature to reach \(150 ^ { \circ } \mathrm { C }\).
  3. Find the rate at which the temperature is increasing at the instant when the temperature reaches \(150 ^ { \circ } \mathrm { C }\).

AnswerMarks Guidance
(i) \(200°C\)B1 For value 200
(ii) \(150 = 200 - 180e^{-0.1t} \Rightarrow e^{-0.1t} = \frac{50}{180}\) Hence \(-0.1t = \ln\frac{5}{18} \Rightarrow t = 12.8\)M1, M1, A1 For isolating the exponential term; For taking logs correctly; For correct value 12.8 (minutes)
(iii) \(\frac{d\theta}{dt} = 18e^{-0.1t}\) Hence rate is \(18e^{-0.1 \times 12.8} = 5.0°C\) per minuteM1, A1, M1, A1 For differentiation attempt; For correct derivative; For using their value from (ii) in their \(\theta\); For value 5.0(0)
Total for Question 5: 8 marks
**(i)** $200°C$ | B1 | For value 200

**(ii)** $150 = 200 - 180e^{-0.1t} \Rightarrow e^{-0.1t} = \frac{50}{180}$ Hence $-0.1t = \ln\frac{5}{18} \Rightarrow t = 12.8$ | M1, M1, A1 | For isolating the exponential term; For taking logs correctly; For correct value 12.8 (minutes)

**(iii)** $\frac{d\theta}{dt} = 18e^{-0.1t}$ Hence rate is $18e^{-0.1 \times 12.8} = 5.0°C$ per minute | M1, A1, M1, A1 | For differentiation attempt; For correct derivative; For using their value from (ii) in their $\theta$; For value 5.0(0)

**Total for Question 5: 8 marks**

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5 At time $t$ minutes after an oven is switched on, its temperature $\theta ^ { \circ } \mathrm { C }$ is given by

$$\theta = 200 - 180 \mathrm { e } ^ { - 0.1 t }$$

(i) State the value which the oven's temperature approaches after a long time.\\
(ii) Find the time taken for the oven's temperature to reach $150 ^ { \circ } \mathrm { C }$.\\
(iii) Find the rate at which the temperature is increasing at the instant when the temperature reaches $150 ^ { \circ } \mathrm { C }$.

\hfill \mbox{\textit{OCR C3  Q5 [8]}}