Edexcel C4 2011 January — Question 2 5 marks

Exam BoardEdexcel
ModuleC4 (Core Mathematics 4)
Year2011
SessionJanuary
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicExponential Functions
TypeRate of change in exponential model
DifficultyModerate -0.3 This is a straightforward differentiation of an exponential function requiring knowledge that d/dx(a^x) = a^x ln(a), followed by direct substitution. While it requires careful algebraic manipulation to express the answer in the required form ln(a), it's a standard C4 technique with no conceptual challenges or multi-step problem-solving.
Spec1.07j Differentiate exponentials: e^(kx) and a^(kx)

2. The current, \(I\) amps, in an electric circuit at time \(t\) seconds is given by $$I = 16 - 16 ( 0.5 ) ^ { t } , \quad t \geqslant 0$$ Use differentiation to find the value of \(\frac { \mathrm { d } I } { \mathrm {~d} t }\) when \(t = 3\).
Give your answer in the form \(\ln a\), where \(a\) is a constant.

Question 2:
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(\frac{dI}{dt} = -16\ln(0.5)\cdot 0.5^t\)M1 A1
At \(t=3\): \(\frac{dI}{dt} = -16\ln(0.5)\cdot 0.5^3\)M1
\(= -2\ln 0.5 = \ln 4\)M1 A1
Total[5]
# Question 2:

| Working/Answer | Marks | Guidance |
|---|---|---|
| $\frac{dI}{dt} = -16\ln(0.5)\cdot 0.5^t$ | M1 A1 | |
| At $t=3$: $\frac{dI}{dt} = -16\ln(0.5)\cdot 0.5^3$ | M1 | |
| $= -2\ln 0.5 = \ln 4$ | M1 A1 | |
| **Total** | **[5]** | |

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2. The current, $I$ amps, in an electric circuit at time $t$ seconds is given by

$$I = 16 - 16 ( 0.5 ) ^ { t } , \quad t \geqslant 0$$

Use differentiation to find the value of $\frac { \mathrm { d } I } { \mathrm {~d} t }$ when $t = 3$.\\
Give your answer in the form $\ln a$, where $a$ is a constant.\\

\hfill \mbox{\textit{Edexcel C4 2011 Q2 [5]}}