Differentiation of exponential and logarithmic functions — Question 155

Original QCE Vault practice · 5 marks

Q155 · Practice questionTechnology-activeComplex unfamiliar5 marks

QUESTION 155 (5 marks)

A contaminant concentration is modelled by

C(t)=Ae^{-kt}

, where

A>0

k>0

and

t

is measured in hours. Measurements give

C(2)=36

and

C(8)=15

Determine

A

and

k

, and hence determine the instantaneous rate of change

C'(5)

WORKED SOLUTION

5 marks

ANSWER

k=\dfrac{\ln(36/15)}6\approx0.1459

A\approx48.20

and

C'(5)\approx-3.39

concentration units/hour.

Worked solution

Dividing the two measurements gives

36/15=e^{6k}

, so

k=\ln(36/15)/6

. Use

36=Ae^{-2k}

to find

A

. Since

C'(t)=-kAe^{-kt}=-kC(t)

, evaluate at

t=5

Eliminates

A

using the two measurements.

[1 mark]

Determines

k

[1 mark]

Determines

A

[1 mark]

Differentiates the concentration model.

[1 mark]

Evaluates

C'(5)

[1 mark]

Practice question aligned to the current QCAA syllabus; review the worked solution and marking criteria.

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