C3 January 2011 Q4

EdexcelOld spec8 marksLogs & Exponentials Modelling

4. Joan brings a cup of hot tea into a room and places the cup on a table. At time \(t\) minutes after Joan places the cup on the table, the temperature, \(\theta\,{}^\circ\mathrm{C}\), of the tea is modelled by the equation

\[\theta = 20 + A\mathrm{e}^{-kt},\]

where \(A\) and \(k\) are positive constants.

Given that the initial temperature of the tea was \(90^\circ\mathrm{C}\),

(a) find the value of \(A\). (2)

The tea takes 5 minutes to decrease in temperature from \(90^\circ\mathrm{C}\) to \(55^\circ\mathrm{C}\).

(b) Show that \(k = \dfrac{1}{5}\ln 2\). (3)

(c) Find the rate at which the temperature of the tea is decreasing at the instant when \(t = 10\). Give your answer, in \(^\circ\mathrm{C}\) per minute, to 3 decimal places. (3)

Mark scheme (a)

Scheme	Marks
\(\theta = 20 + A\mathrm{e}^{-kt}\) (eqn \(*\))
\(\{t = 0, \theta = 90 \Rightarrow\}\ \ 90 = 20 + A\mathrm{e}^{-k(0)}\)	M1
\(90 = 20 + A \Rightarrow \underline{A = 70}\)	A1
	(2)

Notes

M1: Substitutes \(t = 0\) and \(\theta = 90\) into eqn \(*\)

A1: \(\underline{A = 70}\)

Mark scheme (b)

Scheme	Marks
\(\theta = 20 + 70\mathrm{e}^{-kt}\)
\(\{t = 5, \theta = 55 \Rightarrow\}\ \ 55 = 20 + 70\mathrm{e}^{-k(5)}\) \(\dfrac{35}{70} = \mathrm{e}^{-5k}\)	M1
\(\ln\left(\tfrac{35}{70}\right) = -5k\)	dM1
\(-5k = \ln\left(\tfrac{1}{2}\right)\)
\(-5k = \ln 1 - \ln 2 \Rightarrow -5k = -\ln 2 \Rightarrow \underline{k = \tfrac{1}{5}\ln 2}\)	A1 *
	(3)

Notes

M1: Substitutes \(t = 5\) and \(\theta = 55\) into eqn \(*\) and rearranges eqn \(*\) to make \(\mathrm{e}^{\pm 5k}\) the subject.

dM1: Takes ‘lns’ and proceeds to make ‘\(\pm 5k\)’ the subject.

A1 *: Convincing proof that \(k = \tfrac{1}{5}\ln 2\)

Mark scheme (c)

Scheme	Marks
\(\theta = 20 + 70\mathrm{e}^{-\frac{1}{5}t\ln 2}\)
\(\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = -\dfrac{1}{5}\ln 2.(70)\mathrm{e}^{-\frac{1}{5}t\ln 2}\)	M1 A1 oe
When \(t = 10\), \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = -14\ln 2\,\mathrm{e}^{-2\ln 2}\)
\(\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = -\dfrac{7}{2}\ln 2 = -2.426015132\ldots\)
Rate of decrease of \(\theta = 2.426\ {}^\circ C/\mathrm{min}\) (3 dp.)	A1
	(3)
	(8 marks)

Notes

M1: \(\pm\alpha\mathrm{e}^{-kt}\) where \(k = \tfrac{1}{5}\ln 2\)

A1 oe: \(-14\ln 2\,\mathrm{e}^{-\frac{1}{5}t\ln 2}\)

A1: awrt \(\pm 2.426\)