AS June 2018 Q2

EdexcelAS paperCurrent spec7 marksNumerical Methods

2. The temperature, \(\theta^\circ\mathrm{C}\), of coffee in a cup, \(t\) minutes after the cup of coffee is put in a room, is modelled by the differential equation

\[\frac{\mathrm{d}\theta}{\mathrm{d}t} = -k(\theta - 20)\]

where \(k\) is a constant.

The coffee has an initial temperature of \(80^\circ\mathrm{C}\)

Using \(k = 0.1\)

(a) use two iterations of the approximation formula \(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_0 = \dfrac{y_1 - y_0}{h}\) to estimate the temperature of the coffee 3 minutes after it was put in the room. (6)

The coffee in a different cup, which also had an initial temperature of \(80^\circ\mathrm{C}\) when it was put in the room, cools more slowly.

(b) Use this information to suggest how the value of \(k\) would need to be changed in the model. (1)

Mark scheme (a)

Scheme	Marks	AO
\(\dfrac{\mathrm{d}\theta}{\mathrm{d}t} = -k(\theta - 20)\), \(k\) is a constant. \(\theta_0 = 80\)
{Two iterations from \(t = 0\) to \(t = 3 \Rightarrow\)} \(h = 1.5\)
Uses \(h = 1.5,\ \theta_0 = 80,\ k = 0.1\) (condone \(k = -0.1\)) in a complete strategy to find a numerical expression for \(\theta_1 = \ldots\)	M1	3.1b
\(\{\theta_0 = 80,\ k = 0.1 \Rightarrow\}\ \left(\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\right)_{0} = -0.1(80 - 20)\ \{= -6\}\)	M1	3.4
\(\left\{\dfrac{\theta_1 - 80}{1.5} = -6 \Rightarrow\right\}\ \theta_1 = 80 + (1.5)(-6)\)	M1	1.1b
\(\theta_1 = 71\)	A1	1.1b
\(\{\theta_1 = 71 \Rightarrow\}\ \left(\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\right)_{1} = -0.1(\text{“}71\text{”} - 20)\ \{= -5.1\}\)	M1	1.1b
\(\theta_2 = 71 + (1.5)(-5.1) = 63.35\ (^\circ\mathrm{C})\)	A1	2.1
	(6)

Notes

M1: See scheme

M1: Uses the model to evaluate the initial value of \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\) using \(k = 0.1\) (condone \(k = -0.1\)) and the initial condition \(\theta_0 = 80\)

M1: Applies the approximation formula with \(\theta_0 = 80\), \(k = 0.1\) (condone \(k = -0.1\)) and their \(h\) to find a numerical expression for \(\theta_1 = \ldots\)

A1: Finds the approximation for \(\theta\) at 1.5 minutes as 71

M1: Uses their 71 and \(k = 0.1\) (condone \(k = -0.1\)) to find \(\dfrac{\mathrm{d}\theta}{\mathrm{d}t}\)

A1: Applies the approximation formula again to give \(63.35\ (^\circ\mathrm{C})\) or awrt \(63\,(^\circ\mathrm{C})\)

Note: \(h = 0.1 \Rightarrow \theta_1 = 79.4,\ \theta_2 = 78.806\);
\(h = 1 \Rightarrow \theta_1 = 74,\ \theta_2 = 68.6\);
\(h = 0.15 \Rightarrow \theta_1 = 79.1,\ \theta_2 = 78.2135\)

Mark scheme (b)

Scheme	Marks	AO
Decrease \(k\) to become a smaller positive value	B1	3.5c
	(1)
	(7 marks)

Notes

B1: See scheme

Note: Allow B1 for “the value of \(k\) should satisfy \(0 \lt k \lt 0.1\)”

Note: Condone “the value of \(k\) would need to be decreased” for B1

Note: Give B0 for “change \(k\) to become negative”