EXM.1.AHL.TZ0.15

pestleMathematicsAIHLPaper 1EXM· sl-4-4-pearsons-scatter-diagrams-eqn-of-y-on-xsource ↗

Adesh wants to model the cooling of a metal rod. He heats the rod and records its temperature as it cools.

He believes the temperature can be modeled by $T\left( t \right) = a{{\text{e}}^{bt}} + 25$ , where $a{\text{,}}\,\,b \in \mathbb{R}$ .

Hence

Show that ${\text{ln}}\left( {T - 25} \right) = bt + {\text{ln}}\,a$ .

[2]

Find the equation of the regression line of ${\text{ln}}\left( {T - 25} \right)$ on $t$ .

[3]

find the value of $a$ and of $b$ .

[3]

c.i.

predict the temperature of the metal rod after 3 minutes.

[2]

c.ii.

Markscheme / solution

${\text{ln}}\left( {T - 25} \right) = {\text{ln}}\left( {a{{\text{e}}^{bt}}} \right)$ M1

${\text{ln}}\left( {T - 25} \right) = {\text{ln}}\,a + {\text{ln}}\left( {{{\text{e}}^{bt}}} \right)$ A1

${\text{ln}}\left( {T - 25} \right) = bt + {\text{ln}}\,a$ AG

[2 marks]

${\text{ln}}\left( {T - 25} \right) = - 0.00870t + 3.89$ M1A1A1

[3 marks]

$b = - 0.00870$ A1

$a = {e^{3.89...}} = 49.1$ M1A1

[3 marks]

c.i.

$T\left( {180} \right) = 49.1{e^{ - 0.00870\left( {180} \right)}} + 25 = 35.2$ M1A1

[2 marks]

c.ii.

Examiners’ report

[N/A]

c.i.

[N/A]

c.ii.