2019 H2 Maths Paper 1 Q11 — Worked Solution

Part (i)(a)

6 marks

Let $\theta$ be the temperature of the water, in $^{\circ}C$ , $t$ minutes after it is placed in the room. Since the rate of decrease of $\theta$ is proportional to the difference between $\theta$ and the room temperature $16\,{}^{\circ}C$ ,

\frac{d\theta}{dt} = - k(\theta - 16),

where $k$ is a positive constant.

Since the water is cooling towards $16\,{}^{\circ}C$ from above, $\theta > 16$ throughout, so $\theta - 16 > 0$ . Separating variables,

\int\frac{1}{\theta - 16}\, d\theta = \int - k\, dt

\ln(\theta - 16) = - kt + C.

So

\theta - 16 = e^{- kt + C}

= Ae^{- kt},

where $A = e^{C}$ .

Using the initial condition $\theta = 80$ when $t = 0$ ,

80 - 16 = Ae^{0}

A = 64.

So $\theta = 16 + 64e^{- kt}$ . Using the given condition $\theta = 32$ when $t = 30$ ,

32 - 16 = 64e^{- 30k}

e^{- 30k} = \frac{1}{4}

- 30k = \ln\frac{1}{4}

k = \frac{\ln 4}{30} = \frac{\ln 2}{15}.

$\therefore$

\theta = 16 + 64e^{- \frac{\ln 2}{15}t}

as an exact function of $t$ .

Part (i)(b)

1 mark

\theta(45) = 16 + 64e^{- \frac{\ln 2}{15}(45)}

= 16 + 64e^{- 3\ln 2}

= 16 + 64\left( 2^{- 3} \right)

= 16 + 8

= 24.

$\therefore$ the temperature of the water after $45$ minutes is $24\,{}^{\circ}C$ .

Part (ii)

6 marks

Approach

“Rate of change of thickness inversely proportional to thickness” gives $\frac{dT}{dt} = \frac{c}{T}$ . This is separable, and $T = 0$ at $t = 0$ pins down the constant of integration exactly (no logarithm is involved, since the variable being integrated is $T$ itself, not $T - \text{constant}$ ).

Let $T$ cm be the thickness of the ice $t$ minutes after freezing commences. Since the rate of change of $T$ is inversely proportional to $T$ ,

\frac{dT}{dt} = \frac{c}{T},

where $c$ is a positive constant. Separating variables,

\int T\, dT = \int c\, dt

\frac{T^{2}}{2} = ct + D.

Using $T = 0$ when $t = 0$ ,

0 = c(0) + D

D = 0.

So $T^{2} = 2ct$ . Using the given condition $T = 1$ when $t = 60$ ,

1 = 2c(60)

c = \frac{1}{120}.

So

T^{2} = \frac{t}{60},

and since $T \geq 0$ ,

T = \sqrt{\frac{t}{60}}.

The ice is safe to skate on once $T = 3$ :

3 = \sqrt{\frac{t}{60}}

9 = \frac{t}{60}

t = 540.

To check your answer:

T(60)^{2} = 60/60 = 1

, matching the given data point, and

T(540) = \sqrt{540/60} = \sqrt{9} = 3

. ✓

$\therefore$ the ice is first safe to skate on $540$ minutes (i.e. $9$ hours) after freezing commences.

2019 H2 Maths Paper 1, Question 11

Part (i)(a)

Part (i)(b)

Part (ii)

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