Show that $T_0=100$.

Show that $k=\frac{1}{10}\ln \frac{10}{7}$.

Find the temperature of the water when $t=15$.

Sketch the graph of $T$ versus $t$, clearly indicating any asymptotes with their equations and stating the coordinates of any points of intersection with the axes.

Find the time taken for the water to have a temperature of $50 °\text{C}$. Give your answer correct to the nearest second.

The model for the temperature of the water can also be expressed in the form $T=T_0{a}^{\frac{t}{10}}$ for $t\ge 0$ and $a$ is a positive constant.

Find the exact value of $a$.

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

when $t=0, T=100\Rightarrow 100=T_0{\text{e}}^0$         A1

so $T_0=100$         AG

[1 mark]

correct substitution of $t=10, T=70$         M1

$70=100{\text{e}}^{-10k}$  or  ${\text{e}}^{-10k}\mathrm{=}\frac{7}{10}$

EITHER

$-10k=\ln \frac{7}{10}$         A1

$\ln \frac{7}{10}=-\ln \frac{10}{7}$  or  $-\ln \frac{7}{10}=\ln \frac{10}{7}$         A1

OR

${\text{e}}^{10k}\mathrm{=}\frac{10}{7}$         A1

$10k=\ln \frac{10}{7}$         A1

THEN

$k=\frac{1}{10}\ln \frac{10}{7}$         AG

[3 marks]

substitutes $t=15$ into $T$         (M1)

$T=58.6 \left(°\text{C}\right)$         A1

[2 marks]

a decreasing exponential         A1

starting at $\left(0, 100\right)$ labelled on the graph or stated         A1

$T\to 0$ as $t\to \infty$         A1

horizontal asymptote $T=0$ labelled on the graph or stated         A1

Note: Award A0 for stating $y=0$ as the horizontal asymptote.

[4 marks]

$100{\text{e}}^{-kt}=50$  where $k=\frac{1}{10}\ln \frac{10}{7}$        A1

EITHER

uses an appropriate graph to attempt to solve for $t$         (M1)

OR

manipulates logs to attempt to solve for $t$ e.g. $\ln \frac{1}{2}=\left(-\frac{1}{10}\ln \frac{10}{7}\right)t$         (M1)

$t=\frac{\ln 2}{{\displaystyle \frac{1}{10}}\ln {\displaystyle \frac{10}{7}}}=19.433\dots$        A1

THEN

temperature will be $50 °\text{C}$ after $19$ minutes and $26$ seconds        A1

[4 marks]

METHOD 1

substitutes $T_0=100$, $t=10$ and $T=70$ into $T=T_0{a}^{\frac{t}{10}}$         (M1)

$70=100a^{\frac{10}{10}}$        A1

$a=\frac{7}{10}$        A1

METHOD 2

$100a^{\frac{t}{10}}=100{\text{e}}^{-kt}$  where $k=\frac{1}{10}\ln \frac{10}{7}$

EITHER

${\text{e}}^{-k}=a^{\frac{1}{10}}\Rightarrow a={\text{e}}^{-10k}$         (M1)

OR

$a={\left({\text{e}}^{\left(-\frac{1}{10}\ln \frac{10}{7}\right)t}\right)}^{\frac{10}{t}}$         (M1)

THEN

$a={\text{e}}^{-\ln \frac{10}{7}} \left(={\text{e}}^{\ln \frac{7}{10}}\right)$        A1

$a=\frac{7}{10}$        A1

[3 marks]