Ciarán is preparing food for his baby and must use cooled boiled water - Leaving Cert Mathematics - Question 9 - 2014

At $t = 5$ minutes, we have:

Taking the natural logarithm on both sides yields:

We set:

Solving for $t$ gives:

Thus, it takes approximately 31 minutes.

To sketch $f(t)$, we calculate values of $f(t)$ for specific $t$:

• $t = 0$: $y = 77$
• $t = 1$: $y \approx 73.20$
• $t = 5$: $y \approx 65$
• $t = 10$: $y \approx 54.9$
• $t = 20$: $y \approx 39.1$
• $t = 30$: $y \approx 27.9$
• $t = 60$: $y \approx 6.6$
• $t = 90$: $y \approx 2.6$.

Plot these points to form the curve.

For a faster cooling rate, we want $m < k$. A possible value is $m = -0.05$. This shows a steeper decline than $f(t)$, and it can be labeled clearly on the graph.

One possible value for $m$ is $-0.05$. This value is less than $k$, thus reflecting a faster cooling rate than the original function $f(t)$. Any value of $m < k$ would suffice.

To find the rates of change, we differentiate:

$\frac{dy}{dt} = kAe^{kt}.$

Substituting values:

For $t = 1$ minute:

$y = 77e^{-0.0339 \cdot 1} \Rightarrow \frac{dy}{dt} \approx -2.61.$

For $t = 10$ minutes:

$y = 77e^{-0.0339 \cdot 10} \Rightarrow \frac{dy}{dt} \approx -1.86.$

Given:

$\frac{d^{2}y}{dt^{2}} = A \cdot k^{2}e^{kt}$

With $A > 0$ and $k < 0$, we see that:

$A \cdot k^{2} > 0 \\ \text{therefore, } \frac{d^{2}y}{dt^{2}} > 0,$

indicating that the rate of change is indeed increasing over time.