Home / IB DP Math MAI SL : IB Style Mock Exams – Set 3 Paper 2

IB DP Math MAI SL : IB Style Mock Exams – Set 3 Paper 2

Question

The average daytime temperature for a city is given by the function $D(t)=5 \cos (30 t)^{\circ}+20^{\circ} \mathrm{C}$, where $t$ is the time in months after January.

a Sketch the graph of $D$ against $t$ for $0 \leqslant t \leqslant 24$.

b Find the average daytime temperature during May.

c Find the minimum average daytime temperature, and the month in which it occurs.

▶️Answer/Explanation

Sol:

a

For $D(t)=5 \cos (30 t)^{\circ}+20$ :

  • the amplitude is 5
  • the period is $\frac{360}{30}=12$ months
  • the principal axis is $D=20$.

b

May is 4 months after January.
When
$
\begin{aligned}
t=4, \quad D & =5 \times \cos 120^{\circ}+20 \\
& =5 \times\left(-\frac{1}{2}\right)+20 \\
& =17.5
\end{aligned}
$
So, the average daytime temperature during May is $17.5^{\circ} \mathrm{C}$.

c

The minimum average daytime temperature is $20-5=15^{\circ} \mathrm{C}$, which occurs when $t=6$ or 18 .
So, the minimum average daytime temperature occurs during July.

 

Question

As a hot air balloon is inflated, the volume of air inside it after $t$ minutes is given by $V=2 t^3-3 t^2+10 t+2 \mathrm{~m}^3$ where $0 \leqslant t \leqslant 8$.

Find:
a.  the initial volume of air in the balloon

b.  the volume at $t=8$ minutes

c.  $\frac{d V}{d t}$ and explain what it means

d.  the rate of increase in volume at $t=4$ minutes.

▶️Answer/Explanation

Sol:

a

When $t=0, V=2 \mathrm{~m}^3$
Initially there were $2 \mathrm{~m}^3$ of air in the balloon.
b

When $t=8, \quad V=2(8)^3-3(8)^2+10(8)+2$
$
=914 \mathrm{~m}^3
$

After 8 minutes there were $914 \mathrm{~m}^3$ of air in the balloon.

c

$\frac{d V}{d t}=6 t^2-6 t+10 \mathrm{~m}^3 \mathrm{~min}^{-1}$
This function tells us the rate at which the volume of air in the balloon is increasing after $t$ minutes.

d

When $t=4, \frac{d V}{d t}=6(4)^2-6(4)+10$
$
=82 \mathrm{~m}^3 \mathrm{~min}^{-1}
$

Since $\frac{d V}{d t}>0, V$ is increasing.

The volume of air in the balloon is increasing at $82 \mathrm{~m}^3 \mathrm{~min}^{-1}$ at $t=4$ minutes.

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