A large spherical balloon is deflating - Edexcel - A-Level Maths Pure - Question 16 - 2020 - Paper 1

We start with the rearranged differential equation:

$\frac{dr}{dt} = -\frac{k}{r^{2}}$

Separating variables leads to:

$r^{2} dr = -k dt$

Integrating both sides:

$\int r^{2} dr = -k \int dt$

This results in:

$\frac{r^{3}}{3} = -kt + C$

Where C is the constant of integration. We find C using the initial condition:

When (t = 0), (r = 40 \text{ cm}). Thus:

$\frac{(40)^{3}}{3} = C \Rightarrow C = \frac{64000}{3}$

Now substituting C back in gives:

$\frac{r^{3}}{3} = -kt + \frac{64000}{3}$

By multiplying through by 3, we arrive at:

$r^{3} = -3kt + 64000$

From this, we can express the equation linking r and t as:

$r^{3} + 3kt - 64000 = 0$

To find the limits on t, we note that at t = 5 seconds, the radius reduces to 20 cm. Plugging (r = 20\text{ cm}) into the derived equation:

$20^{3} = -3k \cdot 5 + \frac{64000}{3}$

Calculating (20^{3}) gives 8000, thus:

$8000 = -15k + \frac{64000}{3}$

To find k, we can express it in terms of t, ensuring that the balloon remains inflated. The expression must be positive, leading to a condition on t:

The balloon continuously deflates until:

$t \leq \frac{40}{7} \text{ seconds}$

Therefore, the limitation on t must satisfy:

$t < 40/7 ext{ seconds}$