The volume of a spherical balloon of radius $ r $ cm is $ V = \frac{4}{3} \pi r^3 $ - Edexcel - A-Level Maths Pure - Question 1 - 2006 - Paper 7

Using the chain rule:

$\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$

Substituting the expressions we found:

$\frac{dV}{dt} = 4\pi r^2 \cdot \frac{1000}{(2t+1)}$

To solve for $r$:

Integrate:

$r = \int \frac{1000}{(2t + 1)} dt = 500 \ln(2t + 1) + c$

Using the initial condition $V = 0$ when $t = 0$, we can find $c$:

When $t = 0$, $r = 0 \Rightarrow 0 = 500 \ln(1) + c \Rightarrow c = 0$.

Thus, we have:

$r = 500 \ln(2t + 1)$

Next, substitute $r$ back into the volume equation to get:

$V = \frac{4}{3} \pi (500 \ln(2t + 1))^3$

Substituting $t = 5$ into our expression for $r$:

$r = 500 \ln(2(5) + 1) = 500 \ln(11) \approx 500 \times 2.3979 \approx 1198.95 \; \text{cm} \approx 1200 \; \text{cm}$

At $t = 5$:

Substituting into the equation for $\frac{dr}{dt}$:

$\frac{dr}{dt} = \frac{1000}{(2(5) + 1)} = \frac{1000}{11} \approx 90.909 \; \text{cm s}^{-1}$

Now to convert this to the rate of increase of the radius:

$\frac{dr}{dt} \approx 0.029 \; \text{cm s}^{-1}$

Thus, we have:

$\frac{dr}{dt} \approx 2.90 \times 10^{-3} \; \text{cm s}^{-1}$