The volume of a spherical balloon of radius r cm is $V = \frac{4}{3} \pi r^3$ - Edexcel - A-Level Maths Pure - Question 8 - 2006 - Paper 7

Applying the chain rule:

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

Substituting ( \frac{dV}{dr} ) from part (a):

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

To find V, we first integrate the equation:

$V = \int \frac{1000(2t + 1)}{(2t + 1)^2} dt$

This simplifies as:

$V = 1000 \int \frac{1}{(2t + 1)} dt = 1000 \ln(2t + 1) + C$

Given that ( V = 0 ) when ( t = 0 ):

$C = -1000 \ln(1) = 0$

Thus, the final expression is:

$V = 1000 \ln(2t + 1)$

To find the radius r at ( t = 5 ), we calculate:

1. First, substitute ( t = 5 ) into the formula for ( dr ):
   • From part (b): ( \frac{dr}{dt} = \frac{1000}{(2t + 1)} )
   • Therefore, ( \frac{dr}{dt} = \frac{1000}{(2(5) + 1)} = \frac{1000}{11} \approx 90.91 ) cm/s.
2. Now substitute back to find the radius:
   • Given ( V = \frac{4}{3} \pi r^3 ) and substituting ( V = 1000 \ln(11) ) to obtain ( r \approx 4.77 ) cm after performing calculations.

Calculate ( \frac{dr}{dt} ) at ( t = 5 ):

Using the chain rule, we have already established:

$\frac{dr}{dt} = \frac{1000}{(2(5) + 1)}$

First substitute t:

$\frac{dr}{dt} = \frac{1000}{11} \approx 90.91 \, \text{cm/s}$

Now, verify consistency with given values:

1. Calculate for 5 seconds: This gives approximately: $\frac{dr}{dt} = 0.0289 \approx 2.90 \times 10^{-2} \, \text{cm/s}$