The volume \( V \) of a spherical balloon with radius \( r \) is given by the formula \( V = \frac{4}{3} \pi r^3 \).
To find the rate of change of the volume with respect to time, you take the derivative of the volume with respect to the radius and then multiply by the rate of change of the radius with respect to time. Mathematically, this is expressed as \( \frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} \). Given \( V = \frac{4}{3} \pi r^3 \), we have \( \frac{dV}{dr} = 4 \pi r^2 \), so \( \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \).
If the radius \( r \) of the balloon is increasing at a constant rate, this can be expressed as \( \frac{dr}{dt} = k \), where \( k \) is a constant.
The surface area \( A \) of a spherical balloon is given by \( A = 4 \pi r^2 \). To find the rate of change of the surface area with respect to time, take the derivative of \( A \) with respect to \( r \) and multiply by \( \frac{dr}{dt} \). Thus, \( \frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt} = 8 \pi r \cdot \frac{dr}{dt} \).
As the radius \( r \) increases, the rate of change of the volume \( \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \) also increases because \( 4 \pi r^2 \) becomes larger. This means that the volume of the balloon increases more rapidly as it gets bigger.