Matched Problem 4 Helium is pumped into a spherical balloon at a constant rate of \( 800 \mathrm{~cm}^{3} \) per minute. How fast is the radius increasing at the instant the radius is 30 cm ?

To solve this problem, we need to relate the volume of the balloon to its radius and then find the rate at which the radius is increasing when the radius is 30 cm. 1. **Known Conditions:** - The volume \( V \) of a sphere is given by the formula: \[ V = \frac{4}{3} \pi r^3 \] - The rate at which helium is pumped into the balloon is \( \frac{dV}{dt} = 800 \, \text{cm}^3/\text{min} \). - We need to find \( \frac{dr}{dt} \) when \( r = 30 \, \text{cm} \). 2. **Differentiate the Volume Formula:** We will differentiate the volume with respect to time \( t \): \[ \frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3} \pi r^3 \right) \] Using the chain rule, we get: \[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \] 3. **Substituting Known Values:** Now we can substitute \( \frac{dV}{dt} = 800 \, \text{cm}^3/\text{min} \) and \( r = 30 \, \text{cm} \) into the differentiated equation: \[ 800 = 4 \pi (30^2) \frac{dr}{dt} \] 4. **Calculate \( 30^2 \):** \[ 30^2 = 900 \] So the equation becomes: \[ 800 = 4 \pi (900) \frac{dr}{dt} \] 5. **Simplifying the Equation:** \[ 800 = 3600 \pi \frac{dr}{dt} \] 6. **Solving for \( \frac{dr}{dt} \):** \[ \frac{dr}{dt} = \frac{800}{3600 \pi} \] 7. **Calculating the Value:** Now we will calculate \( \frac{dr}{dt} \): \[ \frac{dr}{dt} = \frac{800}{3600 \pi} = \frac{2}{9 \pi} \] Now, let's compute the numerical value of \( \frac{dr}{dt} \). Using the approximation \( \pi \approx 3.14159 \): \[ \frac{dr}{dt} \approx \frac{2}{9 \times 3.14159} \approx \frac{2}{28.27431} \approx 0.0707 \, \text{cm/min} \] Thus, the radius of the balloon is increasing at approximately \( 0.0707 \, \text{cm/min} \) when the radius is 30 cm.