Cuestion 3.9 My score: \( 4 / 6 \mathrm{pts}(66.67 \% \) ) A piston is seated at the top of a cylindrical chamber with radius 6 cm when it starts moving into the chamber at a constant speed of \( 6 \mathrm{~cm} / \mathrm{s} \) (see figure). What is the rate of change of the volume of the cylinder when the piston is 5 cm from the base of the chamber?

To find the rate of change of the volume of the cylinder when the piston is 5 cm from the base, we can use the formula for the volume of a cylinder: \[ V = \pi r^2 h \] where: - \( V \) is the volume, - \( r \) is the radius of the cylinder, - \( h \) is the height of the cylinder (which is the distance from the piston to the base). In this case, the radius \( r \) is given as 6 cm, and the height \( h \) is the distance from the piston to the base of the chamber. When the piston is 5 cm from the base, the height \( h \) is 5 cm. Now, we can differentiate the volume with respect to time \( t \) to find the rate of change of volume: \[ \frac{dV}{dt} = \pi r^2 \frac{dh}{dt} \] Given that the radius \( r = 6 \) cm, we can calculate \( r^2 \): \[ r^2 = 6^2 = 36 \, \text{cm}^2 \] The piston is moving into the chamber at a constant speed of \( 6 \, \text{cm/s} \), which means that the height \( h \) is decreasing at this rate. Therefore, we have: \[ \frac{dh}{dt} = -6 \, \text{cm/s} \] Now we can substitute \( r^2 \) and \( \frac{dh}{dt} \) into the equation for \( \frac{dV}{dt} \): \[ \frac{dV}{dt} = \pi (36) (-6) \] Calculating this gives: \[ \frac{dV}{dt} = -216\pi \, \text{cm}^3/\text{s} \] Thus, the rate of change of the volume of the cylinder when the piston is 5 cm from the base of the chamber is: \[ \frac{dV}{dt} \approx -678.58 \, \text{cm}^3/\text{s} \quad (\text{using } \pi \approx 3.14) \] So, the final answer is: \[ \frac{dV}{dt} = -216\pi \, \text{cm}^3/\text{s} \quad \text{or approximately } -678.58 \, \text{cm}^3/\text{s} \]