To solve the problem of how fast the radius of a hemisphere is changing when its radius is 6 m, given that the volume is increasing at a steady rate of \(18\pi \, \text{m}^3/\text{s}\), we will follow these steps: ### Step 1: Understand the Volume of a Hemisphere The formula for the volume \(V\) of a hemisphere with radius \(r\) is given by: \[ V = \frac{2}{3} \pi r^3 \] ### Step 2: Differentiate the Volume with Respect to Time To find how the radius changes with respect to time, we need to differentiate the volume with respect to time \(t\). Using the chain rule, we have: \[ \frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} \] First, we need to find \(\frac{dV}{dr}\): \[ \frac{dV}{dr} = 2\pi r^2 \] Now substituting this into our differentiation equation gives: \[ \frac{dV}{dt} = 2\pi r^2 \cdot \frac{dr}{dt} \] ### Step 3: Substitute Known Values We know that \(\frac{dV}{dt} = 18\pi \, \text{m}^3/\text{s}\) and we want to find \(\frac{dr}{dt}\) when \(r = 6 \, \text{m}\). Substituting \(r = 6\) into the equation: \[ \frac{dV}{dt} = 2\pi (6^2) \cdot \frac{dr}{dt} \] Calculating \(6^2\): \[ 6^2 = 36 \] So we have: \[ \frac{dV}{dt} = 2\pi (36) \cdot \frac{dr}{dt} = 72\pi \cdot \frac{dr}{dt} \] ### Step 4: Set Up the Equation Now we can set the two expressions for \(\frac{dV}{dt}\) equal to each other: \[ 18\pi = 72\pi \cdot \frac{dr}{dt} \] ### Step 5: Solve for \(\frac{dr}{dt}\) To isolate \(\frac{dr}{dt}\), we can divide both sides by \(72\pi\): \[ \frac{dr}{dt} = \frac{18\pi}{72\pi} \] The \(\pi\) cancels out: \[ \frac{dr}{dt} = \frac{18}{72} = \frac{1}{4} = 0.25 \, \text{m/s} \] ### Conclusion Thus, the rate at which the radius is changing when the radius is 6 m is: **C. 0.25 m/s** ### Explanation of Other Options - **A. 2.50 m/s**: This value is too high given the volume increase rate. The calculations show that the radius cannot change that quickly with the given volume increase. - **B. 2.00 m/s**: Similar to option A, this is also too high. The relationship between volume and radius indicates a much slower change in radius. - **D. 0.20 m/s**: This is lower than the calculated rate of 0.25 m/s. The volume increase rate supports a larger change in radius than this option suggests. ### Revision Summary - The volume of a hemisphere is given by \(V = \frac{2}{3} \pi r^3\). - Differentiate the volume with respect to time to relate volume change to radius change. - Use the chain rule to find \(\frac{dr}{dt}\) when given \(\frac{dV}{dt}\). - The correct answer for the rate of change of the radius when \(r = 6 \, \text{m}\) is \(0.25 \, \text{m/s}\).