Let's break this down step by step: --- ### **Given Data** - **Hemisphere Radius:** \( R = 10 \) ft - **Water pumped in at rate:** \( r \) ft³/min - **Evaporation rate:** Proportional to the surface area, i.e., \( kA \) (where \( k = 0.01 \)) - **Volume of water when height is \( h \):** \( V(h) = 10\pi h^2 - \frac{\pi}{3}h^3 \) --- ## **1. Differential Equation for Rate of Change of Volume** \[ \frac{dV}{dt} = \text{(rate pumped in)} - \text{(rate lost to evaporation)} \] - **Rate pumped in:** \( r \) - **Rate lost to evaporation:** \( kA \), where \( A \) is the surface area of the water (a circle of radius \( a \) at height \( h \)) - **Area of surface:** \( A = \pi a^2 \) For a hemisphere of radius \( R \), at height \( h \): \[ a^2 = 2Rh - h^2 \] So, \[ A = \pi (2Rh - h^2) \] **Plug into differential equation:** \[ \frac{dV}{dt} = r - k\pi(2Rh - h^2) \] But \( V \) is also a function of \( h \): \[ V(h) = 10\pi h^2 - \frac{\pi}{3}h^3 \] So, \[ \frac{dV}{dt} = \frac{dV}{dh} \frac{dh}{dt} \] \[ \frac{dV}{dh} = 20\pi h - \pi h^2 \] Therefore, \[ (20\pi h - \pi h^2)\frac{dh}{dt} = r - k\pi(2Rh - h^2) \] --- ## **2. Express Area of Surface in Terms of \( h \)** Already done above: \[ A = \pi (2Rh - h^2) \] --- ## **3. Solve the Differential Equation** \[ (20\pi h - \pi h^2)\frac{dh}{dt} = r - k\pi(2Rh - h^2) \] \[ \frac{dh}{dt} = \frac{r - k\pi(2Rh - h^2)}{20\pi h - \pi h^2} \] Let’s use the given values: \( R = 10 \), \( k = 0.01 \): \[ \frac{dh}{dt} = \frac{r - 0.01\pi(20h - h^2)}{20\pi h - \pi h^2} \] Let’s denote \( r \) as the input rate. If not specified, leave as \( r \). --- ## **4. No Evaporation: Time to Fill the Tank** If \( k = 0 \), then: \[ \frac{dV}{dt} = r \implies V(t) = rt \] Full volume of hemisphere: \[ V_{full} = \frac{2}{3}\pi R^3 = \frac{2}{3}\pi (10)^3 = \frac{2000}{3}\pi \text{ ft}^3 \] So, \[ rt_{fill} = \frac{2000}{3}\pi \implies t_{fill} = \frac{2000}{3}\frac{\pi}{r} \] --- ## **5. With Evaporation: Depth at Time Found in 4, Will Tank Ever Fill?** At steady state (\(\frac{dh}{dt} = 0\)): \[ r - k\pi(2Rh - h^2) = 0 \implies r = k\pi(2Rh - h^2) \] Solve for \( h \): \[ 2Rh - h^2 = \frac{r}{k\pi} \] \[ h^2 - 2Rh + \frac{r}{k\pi} = 0 \] \[ h = R \pm \sqrt{R^2 - \frac{r}{k\pi}} \] Since \( h \leq 2R \), pick the valid root. **Will the tank ever fill?** The maximum evaporation rate is when \( h = R \): \[ 2Rh - h^2 \leq 2R^2 - R^2 = R^2 \] So, if \( r < k\pi R^2 \), the tank will **never fill** because the evaporation rate will match or exceed the input rate as the surface area grows. **Depth of water at time from question 4:** At \( t_{fill} \), use the solution for \( h(t) \) from the solved differential equation (numerically, if needed) and plug in the time. --- ### **Summary of Answers** 1. **Differential Equation:** \[ (20\pi h - \pi h^2)\frac{dh}{dt} = r - k\pi(2Rh - h^2) \] 2. **Area of Surface:** \[ A = \pi (2Rh - h^2) \] 3. **Solution:** \[ \frac{dh}{dt} = \frac{r - k\pi(2Rh - h^2)}{20\pi h - \pi h^2} \] (Solve/graph numerically for given \( r \), \( k \), \( R \)). 4. **Time to Fill (No Evaporation):** \[ t_{fill} = \frac{2000}{3}\frac{\pi}{r} \] 5. **With Evaporation:** - The tank will **never fill** unless \( r > k\pi(2Rh - h^2) \) for all \( h \leq R \). - **Maximum depth** is: \[ h = R - \sqrt{R^2 - \frac{r}{k\pi}} \] (for valid values; otherwise, max possible for the rate). --- **If you provide a value for \( r \), I can compute the exact numbers and graph as needed!**