Water leaks from a vertical cylindrical tank through a small hole in its base at a rate proportional to the square root of the volume of water remaining. The tank initially contains 125 liters and 17 liters leak out during the first day. A. When will the tank be half empty? \( t= \) B. How much water will remain in the tank after 5 days? volume \( = \) diters

We are told that water leaks from a tank at a rate proportional to the square root of the volume remaining. Denote by \( V(t) \) the volume (in liters) at time \( t \) (in days). Then \[ \frac{dV}{dt} = -k\sqrt{V}, \] where \( k>0 \) is a constant. The negative sign indicates that the volume is decreasing. ### Step 1. Solve the Differential Equation Separate variables: \[ \frac{dV}{\sqrt{V}} = -k\, dt. \] Integrate both sides: \[ \int \frac{dV}{\sqrt{V}} = \int -k\, dt. \] Recall that \[ \int \frac{dV}{\sqrt{V}} = 2\sqrt{V}. \] So we have \[ 2\sqrt{V} = -k\, t + C, \] where \( C \) is a constant. ### Step 2. Use the Initial Condition Initially, at \( t=0 \), the tank contains 125 liters, i.e. \( V(0)=125 \). Substitute: \[ 2\sqrt{125} = C. \] Since \( \sqrt{125} = 5\sqrt{5} \), then \[ C = 10\sqrt{5}. \] Thus the solution becomes \[ 2\sqrt{V} = 10\sqrt{5} - k\, t. \] ### Step 3. Determine the Constant \( k \) We are told that 17 liters leak out during the first day. Therefore, after 1 day the remaining volume is \[ V(1) = 125 - 17 = 108. \] Plug \( t=1 \) and \( V(1)=108 \) into the solution: \[ 2\sqrt{108} = 10\sqrt{5} - k. \] Notice that \[ \sqrt{108} = \sqrt{36\cdot 3} = 6\sqrt{3}, \] so \[ 2\sqrt{108} = 2(6\sqrt{3}) = 12\sqrt{3}. \] Thus \[ 12\sqrt{3} = 10\sqrt{5} - k, \] \[ k = 10\sqrt{5} - 12\sqrt{3}. \] ### Step 4. Part A. When Will the Tank Be Half Empty? “Half empty” means that half of the water has leaked out so that the remaining volume is \[ V = \frac{125}{2} = 62.5 \quad \text{liters}. \] Substitute \( V = 62.5 \) into \[ 2\sqrt{V} = 10\sqrt{5} - k\, t. \] First, compute the left‐hand side: \[ 2\sqrt{62.5} = 2\sqrt{\frac{125}{2}} = 2\frac{\sqrt{125}}{\sqrt{2}} = \frac{2(5\sqrt{5})}{\sqrt{2}} = \frac{10\sqrt{5}}{\sqrt{2}}. \] It is often convenient to rationalize or simplify: \[ \frac{10\sqrt{5}}{\sqrt{2}} = 10\frac{\sqrt{5}}{\sqrt{2}} = 10\sqrt{\frac{5}{2}}. \] So our equation becomes \[ 10\sqrt{\frac{5}{2}} = 10\sqrt{5} - k\, t. \] Now solve for \( t \): \[ k\, t = 10\sqrt{5} - 10\sqrt{\frac{5}{2}}, \] \[ t = \frac{10\sqrt{5} - 10\sqrt{\frac{5}{2}}}{k}. \] Recall \( k=10\sqrt{5}-12\sqrt{3} \). Thus, \[ t = \frac{10\sqrt{5} - 10\sqrt{\dfrac{5}{2}}}{10\sqrt{5}-12\sqrt{3}}. \] This expression represents the time (in days) when the tank is half empty. ### Step 5. Part B. How Much Water Will Remain After 5 Days? Use the general solution already obtained: \[ 2\sqrt{V(5)} = 10\sqrt{5} - k\,(5). \] Substitute \( k=10\sqrt{5}-12\sqrt{3} \): \[ 2\sqrt{V(5)} = 10\sqrt{5} - 5\bigl(10\sqrt{5}-12\sqrt{3}\bigr) = 10\sqrt{5} -50\sqrt{5}+60\sqrt{3}. \] Simplify: \[ 2\sqrt{V(5)} = -40\sqrt{5}+60\sqrt{3}. \] Solve for \( \sqrt{V(5)} \): \[ \sqrt{V(5)} = \frac{-40\sqrt{5}+60\sqrt{3}}{2} = -20\sqrt{5}+30\sqrt{3}. \] Now square both sides to obtain \( V(5) \): \[ V(5) = \bigl(-20\sqrt{5}+30\sqrt{3}\bigr)^2. \] Expanding the square, \[ V(5) = (20\sqrt{5})^2 - 2\,(20\sqrt{5})(30\sqrt{3}) + (30\sqrt{3})^2. \] Compute each term: - \((20\sqrt{5})^2 = 400\cdot 5 = 2000,\) - \(2\,(20\sqrt{5})(30\sqrt{3}) = 1200\sqrt{15},\) - \((30\sqrt{3})^2 = 900\cdot 3 = 2700.\) Thus, \[ V(5) = 2000 - 1200\sqrt{15} + 2700 = 4700 - 1200\sqrt{15}. \] This is an exact expression for the volume in liters after 5 days. ### Final Answers A. The time when the tank is half empty is \[ t = \frac{10\sqrt{5} - 10\sqrt{\frac{5}{2}}}{10\sqrt{5}-12\sqrt{3}}\quad \text{days}. \] B. The volume remaining after 5 days is \[ V(5) = \left(-20\sqrt{5}+30\sqrt{3}\right)^2 = 4700 - 1200\sqrt{15}\quad \text{liters}. \]