1. **Problem Statement:** We have water flowing from an inverted conical tank with the rate of change of water height given by $$\frac{dx}{dt} = -0.6\pi r^2 \sqrt{2g} \frac{\sqrt{x}}{A}$$ where $r=0.1$ ft, $g=32.1$ ft/s$^2$, $x$ is the water height, and $A$ is the cross-sectional area at height $x$. Initial water level is $x_0=8$ ft. 2. **Given:** - $r=0.1$ ft - $g=32.1$ ft/s$^2$ - Initial height $x_0=8$ ft - Initial volume $V=\frac{1}{3}xA=512\frac{\pi}{3}$ ft$^3$ - Step size $h=20$ s - Time to find $x$ after $t=60$ s (1 min) 3. **Find:** (a) Water level after 1 min using RK-4 method with $h=20$ s. (b) Whether the tank empties within 1 min. 4. **Step 1: Express $A$ as a function of $x$** From volume formula $V=\frac{1}{3}xA=512\frac{\pi}{3}$ at $x=8$ ft, $$512\frac{\pi}{3} = \frac{1}{3} \times 8 \times A \implies A = \frac{512\pi}{8} = 64\pi$$ Since $A$ depends on $x$, and volume formula is $V=\frac{1}{3}xA$, for a cone, cross-sectional area varies as $A(x) = kx^2$ for some constant $k$. Using $A(8) = 64\pi$, $$k = \frac{64\pi}{8^2} = \frac{64\pi}{64} = \pi$$ So, $$A(x) = \pi x^2$$ 5. **Step 2: Write the differential equation explicitly:** $$\frac{dx}{dt} = -0.6 \pi (0.1)^2 \sqrt{2 \times 32.1} \frac{\sqrt{x}}{\pi x^2} = -0.6 \pi \times 0.01 \times \sqrt{64.2} \frac{\sqrt{x}}{\pi x^2}$$ Cancel $\pi$: $$= -0.6 \times 0.01 \times \sqrt{64.2} \frac{\sqrt{x}}{x^2} = -0.006 \times 8.012 \times x^{-3/2} = -0.04807 x^{-3/2}$$ 6. **Step 3: Define function $f(t,x)$:** $$f(t,x) = \frac{dx}{dt} = -0.04807 x^{-3/2}$$ 7. **Step 4: Apply RK-4 method with $h=20$ s, initial $x_0=8$ ft, to find $x$ at $t=20,40,60$ s:** For each step: $$k_1 = h f(t_n, x_n)$$ $$k_2 = h f(t_n + \frac{h}{2}, x_n + \frac{k_1}{2})$$ $$k_3 = h f(t_n + \frac{h}{2}, x_n + \frac{k_2}{2})$$ $$k_4 = h f(t_n + h, x_n + k_3)$$ $$x_{n+1} = x_n + \frac{1}{6}(k_1 + 2k_2 + 2k_3 + k_4)$$ **Step 4.1: Step 1 (0 to 20 s):** - $x_0=8$ - $k_1 = 20 \times (-0.04807 \times 8^{-3/2}) = 20 \times (-0.04807 \times 8^{-1.5})$ Calculate $8^{-1.5} = 8^{-1} \times 8^{-0.5} = \frac{1}{8} \times \frac{1}{\sqrt{8}} = \frac{1}{8} \times \frac{1}{2.8284} = 0.0442$ So, $$k_1 = 20 \times (-0.04807 \times 0.0442) = 20 \times (-0.002124) = -0.04248$$ - $k_2 = 20 \times f(10, 8 - 0.04248/2) = 20 \times f(10, 7.97876)$ Calculate $7.97876^{-1.5}$: $7.97876^{-1} = 0.1253$, $\sqrt{7.97876} = 2.825$, so $7.97876^{-1.5} = 0.1253 \times \frac{1}{2.825} = 0.04435$ $$k_2 = 20 \times (-0.04807 \times 0.04435) = -0.0426$$ - $k_3 = 20 \times f(10, 8 - 0.0426/2) = 20 \times f(10, 7.9787)$ Same as $k_2$, so $k_3 = -0.0426$ - $k_4 = 20 \times f(20, 8 - 0.0426) = 20 \times f(20, 7.9574)$ Calculate $7.9574^{-1.5}$: $7.9574^{-1} = 0.1257$, $\sqrt{7.9574} = 2.822$, so $7.9574^{-1.5} = 0.1257 \times \frac{1}{2.822} = 0.0445$ $$k_4 = 20 \times (-0.04807 \times 0.0445) = -0.0428$$ - Update: $$x_1 = 8 + \frac{1}{6}(-0.04248 + 2(-0.0426) + 2(-0.0426) + (-0.0428)) = 8 + \frac{1}{6}(-0.04248 -0.0852 -0.0852 -0.0428)$$ $$= 8 + \frac{1}{6}(-0.25568) = 8 - 0.04261 = 7.9574$$ **Step 4.2: Step 2 (20 to 40 s):** Repeat with $x_1=7.9574$ - $k_1 = 20 \times (-0.04807 \times 7.9574^{-1.5}) = -0.0428$ - $k_2 = 20 \times f(30, 7.9574 - 0.0428/2) = 20 \times f(30, 7.935)$ Calculate $7.935^{-1.5} = 0.0447$ $$k_2 = -0.043$$ - $k_3 = 20 \times f(30, 7.9574 - 0.043/2) = -0.043$$ - $k_4 = 20 \times f(40, 7.9574 - 0.043) = 20 \times f(40, 7.9144)$ Calculate $7.9144^{-1.5} = 0.0449$ $$k_4 = -0.0432$$ - Update: $$x_2 = 7.9574 + \frac{1}{6}(-0.0428 -0.086 -0.086 -0.0432) = 7.9574 - 0.043 = 7.9144$$ **Step 4.3: Step 3 (40 to 60 s):** Repeat with $x_2=7.9144$ - $k_1 = 20 \times (-0.04807 \times 7.9144^{-1.5}) = -0.0432$ - $k_2 = 20 \times f(50, 7.9144 - 0.0432/2) = 20 \times f(50, 7.883)$ Calculate $7.883^{-1.5} = 0.0451$ $$k_2 = -0.0434$$ - $k_3 = 20 \times f(50, 7.9144 - 0.0434/2) = -0.0434$$ - $k_4 = 20 \times f(60, 7.9144 - 0.0434) = 20 \times f(60, 7.871)$ Calculate $7.871^{-1.5} = 0.0452$ $$k_4 = -0.0435$$ - Update: $$x_3 = 7.9144 + \frac{1}{6}(-0.0432 -0.0868 -0.0868 -0.0435) = 7.9144 - 0.0434 = 7.871$$ 8. **Step 5: Conclusion:** (a) Water level after 1 min is approximately $\boxed{7.87}$ ft. (b) Since water level decreased from 8 ft to about 7.87 ft in 1 min, the tank will not be empty during the first minute. **Final answer:** $$x(60) \approx 7.87 \text{ ft}, \quad \text{tank not empty in first minute}$$