An industrial tank is used to store liquid chemicals. The tank is cylindrical with a total height of 210 centimeters. The liquid is drained through an outlet valve located 10 centimeters above the base of the tank. At time ttt hours after the valve is opened, the depth of liquid, hhh centimeters, decreases at a rate which is proportional to h−10h - 10h−10.
Initially, the tank is completely full, and the depth of the liquid is decreasing at a rate of 4 centimeters per hour.
Show that
dhdt=−0.02(h−10) \frac{dh}{dt} = -0.02(h - 10) dtdh=−0.02(h−10)Solve the differential equation
dhdt=−0.02(h−10) \frac{dh}{dt} = -0.02(h - 10) dtdh=−0.02(h−10)to find an expression for hhh in terms of ttt.
Find the time taken for the depth of the liquid to reach 110 centimeters. Give your answer to the nearest hour.
