Question

A swimming pool is 20 ft wide, 40 ft long, 3 ft deep at the shallow end, and 9 ft deep at its deepest point. A cross-section is shown in the figure. If the pool is being filled at a rate of 0.8 ft^3/min, how fast is the water level rising when the depth at the deepest point is 5 ft?

Answer 1

This problem can be solved using differentiation, where two relatedrates of change (here the volume with the depth of water) are usedto find the answer.. let v= volume, t= time, x=depth of water at the deep end. Now, the rate at which the pool fills can be given as a function oftime, where, dv/dt= 0.8 cubic ft per minute. Using the chain rule while differentiating, dv/dx * dx/dt= dv/dt = 0.8 cubic ft per minute. please takethe time to rate me. Thanks! similarly, here, dx/dt is the rate at which the water level isrising... so now we have to find dv/dx. This can be done bydifferentiating the volume of the water in the pool with respect tox. For simplicity, we can just consider the bottom right angledtriangle of the pool, having sides 20 ft and depth of 5ft (assumingthat the pool is a trapezoid), and so the water hasn't reached theshallow end. let A be the angle formed between the vertical wall and the slopingbottom of the pool. (between the base of the triangle and thehypotenuse of the triangle) Using tanA=5/20, we can find that A=tan^-1(1/4)=14.04°(accuracy might be lost, it may be left as tan^-1(1/4)) (tan^-1(1/4) is the inverse function of tan, the one on acalculator to find the angle.) Any volume of water would form another smaller triangle, and sousing the value of the angle found above, A, we can calculate thewidth of the water level at a given depth of x. tanA=x/width therefore, width = x/tanA = x/tan14.04° Now, the volume of water for an given depth x would be v= 1/2 * x * x/tan14.04° * 40 cubic ft. =20(x^2)/ tan14.04° Now, dv/dx= 20 * 2 * x^(2-1) / tan14.04° =40x/tan14.04 Now, when x= 5 ft, then dv/dx=40*5/tan14.04 = 200/tan14.04° (tonot loose accuracy). replacing this value in the equation, dv/dx * dx/dt= dv/dt = 0.8 cubic ft per minute. or, 200/tan14.04° * dx/dt = 0.8 or, dx/dt= 0.8 *tan 14.04°/200= 0.001 ft per minute.

Answer 2

V = 20(12h + h/2*h*tan69.44 + h/2*h*tan45)
V = 20(12h + h^2/2(tan69.44 + tan45))
dV/dt = 20(12*dh/dt + h(tan69.44 + tan45)*dh/dt)
0.8 = 20(12*dh/dt + 5(tan69.44 + tan45)*dh/dt)
0.04 = 12*dh/dt + 18.3305*dh/dt
0.04 = 30.3305*dh/dt
dh/dt = ? ft/min