A rectangular swimming pool is 16 ft wide by 28 ft long by 8 ft deep. Fresh water is being pumped into the pool at the rate of 196 cubic ft/min. Old water is being pumped out of the pool at the rate of 50root(h) cubic ft/min, where h is the current height of the water.
a) Give an expression in terms of h for the rate of change of the pool's height.
b) What is the rate of change of the height when the pool is one quarter full?
c) If the volume is changing at 25 cubic ft/min, what must the height be?
Please show work and answers, THANK YOU!!
a) Give an expression in terms of h for the rate of change of the pool's height.
b) What is the rate of change of the height when the pool is one quarter full?
c) If the volume is changing at 25 cubic ft/min, what must the height be?
Please show work and answers, THANK YOU!!
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V = 16*28*h ft³ = 448h for 0 ≤ h ≤ 8, h in ft.
a)
From the given,
dV/dt = IN - OUT = 196 - 50 √h
b)
dV/dt = (dV/dh)(dh/dt) so dh/dt = (dV/dt) / (dV/dh)
Previously, we obtained V = 448h, so dV/dh = 448, and so
dh/dt = (196 - 50 √h)/448
If the pool is ¼ full, then h = 8/4 = 2 ft. Evaluate dh/dt at this value of h.
c)
Go back to (a) for a relationship between dV/dt and h:
dV/dt = 196 - 50 √h
You are asked to find the value of h when dV/dt = 25. You can do that too.
a)
From the given,
dV/dt = IN - OUT = 196 - 50 √h
b)
dV/dt = (dV/dh)(dh/dt) so dh/dt = (dV/dt) / (dV/dh)
Previously, we obtained V = 448h, so dV/dh = 448, and so
dh/dt = (196 - 50 √h)/448
If the pool is ¼ full, then h = 8/4 = 2 ft. Evaluate dh/dt at this value of h.
c)
Go back to (a) for a relationship between dV/dt and h:
dV/dt = 196 - 50 √h
You are asked to find the value of h when dV/dt = 25. You can do that too.