A rectangular trough is 8ft long, 2ft across the top, and 4 feet deep. if water flows in at a rate of 2 ft 3 /min, how fast is the surface rising when the water is 1 ft deep?
let V be the current volume of the trough and H the current height of the water
then we have
V = 16H
taking derivative in terms of time we get
dV/dt = 16 dH/dt
we know that dV/dt = 2
thus
dH/dt = 2/16 = 1/8
thus the water is rising at
1/8 ft per minute
or
1.5 inches per minute
The rising rate of water is \(\frac{1}{8} ft/min\)
Let us consider that V is the volume of rectangular tank and h is the current height of the water
Given that, A rectangular trough is 8ft long, 2ft across the top, and 4 feet deep.
So that,
\(V = 8*2*h\\\\V=16h\)
Differentiate volume with respect to time.
\(\frac{dV}{dt}=16\frac{dh}{dt} \\\\\frac{dV}{dt}=2ft^{3}/min\\ \\ \frac{dh}{dt}=\frac{2}{16}=\frac{1}{8} ft/min.\)
Thus, the rising rate of water is \(\frac{1}{8} ft/min\)
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