Question

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?

Answer 1

 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

Answer 2

The rising rate of water is $\frac{1}{8} ft/min$

Differentiation :

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 $\frac{1}{8} ft/min$

Learn more about the differentiation here:

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