A rectangular water tank is being filled at a constant rate of 101t/s.The base of the tank has width,w=10m and length,l=15m. If the volume of the tank is given by v=w*l*h where h represents the height of the tank,what is the rate of the height of water in the tank?
A rectangular water tank is being filled at the constant rate of 35lt/s. The base of the tank has width,w= 3m and length,l= 5m. If the volume of the tank is given by V=w×l×h where h representsthe height of the tank, what is the rate of change of the height of water in the tank?.
can this be calculated further
To find the rate of the height of water in the tank, we need to differentiate the volume equation with respect to time.
Given:
Rate of filling the tank = 101 m^3/s
Width of the tank, w = 10 m
Length of the tank, l = 15 m
Volume of the tank, V = w * l * h
Differentiating the volume equation with respect to time(t), we get:
dV/dt = (d/dt)(w * l * h)
Since the width (w) and length (l) of the tank are constant values, their derivatives with respect to time will be zero.
Therefore, the equation becomes:
dV/dt = (d/dt)(h * w * l)
Now, substituting the values:
101 = (d/dt)(h * 10 * 15)
We have the constant values of width and length, so we can simplify further:
101 = 150 * (d/dt)h
Finally, solving for (d/dt)h, the rate of the height of water in the tank, we get:
(d/dt)h = 101 / 150
= 0.6733 m/s
So, the rate of the height of water in the tank is approximately 0.6733 m/s.
not sure what "101t/s" is supposed to mean. If that's tons/second, then you will have to figure the volume per ton. I'll give you a sample calculation.
h = v/(lw) = v/150
so if v is increasing by 101 m^3/s, h is increasing by 101/150 m/s