Question

the bottom of a rectangular swimming tank is 25 m by 40 m water is pumped into the tank at the rate of 500 cubic metres per minute find the rate at which the level of water in the tank is rising​

Answer 1

Given:

The dimensions of the bottom of the rectangular swimming tank:

Length = 40 m

Breadth = 25 m

The water is pumped into the tank at the rate of 500 cubic meters per minute

To find:

The rate at which the level of water in the tank is rising​

Solution:

We know the formula to calculate the volume of a cuboid is given as:

$\boxed{\boxed{\bold{Volume\:of\: a \: cuboid = length \times breadth \times height}}}$

Let "h" represent the height of the rectangular swimming tank.

Since the water being pumped in the tank is constantly changing and rising, so we can say that

The rate of change of volume of water in the tank = $\frac{dV}{dt} = 500 \:m^3/min$

and

Also, the rate of change of the level of rising water in the tank = $\frac{dh}{dt}$

Now, using the formula of the cuboid above, we can write

The volume of the rectangular tank = length × breadth × height

⇒ $V = 40 \times 25 \times h$

differentiating w.r.t "t" on both sides

⇒ $\frac{dV}{dt} = \frac{d(40\times 25\times h)}{dt}$

here the length & breadth of the tank remains constant

⇒ $\frac{dV}{dt} = 40\times 25\times\frac{dh}{dt}$

substituting $\frac{dV}{dt} = 500 \:m^3/min$

⇒ $500 = 40 \times 25 \times \frac{dh}{dt}$

⇒ $\frac{dh}{dt} = \frac{500}{40 \times 25}$

⇒ $\frac{dh}{dt} = \frac{20}{40}$

⇒ $\frac{dh}{dt} = \frac{1}{2}$

⇒ $\bold{\frac{dh}{dt} = 0.5\: m/min}$

Thus, the rate at which the level of water in the swimming tank is rising​ is 0.5 m/min.

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