Question

A water tank has vertical sides and a horizontal rectangular base, as shown in the diagram. The area of the base is $2{m^3}$. At time $t = 0$ the tank is empty and water begins to flow at a rate of $1{m^3}$ per hour. At the same time water begins to flow out from the base at the rate of $0.2\sqrt h {m^3}$ per hour where $hm$is the depth of the water in the tank at time t hours. Form a differential equation satisfied by h and t, and show that the time T hours taken for the depth of water to reach hm is given by $T = \int\limits_0^4 {\dfrac{{10}}{{5 - \sqrt h }}dh}$.

Answer 1

We will write volume as $V = l \times b \times h$. Putting the value of area from the question and we suppose the height to be h. Then we will use the given information of the water tank and the water flowing. Then, we will calculate the change in volume of water and put the value of V. Then, we will integrate and on simplification, we will get the answer.

Complete step by step solution:
The water tank has a rectangular base and vertical sides. Therefore, it has a cuboid shape.
Here, V is the volume of the tank.
l is the length; b is the breadth of the tank and h is the height of the tank.
The volume of the tank is given by
$V = l \times b \times h$
Given that area of the rectangular base is $2{m^3}$.
Therefore, $l \times b = 2{m^3}$
$\Rightarrow V = 2{m^3} \times h$
Hence, we have
$\Rightarrow V = 2h{m^3}$
At time $t = 0$, the tank is empty and water begins to flow at a rate of $1{m^3}$ per hour.
At the same time water begins to flow out from the base at the rate of $0.2\sqrt h {m^3}$ per hour. Now, we will determine the value of $\dfrac{{dV}}{{dt}}$.
Here, $\dfrac{{dV}}{{dt}}$ is the change in volume of water with respect to time.
$\Rightarrow \dfrac{{dV}}{{dt}} = 1 - 0.2\sqrt h$ ....(1)
Given that $V = 2h$
Putting the value of V in equation (1)
$\Rightarrow \dfrac{{d(2h)}}{{dt}} = 1 - 0.2\sqrt h$
On simplification we get,
$\Rightarrow 2\dfrac{{dh}}{{dt}} = 1 - 0.2\sqrt h$
On dividing the equation by 2 we get,
$\Rightarrow \dfrac{{dh}}{{dt}} = \dfrac{{1 - 0.2\sqrt h }}{2}$
On multiplying the equation by dt we get,
$\Rightarrow dh = \dfrac{{1 - 0.2\sqrt h }}{2}dt$
Integrating both sides
$\Rightarrow \int {dt} = 2\int {\dfrac{1}{{1 - 0.2\sqrt h }}dh}$ ....(2)
Let $T = \int {dt}$
Putting the value of T in equation (2)
$\Rightarrow T = 2\int\limits_0^4 {\dfrac{1}{{1 - 0.2\sqrt h }}dh}$
We can also write 0.2 as $\dfrac{1}{5}$ .
$\Rightarrow T = 2\int\limits_0^4 {\dfrac{1}{{1 - \dfrac{1}{5}\sqrt h }}dh}$
On simplification,
$\Rightarrow T = 2\int\limits_0^4 {\dfrac{1}{{\dfrac{{5 - \sqrt h }}{5}}}dh}$
On multiplying the numerator and denominator by 5 we get,
$\Rightarrow T = \int\limits_0^4 {\dfrac{{10}}{{5 - \sqrt h }}dh}$
Hence, proved.

Note:
The tank has cuboid shape, and the volume of cuboid is given by multiplying length, breadth and height of the cuboid.
Therefore, $V = l \times b \times h$.
Here, l is the length, b is the breadth and h are the height of the cuboid.