Answer:
[tex]\frac{dh}{dt} = -3.48 \times 10^{-3}\sqrt{h}[/tex] feet per second is the differential equation
Step-by-step explanation:
Given:
The radius of the cylindrical tank= 2 feet
The height of the cylindrical tank = 10 feet
The radius of the circular hole = 3/4 inches
To Find:
The differential equation for the height h of the water at time t.
Solution:
Finding the surface area(A) of the tank
Surface area = [tex]\pi R^2[/tex]
On substituting the values
Surface area =[tex]\pi(2)^2[/tex]
= [tex]4\pi[/tex]square feet
Finding the surface area(a) of the hole
The radius is given in inches, so converting into feet we have
1 inch = 0.083 foot
similarly
[tex]\frac{3}{4} = 0.75 inches = 0.75 \times 0.083[/tex] = 0.0625 feet.
Now the surface area,
= [tex]\pi \times 0.0625[/tex]
= [tex]0.0625 \pi[/tex] square feet
Now let the velocity of water through the hole is v
According law of conservation of energy, the penitential energy due to the height h of the water gets converted into kinetic energy.
[tex]\frac{1}{2}mv^2 =mgh[/tex]
[tex]v^2 = \frac{2mgh}{m}[/tex]
[tex]v^2 = 2gh[/tex]
[tex]v= \sqrt{2gh}[/tex]
The rate of water flowing through the hole is = [tex]a\times v[/tex]
= >[tex]a \times \sqrt{2gh}[/tex]
At any time t
[tex]V(t) = A \times h(t)[/tex]
[tex]\frac{dV}{dt} = -a\sqrt{2gh}[/tex]
[tex]\frac{d(Ah(t))}{dt} = -a\sqrt{2gh}[/tex]
[tex]A \frac{d(h(t))}{d(t)} = -a\sqrt{2gh}[/tex]
[tex]\frac{dh}{dt} = -\frac{a}{A} \sqrt{2gh}[/tex]
On substituting the values, we get
[tex]\frac{dh}{dt} = -\frac{0.0625\pi}{4\pi}\sqrt{2\times 32 \times 10[/tex]
[tex]\frac{dh}{dt} = -3.48 \times 10^{-3}\sqrt{h}[/tex] feet per second
