Toricelli’s Law. Suppose that a water tank has a hole with
area a at its bottom and that water is draining from the hole.
Let y(t) (in feet) and V (t ) (in cubic feet) denote the depth
and the volume of water in the tank at time t (in seconds).
Then (under ideal conditions) the velocity of the stream of
water exiting the tank will be
v=
2 gy
( g ≈ 32 ft / s2 )
and so
dV
= − a 2 gy
dt
If A(y) is the cross-sectional area of the tank at depth y, then
this equation takes the form
A( y )
dy
= − a 2 gy
dt
Note. If, as indicated above, we take g ≈ 32 ft / s2 then the
equation becomes
A( y )
dy
= −8a y
dt
#27, p419 A tank shaped like a vertical cylinder contains water to a depth of 9
ft (see figure). A bottom plug is pulled at time t = 0. After 1 h the depth has
dropped to 4 ft. How long will it take for all of the water to drain from the
tank?
Solution. In this problem the cross-sectional
area of the tank is constant
A( y ) = πr 2
and Toricelli’s equation becomes
(π r 2 )
dy
= − 8a y
dt
This is a separable differential equation, which can be solved by first
separating the variables
(π r 2 )
dy
= −8adt
y
and then integrating both sides independently:
(π r 2 ) ∫
dy
= − 8a ∫ dt
y
(π
π r 2 )2 y = − 8at + C
Note. Since in this problem we are only concerned with the value of t
corresponding to y = 0 (the time when the tank is empty), there is little to be
gained by writing the solution in explicit form.
y= 0 ⇔
− 8a + C = 0 ⇔ t =
C
8a
What we need are the values of the constants C and 8a, and this information is
obtained by “fitting” the solution
2π r 2 y = −8at + C
to the conditions of the problem.
πr 2
t = 0, y = 9 ⇒ 2π r 2 9 = −0 + C ⇒ C = 6π
2π
π r 2 y = − 8at + 6π
πr 2
2
To maintain consistency of units (since g ≈ 32 ft / s ) we should use time
measured in seconds rather than hours. Thus the second condition should be
entered as follows.
π r 2 4 = − 8a( 3600) + 6π
πr 2
t = 3600 s , y = 4 ⇒ 2π
⇒ 8a(3600) = 2π
πr 2
πr 2
⇒ 8a =
1800
Finally, as indicated previously, the tank will be empty ( y
t=
= 0) when
C
2 1800
= (6πr )( 2 ) = 6(1800) = 10, 800 s
8a
πr
(or
t = 3h )