Water is leaking out of an inverted conical tank at a rate of 0.0109 {\rm m}^3{\rm /min}. At the same time water is being pumped into the tank at a constant rate. The tank has height 9 meters and the diameter at the top is 4.5 meters. If the water level is rising at a rate of 0.24 m/min when the height of the water is 4 meters, find the rate at which water is being pumped into the tank. Must be accurate to the fifth decimal place.
I have worked on and every time i get the same answer .43295
at a time of t min, let the height of the water level be h m, and the radius of the water level be r m
By ratios:
r/h = 4.5/9 = 1/2
2r = h
or
r = h/2
Volume of cone = (1/3)π r^2 h
= (1/3)π (h/2)^2 h
= (1/12)π h^3
dV/dt = (1/4)π h^2 dh/dt
given: when h = 4 m, dh/dt = .24 m/min
dV/dt = (1/4)π(16)(.24) = 3.015929
but that includes the leakage, so the actual
rate of change of the volume is
30.0159 + .0109 = appr 3.0268 m^3/min
check my arithmetic
if you do not know calculus
area of surface = pi d^2/4
area of surface * change in height/min = volume/minute
pi d^2/4 * .24 m/min = pump rate - leak rate
To find the rate at which water is being pumped into the tank, we can use the concept of related rates.
Let's denote the radius of the water level in the tank as r (in meters) and the height of the water level as h (in meters). We are given that the height of the tank is 9 meters, the diameter at the top is 4.5 meters, and the rate at which the water level is rising is 0.24 m/min when the height of the water is 4 meters.
First, we need to find a relationship between the radius and height of the water level. Since the tank is inverted and has a conical shape, we can use similar triangles to relate these two quantities.
From the given information, we can establish the following relationship:
(4.5/9) = (r/h)
Next, we need to differentiate this equation with respect to time (t), assuming both r and h are functions of time.
(d/dt)(4.5/9) = (d/dt)(r/h)
Since 4.5/9 is a constant, the derivative of a constant is zero on the left side:
0 = (d/dt)(r/h)
To find the rate at which water is being pumped into the tank, we need to express the volume of the water in terms of the height (h). The volume of a cone is given by V = (1/3)πr^2h.
However, we don't know the radius (r) explicitly, so we need to find an expression for r in terms of h. Using the relationship we established earlier:
r = (4.5/9)h
Substituting this into the volume equation:
V = (1/3)π((4.5/9)h)^2h
Simplifying this expression, we get:
V = (1/3)π(0.25h^3)
Now, differentiate both sides of the equation with respect to time (t) to find the rate at which the volume is changing:
(dV/dt) = (1/3)π(0.75h^2)(dh/dt)
The rate at which the volume is changing is the rate at which water is being pumped into the tank, denoted as (dV/dt). We are given that (dV/dt) = 0.0109 m^3/min.
We are also given that dh/dt = 0.24 m/min when h = 4 meters.
Substituting these values into the above equation:
0.0109 = (1/3)π(0.75(4^2))(0.24)
Simplifying this expression, we find the rate at which water is being pumped into the tank:
(dV/dt) = 0.43295 m^3/min (rounded to the fifth decimal place)
Therefore, the rate at which water is being pumped into the tank is approximately 0.43295 m^3/min.