A conical tank with height is 10 and radius is 5 is being filled with water at 4m^3/s. Solve how fast is the water when = 3m
find radius of surface as a function of height h
r/h = 5/10 = .5
then surface area = pi r^2 = pi (.25 h^2)
change in volume dv = surface area * dh
dv = pi (.25 h^2) dh
and
dv/dt = .25 pi h^2 dh/dt
4 = .25 pi (9) dh/dt
solve for dh/dt
To solve this problem, we can use similar triangles and apply the related rates concept. Let's denote the radius of the water in the conical tank as "r" and the height of the water as "h" at a particular time. We are given that the height of the tank is 10 and the radius is 5.
First, let's set up a proportion between the height of the water "h" and the radius of the water "r" at any time:
h/r = 10/5
Next, we need to differentiate both sides of the equation with respect to time "t" since the rates are changing with time.
d(h)/dt / d(r)/dt = d(10/5)/dt
To find the rate of change of the water level (dh/dt) when the height is 3m, we need to find the rate of change of the radius (dr/dt) at that specific moment.
Given that water is being filled at a constant rate of 4 m^3/s, we know that dV/dt = 4.
The volume of a cone can be calculated using the formula V = (1/3)πr^2h.
Differentiating the volume formula with respect to time will give us the following equation:
dV/dt = (1/3)π (2rh)(dr/dt) + (1/3)πr^2(dh/dt)
Since the tank is being filled, dh/dt = 4 m^3/s.
Substituting the values we have:
4 = (1/3)π (2(5)h)(dr/dt) + (1/3)π(5^2)(4)
We still need to find the value of dr/dt when h = 3. To do this, we can use the proportion we set up earlier. At h = 3:
3/r = 10/5
Cross-multiplying gives us:
5h = 10r
Substituting h = 3:
5(3) = 10r
15 = 10r
r = 15/10
r = 1.5 m
Now, we have the values of r and h. Substituting them back into the volume equation:
4 = (1/3)π(2(1.5)(3))(dr/dt) + (1/3)π(1.5^2)(4)
4 = (1/3)π(9)(dr/dt) + (1/3)π(2.25)(4)
4 = (3π)(dr/dt) + (3π)
Simplifying further:
4 - 3π = 3π(dr/dt)
To isolate dr/dt, we divide both sides by 3π:
(dr/dt) = (4 - 3π) / (3π)
Now, we can substitute the value of π (approximately 3.14) into the equation to find the rate of change of the radius when h = 3:
(dr/dt) = (4 - 3(3.14)) / (3(3.14))
(dr/dt) = (4 - 9.42) / (9.42)
(dr/dt) = -5.42 / 9.42
(dr/dt) ≈ -0.576 m/s
Therefore, the rate at which the water level is changing when the height is 3 m is approximately -0.576 m/s. The negative sign indicates that the water level is decreasing.