A hemispherical tank with a radius of 10m is filled from an input pipe at a rate of 3m^3/min. How fast is the water level rising when the water level is 5m from the bottom of the tank? (hint: the volume of the cap of thickness of h sliced from a sphere of radius r is pih^2(3r-h)/3)
v = π/3 h^2 (3r-h) = πrh^2 - π/3 h^3
dv/dt = 2πrh dh/dt - πh^2 dh/dt
= πh(2r-h) dh/dt
Pluggin in your numbers, then, we have
π*5(20-5) dh/dt = 3
dh/dt = 3/(75π) = 0.021 m/min
Well, well, well, what do we have here? A question about a hemispherical tank? That's quite spherical of you to ask!
Now, let's approach this problem with a splash of humor, shall we? We have a hemispherical tank with a radius of 10m. So, picture a giant half of a watermelon, but a bit larger and not quite as tasty.
We know that the tank is being filled at a rate of 3m^3/min. That's like pouring a gallon of water into your tank every second – quite a flow, I must say!
The hint gives us a formula for the volume of the cap sliced from a sphere. But instead of dissecting spheres like a science experiment gone horribly wrong, let's focus on the task at hand.
We want to find how fast the water level is rising when it is 5m from the bottom of the tank. Essentially, we want to find dh/dt, which represents the rate at which the height changes over time.
To tackle this task, we can start by noting that the volume of the water in the tank is changing over time. And thankfully, we have a formula for the volume of a cap! It's πh^2(3r - h)/3. So, let's take the derivative of this volume formula with respect to time, t.
dV/dt = π(2h)(dh/dt)(3r - h)/3
Now, we know that the water level is 5m from the bottom of the tank, so our h is 10 - 5 = 5m. And we already know that dV/dt is 3m^3/min.
Plugging in these values, we get:
3 = π(2(5))(dh/dt)(3(10) - 5)/3
Math, math, math... cancel, cancel, cancel... simplify, simplify, simplify...
3 = 10π(dh/dt)(25)
3/250π = dh/dt
And there you have it! The rate at which the water level is rising when it is 5m from the bottom of the tank is approximately 0.0038 m/min. So, the water level is rising hilariously slowly, roughly like a snail doing the moonwalk.
I hope this amusing answer has quenched your thirst for knowledge about hemispherical tanks!
To find how fast the water level is rising, we need to determine the rate at which the volume of water in the tank is changing with respect to time.
Let's denote:
- V as the volume of water in the tank
- r as the radius of the tank (which is 10m)
- h as the distance between the water level and the bottom of the tank
We know that the rate of change of volume with respect to time is given by dV/dt.
To find dV/dt, we can use the volume formula of the cap sliced from a sphere:
V = (πh^2(3r - h))/3
Differentiate both sides of the equation with respect to time (t):
dV/dt = d/dt [(πh^2(3r - h))/3]
Now, we need to consider the relationship between the radius, height, and the rate at which the water level changes.
Since the tank is hemispherical, the radius and height are related by the equation:
r^2 = h(2r - h)
Differentiate both sides of this equation with respect to time (t), remembering that r is constant:
2r(dr/dt) = (dh/dt)(2r - h) + h(d(2r - h)/dt)
We are looking for the rate of change of the water level, dh/dt, when h = 5m and dr/dt = 3m^3/min.
Plug in the values: r = 10m, h = 5m, and dr/dt = 3m^3/min.
2(10)(3) = (dh/dt)(2(10) - 5) + 5(d(2(10) - 5)/dt)
60 = (dh/dt)(15) + 5(d(20 - 5)/dt)
60 = 15(dh/dt) + 5(15)
60 = 15(dh/dt) + 75
15(dh/dt) = -15
(dh/dt) = -1 m/min
Therefore, the water level is decreasing at a rate of 1 meter per minute when the water level is 5 meters from the bottom of the tank.
To find the rate at which the water level is rising, we need to calculate the derivative of the water volume with respect to time. Let's break down the problem step by step.
1. First, let's identify the known values:
- Radius of the hemispherical tank (r) = 10m
- Rate of water input (V') = 3m^3/min
- Distance from the bottom of the tank to the water level (h) = 5m
2. To find the volume of the water in the tank, we need to subtract the volume of the empty space (air) from the total volume of the hemisphere. This can be calculated using the formula you provided:
- Volume of the empty space (air) = (πh^2(3r - h))/3
- Volume of the water (V) = Total volume of the hemisphere - Volume of the empty space (air)
= (2/3)πr^3 - (πh^2(3r - h))/3
3. Now, we differentiate the volume of the water (V) with respect to time (t) to find the rate of change of the water volume:
dV/dt = d/dt[(2/3)πr^3 - (πh^2(3r - h))/3]
dV/dt = (2/3)π(3r^2)(dr/dt) - π(2h(3r - h))(dh/dt)/3
4. Since we are interested in finding the rate at which the water level is rising (dh/dt), we can rearrange the equation and solve for it:
dh/dt = [3(2/3)πr^2(dr/dt) - (2h(3r - h))(dV/dt)] / [-π(2(3r - h))/3]
dh/dt = (2r^2(dr/dt) - h(3r - h)(dV/dt)) / (2(3r - h))
5. Plugging in the given values:
- r = 10m
- dr/dt = 0 (The radius of the tank is constant)
- h = 5m
- dV/dt = 3m^3/min
6. Substituting these values, we can calculate the rate at which the water level is rising:
dh/dt = (2(10^2)(0) - 5(3(10) - 5)(3))/(2(3(10) - 5))
dh/dt = (-75)/(50)
dh/dt = -1.5 m/min
Therefore, the water level is falling at a rate of 1.5 meters per minute when the water level is 5 meters from the bottom of the tank. Note that the negative sign indicates a downward direction.