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)
To find the rate at which the water level is rising, we need to determine how the volume of water in the tank is changing with respect to time.
Let's denote the radius of the hemispherical tank as "r" and the height of the water level as "h". In this case, r = 10m and h = 5m (since the water level is 5m from the bottom of the tank).
First, let's derive the formula for the volume of the water in the tank as a function of h. We can represent the volume of the water as a cap sliced from a sphere:
V(h) = πh^2(3r - h)/3
Now, let's differentiate the volume function with respect to time (t) using implicit differentiation, as we want to find how the volume is changing over time:
dV/dt = (dV/dh) * (dh/dt)
Next, let's find the values of dV/dh and dh/dt.
For dV/dh, we differentiate the volume function V(h) with respect to h:
dV/dh = π(2h)(3r - h)/3 + πh^2(-1)/3
= πh(6r - 2h - 1)/3
For dh/dt, we are given that the water is being filled at a constant rate of 3m^3/min. Therefore, dh/dt = 3m^3/min.
Now we can substitute the values of dV/dh and dh/dt into the equation for dV/dt:
dV/dt = (πh(6r - 2h - 1)/3) * (3)
Simplifying this expression, we get:
dV/dt = πh(6r - 2h - 1)
Now let's substitute the known values of r = 10m and h = 5m into the equation:
dV/dt = π(5)(6(10) - 2(5) - 1)
Finally, calculate the value of dV/dt:
dV/dt = π(5)(60 - 10 - 1)
= π(5)(49)
= 245π
Thus, the water level is rising at a rate of 245π cubic meters per minute when the water level is 5m from the bottom of the tank.