A water tank has a shape of an inverted cone with a base radius of 2m and a height of 4 m. If water is being pumped into the tank at a rate of 2m3/min, then find the rate at which the water level is rising when the water is 3m deep.
Attempted solution:
V= 1/3πr^2h
V= 1/3π(2h/4)^2 h
V= 1/3π (4h^2/16) h
V= 4/48 π h^3
Dv/Dt = 4/48 π h^3 dh/dt
2= 4/48 π 2^3dh/dt
2(48)/4π x 8 = dh/dt
dh/dt = 3π m/min
Is this the correct solution, or did I go wrong?
I do not understand that.
Your attempted solution is almost correct, but there is a minor error in your calculation.
Let's go through the solution step by step to find the rate at which the water level is rising when the water is 3m deep.
Given data:
Base radius of the inverted cone = 2m
Height of the inverted cone = 4m
Rate of water being pumped into the tank = 2m^3/min
First, let's find the volume of the water in the tank:
Volume of an inverted cone = (1/3) * π * r^2 * h
Substituting the values:
V = (1/3) * π * (2^2) * h
V = (4/3) * π * h
Now, we need to find the rate at which the water level is rising, i.e., dh/dt, when the water depth is 3m. To find this, we need to differentiate the volume equation with respect to time:
dV/dt = (4/3) * π * dh/dt
Given that dV/dt = 2m^3/min (as water is being pumped into the tank at that rate), and the water depth h = 3m, we can substitute these values into the equation:
2 = (4/3) * π * dh/dt * 3
Now, solve for dh/dt:
dh/dt = 2 * 3 / ((4/3) * π)
dh/dt = 6 / (4/3) * π
dh/dt = (6 * 3) / 4π
dh/dt = 18 / (4π)
dh/dt ≈ 18 / 12.57
Therefore, the rate at which the water level is rising when the water is 3m deep is approximately 1.43 m/min.
So, your attempted solution was almost correct, but there was an error in calculating the value of dh/dt.
You were ok to here:
V= 4/48 π h^3
V = π/12 h^3
But that means
dV/dt = π/12 * 3h^2 dh/dt
= π/4 h^2 dh/dt
Now plug in dV/dt=2, h=3, and you get
π/12 * 9 dh/dt = 2
dh/dt = 8/3π ft/min
if the point of the cone is up, then
volume= PI*2^2 *4 when full
so the volume of the cone filling up is
volume=full-PI* radius of top^2*distance to top^2, and if h is the distance from the bottom to the top of the water,
V = full- PI ((4=h)/2)^2* (4-h)
= full-PI/4 * (4-h)^3 check that
then figure out
dV/dt from this function, you know dv/dt, solve for dh/dt