Water is flowing at a rate of 10 m3/s into the cylindrical tank shown below.
a) Find the volume V of the water as a function of the water level h.
b) Find h as a function of the time t during which water has been flowing into the tank.
height of cylinder: 80m
height of water: ?
diameter of cylinder: 40m
V = pi r^2 h
dV/dt = pi r^2 dh/dt = 10 m^3/s
so
dh/dt = 10/(pi r^2)
h = 0 + [ 10/(pi r^2) ] t
Andrew’s car starting at 12:00 noon travels west at a speed of 30 kph. Andrei’s car starting from the same point at 2:00 pm travels north at 45 kph. Find how fast the two are separating at 4:oopm.
To find the volume of the water as a function of the water level, we need to determine the relationship between the height of the water level and the volume of water in the cylindrical tank.
a) The volume V of a cylinder is given by V = πr^2h, where r is the radius of the cylinder and h is the height. In this case, we are given the diameter of the cylinder, which is 40m. The radius can be calculated by dividing the diameter by 2, so the radius (r) is 40m / 2 = 20m.
Now we can express the volume V as a function of the water level h. Since the water level can vary from 0 to the height of the cylinder (80m), we have:
V(h) = πr^2h
Substituting the value of r = 20m, the function becomes:
V(h) = π(20m)^2h
Simplifying further, we get:
V(h) = 400πh
b) To find the water level h as a function of the time t during which water has been flowing into the tank, we need to relate the rate of water flow to the change in volume over time. Since the water is flowing at a rate of 10 m^3/s, we can express this as:
dV/dt = 10 m^3/s
We know that the volume V is a function of the water level h, so we can write dV/dt as dh/dt * dV/dh using the chain rule of differentiation:
dh/dt * dV/dh = 10 m^3/s
We can rearrange this equation to solve for dh/dt:
dh/dt = 10 / dV/dh
We can differentiate the function for V(h) with respect to h to find dV/dh:
dV/dh = 400π
Substituting this value into the equation:
dh/dt = 10 / 400π
Simplifying further:
dh/dt = 1 / (40π)
Therefore, the height of the water level, h, as a function of the time t is given by:
h(t) = (1 / (40π)) * t + C
Where C is the constant of integration, representing the initial height of the water level.
To find the volume V of the water as a function of the water level h, we can use the formula for the volume of a cylinder. The volume of a cylinder is given by the formula V = πr^2h, where r is the radius of the cylinder and h is the height of the water.
In this case, the diameter of the cylinder is given as 40m, so the radius (r) can be calculated by dividing the diameter by 2. So, r = 40m / 2 = 20m.
Now, the height of the water (h) is the variable we want to find. We know that the height of the cylinder is 80m, so we can write the equation as:
V = π(20^2)h
Since the water is flowing into the tank at a rate of 10 m^3/s, we need to express the volume as a function of time (t). The volume is increasing at a constant rate, so we can write:
V = 10t
Equating both equations, we get:
π(20^2)h = 10t
Simplifying further, we have:
h = (10t) / (π * 400)
Therefore, the volume V of the water as a function of the water level h is given by V = π(20^2)h, and the height of the water h as a function of the time t is h = (10t) / (π * 400).
To find h as a function of the time t during which water has been flowing into the tank, we can use the equation h = (10t) / (π * 400). By plugging in the value of t, we can calculate the corresponding value of h.