Water is flowing at a rate of 50 cubic meters per minute into a holding tank shaped like a cone, sitting vertex down. The tank's base diameter is 40 m and a height of 10 m.
Write an expression for the rate of change of the water level with respect to time, in terms of h (the water's height in the tank).
using similar triangles, the radius of the water surface at height h can be found
r/h = (40/2)/10 = 2
so, r = 2h
v = 1/3 π r^2 h
= π/3 (2h)^2 h
= 4π/3 h^3
dv/dt = 4π h^2 dh/dt
50 = 4π h^2 dh/dt
dh/dt = 25/(2πh^2)
To find the expression for the rate of change of the water level with respect to time in terms of h, we can use the concept of similar triangles.
Let's first find the volume of the cone at any given height h. The volume of a cone is given by the formula:
V = (1/3) * π * r^2 * h,
where V is the volume, π is a constant (approximately 3.14159), r is the radius of the base, and h is the height.
In this case, the base diameter is 40 m, so the radius, r, is half of that, 20 m.
Substituting the values into the formula, we get:
V = (1/3) * π * (20^2) * h,
V = (1/3) * π * 400 * h,
V = (400/3) * π * h.
Now, since the water is flowing into the tank at a rate of 50 cubic meters per minute, we can differentiate both sides of the equation with respect to time t:
dV/dt = d/dt [(400/3) * π * h],
dV/dt = (400/3) * π * (dh/dt).
Therefore, the expression for the rate of change of the water level with respect to time in terms of h is:
dh/dt = (3/400π) * dV/dt.
Since dV/dt is 50 cubic meters per minute, the final expression becomes:
dh/dt = (3/400π) * 50.
Simplifying further:
dh/dt = (3/8π) cubic meters per minute.
So, the expression for the rate of change of the water level with respect to time, in terms of h (the water's height in the tank), is (3/8π) cubic meters per minute.
To find the expression for the rate of change of the water level with respect to time, we can use the concept of similar triangles and the formula for the volume of a cone.
First, let's consider a small change in the height of the water level (∆h) and the corresponding change in the radius of the water level (∆r). Since the tank is shaped like a cone, the change in height and radius would be proportional. Specifically, we can say that ∆r/∆h is equal to the ratio of the radius of the tank base (r) to the height of the tank (h).
Using similar triangles, we know that the ratio of the radius of the water level to the height of the water level will remain constant. Therefore, we have:
∆r/∆h = r/h
To find the rate of change of the water level with respect to time (∆h/∆t), we need to express the volume of the water (∆V) in terms of the rate of change of the water level (∆h/∆t) and the radius of the tank base (r).
The volume of a cone is given by the formula:
V = 1/3 * π * r^2 * h
Taking the derivative with respect to time (t) on both sides of the equation, we get:
dV/dt = 1/3 * π * (2rh * dh/dt + r^2 * dh/dt)
Since the rate of change of the water level (∆h/∆t) can be represented as dh/dt, we can rewrite the equation as:
dV/dt = π * (2rh * dh/dt + r^2 * dh/dt)/3
Substituting the value of r/h from our earlier equation (∆r/∆h = r/h), we have:
dV/dt = π * (2(r/h) * h * dh/dt + r^2 * dh/dt)/3
Simplifying further, we get:
dV/dt = π * [(2r + r^2/h) * dh/dt]/3
Finally, since the rate of change of the water level with respect to time (∆h/∆t) is equal to dh/dt, the expression becomes:
dV/dt = π * [(2r + r^2/h) * ∆h/∆t]/3
Therefore, the expression for the rate of change of the water level with respect to time, in terms of h (the water's height in the tank), is:
dV/dt = π * [(2r + r^2/h) * ∆h/∆t]/3