Along reservoir has the shape of a right circular cone having a radius of 40m at the top and a height of 10m at the centre. it is being filled at a constant rate of 40 m^3/min.Find the rate at which the water level is rising when the height is 5m?
let the height of the water be h m
let the radius of the water level be r m
Make a sketch of the cross-section.
By similar triangles:
r/h = 40/10
r = 4h or h = r/4
V = (1/3)π r^2 h
= (1/3)π(4h)^2 h
= (16/3)π h^3
dV/dt = 16π h^2 dh/dt
plug in the given values and calculate
To find the rate at which the water level is rising when the height is 5m, we need to use related rates and the formula for the volume of a cone.
The formula for the volume of a cone is given by:
V = (1/3) π r^2 h
Given:
The radius of the top of the cone, r = 40m
The height of the cone, h = 10m
We are also given that the reservoir is being filled at a constant rate of 40 m^3/min, which means that the volume, V, of the cone is increasing at a constant rate of 40 m^3/min.
To find the rate at which the water level is rising when the height is 5m, we need to find dh/dt when h = 5m.
Now, we can differentiate both sides of the volume formula with respect to time, t, using the chain rule:
dV/dt = (1/3) π [(d/dt) (r^2) h + r^2 (d/dt) h]
Since the radius, r, is constant, we can simplify the equation:
dV/dt = (1/3) π (2r) (dh/dt)
We know that dV/dt = 40 m^3/min and r = 40m.
So, we have:
40 = (1/3) π (2*40) (dh/dt)
Now we can solve for dh/dt by rearranging the equation:
dh/dt = (40 * 3) / (π * 80)
Simplifying further, we get:
dh/dt = 3/2π
Therefore, when the height is 5m, the rate at which the water level is rising is 3/2π m/min.