reservoir has the shape of a right-circular cone. The altitude is 10 feet, and the radius of the base is 4 ft. Water is poured into the reservoir at a constant rate of 5 cubic feet per minute. How fast is the water level rising when the depth of the water is 5 feet?
since h = (10/4) r,
v = pi/3 r^2 (10/4 r)
= 5pi/6 r^3
now differentiate and solve for dr/dt
To find how fast the water level is rising, we need to use related rates. Let's represent the depth of the water as "h" feet, where h is changing with respect to time t (dh/dt represents the rate at which h is changing).
Given:
Altitude (height of the cone) = 10 ft
Radius of the base (r) = 4 ft
Rate of pouring water (dV/dt) = 5 cubic ft/min
Depth of water (h) = 5 ft
We can use the formula for the volume of a cone to relate the variables:
V = (1/3)πr²h
Where V is the volume of water in the cone.
We differentiate both sides of this equation with respect to time (t):
dV/dt = (1/3)π * (2rh * dh/dt) + (1/3)πr² * dh/dt
Now, we know the rate of pouring water (dV/dt), which is 5 cubic ft/min. Substituting this value in the equation and the given values of h and r, we can solve for dh/dt (the rate at which the depth is changing) when h = 5 ft.
5 = (1/3)π * (2 * 4 * 5 * dh/dt) + (1/3)π * 4² * dh/dt
We can simplify this equation and solve for dh/dt:
5 = (40/3)π * dh/dt + (16/3)π * dh/dt
5 = (56/3)π * dh/dt
dh/dt = 5 / [(56/3)π]
dh/dt = 5 * 3 / (56π)
dh/dt = 15 / (56π)
So, when the depth of the water is 5 feet, the water level is rising at a rate of approximately 15 / (56π) ft/min.
To find how fast the water level is rising when the depth of the water is 5 feet, we can use related rates.
Let's denote the radius of the water surface as r (in feet) and the height (depth) of the water as h (in feet).
From the given information, we know that the altitude (slant height) of the cone is 10 feet, and the radius of the base is 4 feet.
We can use the similarity of triangles to relate the radius of the water surface to the height of the water. Since the altitude of the cone and a vertical line from the tip of the cone to the water surface are perpendicular, we can create two similar triangles:
The larger triangle: has height 10 ft and base diameter 8 ft (twice the radius of the base).
The smaller triangle: has height h ft and base diameter 2h ft (twice the radius of the water surface).
Using the similarity of these triangles, we can derive the relationship between r and h:
h/10 = 2h/8
Simplifying, we get:
10h = 20h/8
80h = 20h
h = 0.25
So, when the depth of the water is 5 feet, the radius of the water surface is 0.25 feet.
To find how fast the water level is rising (dh/dt) when h = 5 ft, we need to express the volume of the water in terms of h and differentiate it with respect to time.
The volume of a cone can be calculated using the formula:
V = (1/3) * π * r^2 * h
Since we have a right-circular cone, we can substitute r in terms of h using the relationship we derived earlier:
r = (1/4)h
Plugging this into the volume formula, we get:
V = (1/3) * π * [(1/4)h]^2 * h
Simplifying:
V = (1/3) * π * (1/16) * h^3
V = (π/48) * h^3
To find the rate at which the volume is changing, we differentiate both sides with respect to time (t):
dV/dt = (π/48) * 3h^2 * dh/dt
Since the rate at which water is poured into the reservoir is 5 cubic feet per minute, we know that dV/dt = 5. Also, when h = 5 ft, we need to find dh/dt.
Plugging the given values into the equation, we have:
5 = (π/48) * 3(5^2) * dh/dt
5 = (π/48) * 3(25) * dh/dt
5 = (25π/16) * dh/dt
Simplifying, we get:
dh/dt = (16/25π) * 5
Calculating this expression, we find:
dh/dt ≈ 1.019 ft/min
Therefore, when the depth of the water is 5 feet, the water level is rising at a rate of approximately 1.019 feet per minute.