A container in the shape of a right circular cone with vertex angle a right angle is partially filled with water.
a) Suppose water is added at the rate of 3 cu.cm./sec. How fast is the water level rising when the height h = 2cm.?
b) Suppose instead no water is added, but water is being lost by evaporation. Show the level falls at a constant rate.
Since the vertex angle is 90°, the radius of the cone must be equal to the height of the cone, so the radius of the water level = height of the water level , so
r = h
Vol = (1/3)πr^2h
= (1/3)πh^3
d(vol)/dt = πh^2 dh/dt
3 = π(2^2) dh/dt
dh/dt = 3/(2π) cm/sec
b) was there no rate of evaporation given?
a) To solve this problem, we can use the formula for the volume of a cone:
V = (1/3) * π * r^2 * h
Where V is the volume, r is the radius of the cone's base, and h is the height.
We are given that the water is being added at a rate of 3 cu.cm./sec, which means the volume of water is increasing at a rate of 3 cu.cm./sec.
We want to find how fast the water level is rising, which is the rate at which the height is changing, dh/dt.
To find this, we can differentiate the volume equation with respect to time:
dV/dt = (1/3) * π * (2r * dr/dt) * h + (1/3) * π * r^2 * dh/dt
Since the vertex angle is a right angle, we can use the Pythagorean theorem to relate the radius r and the height h:
r^2 + h^2 = l^2
Where l is the slant height of the cone.
At h = 2 cm, we can find the value of l. Let's assume l = 10 cm (any value greater than 2 will work):
r^2 + 4 = 100
r^2 = 96
r ≈ 9.798 cm
Now, we can use the given information that dr/dt = 0 (since the radius is not changing):
dV/dt = (1/3) * π * (2 * 9.798 * 0) * 2 + (1/3) * π * (9.798)^2 * dh/dt
3 = (1/3) * π * (9.798)^2 * dh/dt
dh/dt = 3 * (3/π * (9.798)^2)
dh/dt ≈ 0.296 cm/sec
Therefore, the water level is rising at a rate of approximately 0.296 cm/sec when h = 2 cm.
b) If no water is being added, but water is being lost through evaporation, the volume of water is decreasing at a certain rate.
Since the cone is in the shape of a right circular cone, the rate of change of the volume is proportional to the rate of change of the height.
In other words, the volume is decreasing uniformly with respect to time.
Therefore, the water level falls at a constant rate.
To solve these problems, we can use the concepts of related rates. Related rates problems involve determining how variables change with respect to each other. In this case, we want to determine how the water level is changing with respect to time.
a) To find the rate at which the water level is rising, we need to find the derivative of the height with respect to time (dh/dt) when h = 2 cm.
Given information:
- The container is in the shape of a right circular cone with a vertex angle of a right angle.
- Water is added at a rate of 3 cu.cm./sec.
To start, we need to find a relationship between the height (h) and the volume (V) of water in the container.
The volume of a cone can be calculated using the formula V = (1/3)πr²h, where r is the radius of the circular base of the cone.
Since the vertex angle is a right angle, the slope from the vertex to a point on the circumference of the circular base is the radius (r). Let's call the radius of the circular base "r".
From similar triangles, we know that r/h = x/2, where x is the distance from the vertex to a point on the water surface.
Rearranging the equation, we get x = (2r/h)h = 2r.
The volume V can be expressed as V = (1/3)πr²h.
Now we have the relationship between V, r, and h. We can differentiate this equation with respect to time to find dV/dt, the rate at which the volume is changing with respect to time:
dV/dt = (1/3)π(2rh)(dh/dt) + (1/3)πr²(dh/dt).
We can substitute the given values, h = 2 cm, and dV/dt = 3 cu.cm./sec, into the equation to solve for dh/dt:
3 = (1/3)π(2rh)(dh/dt) + (1/3)πr²(dh/dt).
Now we solve for dh/dt:
dh/dt = 3 / [(1/3)π(2rh) + (1/3)πr²].
Substituting h = 2 cm and x = 2r, we can solve for dh/dt.
b) If no water is added, but water is lost by evaporation, the volume of water decreases over time. In this case, the rate at which the volume of water decreases is constant, which means the water level falls at a constant rate.
To find the rate at which the water level falls, we need to find the derivative of the height with respect to time (dh/dt) when the volume (V) is decreasing at a constant rate.
The volume V can still be expressed as V = (1/3)πr²h.
Differentiating both sides with respect to time, we get:
dV/dt = (1/3)πr²(dh/dt).
Since dV/dt is constant, dh/dt is also constant. Therefore, the water level falls at a constant rate when no water is added, but there is evaporation.