An inverted conical water tank with a height of 16 ft and a radius of 8 ft is drained through a hole in the vertex at a rate of 5 ft^3/s. What is the rate of change of the water depth when the water is 4 ft
To find the rate of change of the water depth, we need to find the derivative of the depth with respect to time.
Let's denote the depth of the water as h and the time as t.
Given:
Height of the inverted conical water tank = 16 ft
Radius of the inverted conical water tank = 8 ft
Drainage rate = 5 ft^3/s
We can use similar triangles to relate the changing water depth to the changing height of the water in the inverted cone. Let's call the changing height of the water in the inverted cone as x.
Since we are given that the water depth is 4 ft, we can determine the changing height of the water in the inverted cone as follows:
x = 16 - h
Now, we need to express the volume of the water in terms of h.
The formula for the volume of a cone is V = (1/3) × π × r^2 × h, where r is the radius of the cone and h is the height.
Using similar triangles, we know that the ratio of the small cone to the large cone is the same as the ratio of their corresponding heights.
So, the ratio of x to h is equal to the ratio of the radius of the small cone to the radius of the large cone:
x/h = r/8
Since h = 16 - x, we can solve for h:
h = (16x)/(x+8)
The volume of the water in the inverted cone is given by V = (1/3) × π × r^2 × h:
V = (1/3) × π × 8^2 × (16x)/(x+8)
Knowing that the drainage rate is 5 ft^3/s, we can differentiate the volume with respect to time t using implicit differentiation.
dV/dt = (1/3) × π × 8^2 × d/dt[(16x)/(x+8)]
To find d/dt[(16x)/(x+8)], we differentiate numerator and denominator separately using the quotient rule.
d/dt[(16x)/(x+8)] = [(16 * d(x)/dt * (x+8) - 16x * d(x+8)/dt)] / (x+8)^2
= [(16 * dx/dt * (x+8) - 16x * dx/dt)] / (x+8)^2
Since we know dx/dt (the rate of change of height) is equal to -5 ft^3/s (negative because water is being drained), we substitute it into the equation:
= [(16 * (-5) * (x+8) - 16x * (-5))] / (x+8)^2
Now, let's substitute x = 4 into the equation to find the rate of change of water depth when the water level is 4 ft:
[(16 * (-5) * (4+8) - 16 * 4 * (-5))] / (4+8)^2
Simplify the expression:
= [(16 * (-5) * 12 - 16 * 4 * (-5))] / (12)^2
= [(-960 + 320)] / 144
= (-640) / 144
= -40/9 ft/s
Therefore, the rate of change of the water depth when the water level is 4 ft is -40/9 ft/s.
To find the rate of change of the water depth, we need to use related rates.
Let's denote the height of the water as "h" and the radius of the water surface as "r" (which is also the radius of the conical tank).
We are given that the height of the water, h, is changing with respect to time at a rate of dh/dt = -5 ft^3/s (negative because the water is being drained).
We need to find the rate at which the water depth is changing when the water is 4 ft, which means we need to find dh/dt when h = 4 ft.
We can use the formula for the volume of a cone:
V = (1/3) * π * r^2 * h
Differentiating both sides of the equation with respect to time (t) using the chain rule, we get:
dV/dt = (1/3) * π * (2r * dr/dt * h + r^2 * dh/dt)
Since the hole at the vertex is the only path through which the water can be drained, the change in volume of the water is equal to the rate at which the water is being drained:
dV/dt = -5 ft^3/s
Now, let's substitute the known values into the equation:
-5 ft^3/s = (1/3) * π * (2 * 8 ft * dr/dt * 4 ft + (8 ft)^2 * (-5 ft/s))
Simplifying the equation gives:
-5 = (1/3) * π * (64 ft * dr/dt - 320 ft^2/s)
Divide both sides of the equation by (1/3) * π * 64 ft to isolate dr/dt:
-5 / [(1/3) * π * 64 ft] = dr/dt - 5 ft/ft
Finally, we have:
dr/dt = -5 / [(1/3) * π * 64 ft] + 5 ft/ft
Now, we can calculate the value of dr/dt.
using similar triangles, it is easy to see that when the water has depth y, the radius of the water surface is y/2
so, the volume of water is
v = pi/3 (y/2)^2 y = pi/12 y^3
now, use the fact that
dv/dt = pi/4 y^2 dy/dt
to find your answer.