An inverted pyramid is being filled with water at a constant rate of 60 cubic centimeters per second. The pyramid, at the top, has the shape of a square with sides of length 8 cm, and the height is 12 cm.
Find the rate at which the water level is rising when the water level is 8 cm.
the width (w) of the base is 2/3 of the height (h) ... w = 2 h / 3
the filled volume is ... 1/3 * h * w^2
substituting ... v = 1/3 * h * (2 h / 3)^2 = 4/27 * h^3
differentiating with respect to time ... dv/dt = 4/9 h^2 dh/dt
dh/dt = (9 dv/dt) / (4 h^2) = 9 * 60 / (4 * 8^2) ... cm/sec
To find the rate at which the water level is rising when the water level is 8 cm, we can use related rates.
Let's define the variables:
- A: the area of the water surface in the pyramid (cm^2)
- V: the volume of the water in the pyramid (cm^3)
- h: the height of the water in the pyramid (cm)
- t: time (s)
Given:
- The area of the square base at the top of the pyramid is 8 cm * 8 cm = 64 cm^2.
- The height of the pyramid is 12 cm.
- The rate at which the pyramid is being filled is constant at 60 cm^3/s.
Since the pyramid is being filled at a constant rate, the change in volume with respect to time can be represented by:
dV/dt = 60 cm^3/s
We want to find dh/dt, the rate at which the water level is rising when the water level is 8 cm.
First, let's express the volume of the pyramid in terms of the height h:
V = (1/3) * 64 cm^2 * h
Next, differentiate both sides of this equation with respect to time t:
dV/dt = (1/3) * 64 cm^2 * dh/dt
Substituting the known values:
60 cm^3/s = (1/3) * 64 cm^2 * dh/dt
Now, solve for dh/dt:
dh/dt = (60 cm^3/s) / ((1/3) * 64 cm^2)
= (60 cm^3/s) * (3 / 64 cm^2)
≈ 2.81 cm/s
Therefore, the rate at which the water level is rising when the water level is 8 cm is approximately 2.81 cm/s.
To find the rate at which the water level is rising when the water level is 8 cm, we can use a geometric approach.
First, let's define the variables:
- V: Volume of water in the pyramid (in cubic centimeters)
- h: Height of the water level (in centimeters)
- A: Area of the cross-section of the water (in square centimeters)
Since the pyramid has the shape of a square at the top, the area of the cross-section is given by the formula A = side length^2. Therefore, when the water level is h centimeters, the side length of the cross-section is 8 - (8/12) * h, using similar triangles.
To find the volume of water in the pyramid, we can use the formula V = (1/3) * A * h, as the pyramid has a volume of one-third the volume of a rectangular prism with the same cross-sectional area and height.
Now, we can differentiate both sides of the equation with respect to time (t) using the chain rule:
dV/dt = (1/3) * d/dt (A * h)
Since the rate of change of volume with respect to time is given as 60 cubic centimeters per second, dV/dt = 60.
To find the rate at which the water level is rising (dh/dt), we need to solve for dh/dt.
dV/dt = (1/3) * d/dt (A * h)
60 = (1/3) * (d/dt (A) * h + A * (d/dt (h)))
60 = (1/3) * ((2 * side length * (d/dt (side length)) * h) + (A * (d/dt (h))))
Substituting the values we know:
60 = (1/3) * ((2 * (8 - (8/12) * h) * (-8/12) * dh/dt) * h + ((8 - (8/12) * h)^2 * (d/dt (h))))
Now, we can substitute h = 8 and solve for dh/dt.
60 = (1/3) * ((2 * (8 - (8/12) * 8) * (-8/12) * dh/dt) * 8 + ((8 - (8/12) * 8)^2 * (d/dt (8))))
60 = (1/3) * ((2 * (8 - (8/12) * 8) * (-8/12) * dh/dt) * 8 + (8 - (8/12) * 8)^2 * 0)
60 = (1/3) * ((2 * (8 - (8/12) * 8) * (-8/12) * dh/dt) * 8)
60 = (1/3) * ((2 * (8 - (8/12) * 8) * (-8/12) * dh/dt) * 8)
60 = (1/3) * ((2 * (8 - 8/3) * (-8/12) * dh/dt) * 8)
60 = (1/3) * ((2 * (24/3 - 8/3) * (-8/12) * dh/dt) * 8)
60 = (1/3) * ((2 * (16/3) * (-8/12) * dh/dt) * 8)
60 = (1/3) * ((2 * (-128/36) * dh/dt) * 8)
60 = (1/3) * (-256/9 * dh/dt)
180 = -256/9 * dh/dt
dh/dt = -180 * 9/256
dh/dt = -45/4 cm/s
Therefore, when the water level is 8 cm, the water level is decreasing at a rate of 45/4 cm/s.