A trough is 16 ft long and its ends have the shape of isosceles triangles that are 5 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 11 ft3/min, how fast is the water level rising when the water is 7 inches deep?
1.9
To solve this problem, we can use similar triangles and the relationship between volume and height of a trough. Let's denote the depth of the water as "h" (in feet) and the volume of water in the trough as "V" (in cubic feet).
It is given that the trough is 16 ft long, and the ends have the shape of isosceles triangles that are 5 ft across at the top and have a height of 1 ft. Thus, at any given height "h" (in feet), the width of the trough is (5 - 2h) ft.
The volume of the trough can be calculated as follows:
V = (1/2) * (5 - 2h) * h * 16
Now, we need to find how fast the water level is rising when the water is 7 inches deep, i.e., h = 7/12 ft.
To find the rate at which the water level is rising, we need to differentiate the volume equation with respect to time (t) and then find dh/dt (the rate of change of height with respect to time).
Differentiating the volume equation with respect to time (using the product rule and chain rule):
dV/dt = (1/2) * (5 - 2h) * dh/dt * 16 + 16 * h * (-2) * dh/dt
Substituting h = 7/12 ft and dV/dt = 11 ft^3/min:
11 = (1/2) * (5 - 2 * 7/12) * (7/12) * 16 + 16 * (7/12) * (-2) * dh/dt
Simplifying the equation:
11 = (1/2) * (5 - 7/6) * (7/12) * 16 - (28/12) * dh/dt
Now, we can solve for dh/dt (the desired rate of change of height with respect to time):
11 = (1/2) * (25/6) * (7/12) - (28/12) * dh/dt
Multiply through by (12/28):
(12/28) * 11 = (1/2) * (25/6) * (7/12) - dh/dt
(12 * 11) / 28 = (25/6) * (7/12) - dh/dt
Simplifying:
4.7143 = 2.9167 - dh/dt
dh/dt = 2.9167 - 4.7143
dh/dt ≈ -1.7976 ft/min
Therefore, the water level is decreasing at a rate of approximately 1.7976 ft/min when the water is 7 inches deep.
To find the rate at which the water level is rising, we need to use related rates. We know the trough is being filled with water at a constant rate of 11 ft³/min. We need to find the rate at which the water level is rising when the water is 7 inches deep.
Let's break down the problem step by step:
Step 1: Convert the dimensions of the trough and the water depth to the same unit.
The trough is given in feet, and the water depth is given in inches. There are 12 inches in 1 foot, so we need to convert the 7 inches to feet.
7 inches = 7/12 feet
Step 2: Determine the volume of water currently in the trough at the given depth.
The trough has the shape of an isosceles triangle at each end. The volume of water in such a trough can be calculated using the formula for the volume of a triangular prism:
Volume = (1/2) * base * height * length
The length of the trough is given as 16 feet. The base and height of the triangles can be calculated using similar triangles:
The small triangle at the bottom has a base of 5 feet and height of 1 foot.
The larger triangle at the top will have a base that corresponds to the width of the trough at that particular depth, which we'll call "b". The height of the larger triangle will be the same as the depth of water, which is 7/12 feet.
Using similar triangles, we can set up a proportion to find the width "b":
(base of small triangle)/(height of small triangle) = (base of large triangle)/(height of large triangle)
5/1 = b/(7/12)
Cross-multiplying, we have:
5*(7/12) = b
b = 35/12 feet
Substituting the values into the volume formula:
Volume = (1/2) * (35/12) * (7/12) * 16 ft³
Step 3: Calculate the rate at which the volume is changing with respect to time.
The volume of water in the trough is changing at a rate of 11 ft³/min. Therefore, the calculated volume is also changing at this rate:
dV/dt = 11 ft³/min
Step 4: Differentiate the volume equation with respect to time.
We are looking for the rate at which the water level is rising, so we need to find (dh/dt), the derivative of the height (h) with respect to time (t).
Differentiating the volume equation with respect to time, we get:
(dV/dt) = (1/2) * (35/12) * (7/12) * (dh/dt) ft³/min
Step 5: Plug in the known values into the equation and solve for (dh/dt).
We know the volume is changing at a rate of 11 ft³/min, so we can substitute the values:
11 = (1/2) * (35/12) * (7/12) * (dh/dt)
Solving for (dh/dt), we get:
(dh/dt) = (2 * 12 * 11) / (35 * 7/12) ft/min
Simplifying, we have:
(dh/dt) = 8.57 ft/min
So, the water level is rising at a rate of approximately 8.57 ft/min when the water depth is 7 inches.