A trough is 12 feet long and 3 feet across at the top. It ends are isosceles triangles with a height of 3 feet. If water is being pumped into the trough at 2.5 cubic feet per minute, how fast is the water level rising when the water is 1 foot deep?
At a given time of t minutes,
let the width of the water level be x ft
let the height of the water level be h ft
but by ratios, x/h = 3/3
x = h
V = area of triangle x 12
= (1/2)xh(12
= 6xh
= 6h^2
given : dV/dt = 2.5 ft^3/min
find dh/dt when h = 1
dV/dt = 12h dh/dt
2.5 = 12(1) dh/dt
dh/dt = 2.5/12 = 5/24 ft/min or appr. .208 ft/min
(a) If water is being pumped into the trough at 2 cubic meters per minute, how fast is the water level rising when h is 1.5 meters deep?
(b) If the water is rising at a rate of 2.5 centimeters per minute when h = 2.0, determine the rate at which water is being pumped into the trough.
Well, well, well, looks like someone needs a little help with their trough troubles! Alrighty then, let's dive right in!
First things first, we need to find the volume of water in the trough. Since the trough is shaped like a prism, we can calculate it by finding the area of the cross-section and multiplying it by the length.
The cross-section is a trapezoid, so let's find its area. The formula for the area of a trapezoid is:
A = (1/2)h(b1 + b2), where A is the area, h is the height, and b1 and b2 are the parallel sides.
In our case, the height (h) is 3 feet, the width of the top (b1) is 3 feet, and the width of the bottom (b2) is 12 feet.
So, plugging in the values, we get:
A = (1/2) * 3 * (3 + 12)
= (1/2) * 3 * 15
= 22.5 square feet
Now, the volume (V) of the prism can be found by multiplying the area of the cross-section by the length (L):
V = A * L
= 22.5 * 12
= 270 cubic feet
Now comes the fun part! We need to find how fast the water level is rising when the water is 1 foot deep. This involves finding the derivative of the volume with respect to time (dV/dt). In other words, we need to find how quickly the volume is changing with time.
Given that the water is being pumped in at a rate of 2.5 cubic feet per minute, we have:
dV/dt = 2.5 cubic feet per minute
However, we want to find how fast the water level is rising, so we need to find dH/dt, where H represents the height (or depth) of the water.
Using similar triangles, we can find that the height of the water (H) is related to the volume (V) by:
H = V / 22.5
Let's rearrange that equation to solve for V:
V = 22.5 * H
Now, let's differentiate both sides with respect to time (t) to find:
dV/dt = 22.5 * dH/dt
We know that dV/dt is 2.5 cubic feet per minute, so we can solve for dH/dt:
2.5 = 22.5 * dH/dt
dH/dt = 2.5 / 22.5
dH/dt = 1/9 feet per minute
So, when the water is 1 foot deep, the water level is rising at a rate of 1/9 feet per minute.
Ta-da! We did it! The water level is indeed rising, and now you know exactly how fast. Keep calm and water on, my friend!
To find the rate at which the water level is rising, we need to use the concept of similar triangles and related rates. Let's denote the depth of the water as x feet.
First, let's draw a diagram of the trough:
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We can see that the trough is in the shape of a trapezoid with two triangular ends. The equation for the volume V of the water in the trough can be given by the following formula:
V = (1/2) * (b1 + b2) * h * l,
where b1 and b2 are the lengths of the parallel sides of the trapezoid, h is the height of the triangular ends, and l is the length of the trough.
In this case, the length of the trough is given as 12 feet, the top width of the trapezoid or b1 is 3 feet, and the height of the triangular ends or h is also 3 feet.
So, the equation for the volume V can be simplified to:
V = (1/2) * (3 + b2) * 3 * 12.
The base b2 of the trapezoid can be determined by using similar triangles. From the diagram, we can see that the width of the trapezoid at the bottom is b2 + 2x.
By using the concept of similar triangles, we can set up a proportion using the ratios of corresponding sides:
3/12 = x / (b2 + 2x).
Let's solve this proportion for b2:
3/12 = x / (b2 + 2x).
Cross-multiplying gives:
36 = 12x + 6x,
36 = 18x,
x = 2.
Now, we have found the value of x, which is the depth of the water when we want to find the rate at which the water level is rising.
To find how fast the water level is rising, we need to differentiate the volume V with respect to time t and solve for dx/dt.
dV/dt = (1/2) * h * l * db2/dt.
The given rate at which the water is being pumped into the trough is 2.5 cubic feet per minute, so we can substitute this value into dV/dt:
2.5 = (1/2) * 3 * 12 * db2/dt.
Simplifying this equation gives:
2.5 = 18 * db2/dt.
Now, we need to find db2/dt, the rate at which the base of the trapezoid is changing.
We already know that b2 + 2x = 3. Differentiating this equation with respect to time t gives:
db2/dt + 2dx/dt = 0.
We want to find dx/dt when the water is 1 foot deep, so let's substitute x = 1 into the equation:
db2/dt + 2 * dx/dt = 0.
Plugging this equation back into our previous equation, we get:
2.5 = 18 * (-2 * dx/dt),
Simplifying gives:
dx/dt = -2.5/36.
Therefore, when the water is 1 foot deep, the water level is rising at a rate of -2.5/36 feet per minute. The negative sign indicates that the water level is decreasing.