A trough is 9 feet long and 1 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of y=x4 from x=−1 to x=1 . The trough is full of water. Find the amount of work in foot-pounds required to empty the trough by pumping the water over the top. Note: The weight of water is 62 pounds per cubic foot.
Well, well, well, looks like we have a mathematical conundrum involving a trough and some pumping action! Don't worry, I've got this covered, and I'll throw in a dash of my magnificent humor to keep things amusing.
First things first, let's calculate the volume of water in the trough. Since the cross-section of the trough is shaped like the graph of y = x^4, we can use integration to find the volume.
We'll integrate from x = -1 to x = 1, which are the limits of the cross-section. The height of the trough is 1 foot, and the width changes depending on the value of x.
So, the volume of water in the trough is given by:
V = ∫[-1 to 1] (1)*(x^4) dx
Now, let's integrate that bad boy:
V = [x^5/5] (from -1 to 1)
V = [(1^5/5) - (-1^5/5)]
V = [1/5 - (-1/5)]
V = [1/5 + 1/5]
V = 2/5 cubic feet
Now comes the fun part - calculating the amount of work required to pump the water over the top! We know that the weight of water is 62 pounds per cubic foot, so we can multiply the volume by the weight to find the total weight of the water.
Weight = (2/5) * 62 pounds
Weight = 124/5 pounds
Weight = 24.8 pounds (approximately)
Since we're calculating the work in foot-pounds, we need to multiply the weight by the height of the trough, which is 1 foot.
Work = 24.8 * 1 foot-pounds
Work = 24.8 foot-pounds
Voila! The amount of work required to empty the trough by pumping the water over the top is approximately 24.8 foot-pounds.
I hope my mathematical antics brought a smile to your face while solving this problem! If you have any more mathematical (or non-mathematical) questions, feel free to throw them my way. I'm always here to sprinkle some humor into your life!
To find the amount of work required to empty the trough by pumping the water over the top, we need to calculate the volume of water in the trough and then determine the work done to lift it.
Step 1: Calculate the volume of the trough
The trough has a length of 9 feet and a height of 1 foot. However, the cross-section of the trough is shaped like the graph of y = x^4 from x = -1 to x = 1.
To find the volume, we need to integrate the cross-sectional area over the length of the trough.
The cross-sectional area at any point x is given by the equation A(x) = y(x) * length = x^4 * 9.
Step 2: Integrate the cross-sectional area
∫[from -1 to 1] x^4 * 9 dx
Step 3: Evaluate the integral
Let's calculate the integral:
∫[from -1 to 1] x^4 * 9 dx = 9 * ∫[from -1 to 1] x^4 dx
Using the power rule of integration, we can integrate x^4 as (1/5) * x^5:
= 9 * (1/5) * [x^5] [from -1 to 1]
= 9 * (1/5) * [1^5 - (-1)^5]
= 9 * (1/5) * [1 - (-1)]
= 9 * (1/5) * 2
= 18/5
= 3.6 cubic feet
Step 4: Calculate the work done to lift the water
The weight of water is given as 62 pounds per cubic foot.
So, the work done to lift the water is given by the product of the weight per cubic foot and the volume:
Work = 62 * 3.6
= 223.2 foot-pounds
Therefore, the amount of work required to empty the trough by pumping the water over the top is 223.2 foot-pounds.
To find the amount of work required to empty the trough by pumping the water over the top, we need to calculate the weight of water in the trough and then multiply it by the distance it needs to be lifted.
First, let's find the volume of water in the trough. Since the trough is shaped like the graph of y = x^4, we can imagine slicing it into infinitely small horizontal strips. Each strip has a width dx and a height y = x^4.
The length of the trough is given as 9 feet, so the limits of integration for x are -1 to 1. The volume can be found by integrating the cross-sectional area over this interval:
V = ∫[from -1 to 1] x^4 dx
Integrating x^4 with respect to x gives us:
V = [x^5/5] [from -1 to 1] = (1^5/5) - (-1^5/5) = 2/5
Now, we can calculate the weight of water in the trough. The weight of water is given as 62 pounds per cubic foot, and we have found the volume to be 2/5 cubic feet.
Weight of water = (62 pounds/ft³) * (2/5 ft³) = 24.8 pounds
Finally, we need to calculate the work required to lift this weight of water to the top of the trough, which is 1 foot.
Work = Weight * Distance = 24.8 pounds * 1 foot = 24.8 foot-pounds
Therefore, the amount of work required to empty the trough by pumping the water over the top is 24.8 foot-pounds.
the weight of a thin sheet of water of thickness dy and area 9x is
9x*62 dy
At depth y, the width across the water surface is 2∜y, so the weight of the sheet of water is
1116∜y dy
Now, since work = force * distance, we have to add up all the work lifting thin sheets of water to a height of 1
W = ∫[0,1] 1116∜y(1-y) dy = 396.8 J