OK,skeeter, if you read this, I need your help one last time.
A 5 lb bucket that is leaking is lifted from the ground into the air by a worker pulling 20 ft of rope at a constant speed. The rope weighs 0.08 lb/ft.
The bucket starts with 2 gallons of liquid. Each gallon of liquid weighs 16 pounds. The liquid leaks at a constant rate. It finishes leaking (becomes empty) just as it reaches the top. Find the work done.
Tell me what parts I have correct and correct me where I need correcting.
The interval that is sliced is [0,20] because it is the distance the bucket travels?
Since the weight of the rope changes as it is pulled up, we would have
.08(20-yi)
How do I also factor in the liquid in the bucket?
Can I look at it this way.
At the ground the bucket has 2 gallons of water therefore (0,2)
When it reaches the top at height 20 the water is gone (20,0)
So, 2-0/0-20 = -.1.
So 16, the weight of each gallon and 2 the number of gallons we started with, becomes
16(-.1y+2).
So the final integral would be
Woops. My typo. The only thing I am unsure of is the +2 at the end. Is that really just the number of gallons we started with? Or is it found another way?
I feel a little uncomfortable doing it the way you did, only because the what you did it feels more physics based. I am trying to look at it purely from a calculus perspective where W=Fi*Di, and a limit is used to find the work done.