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[Argembarger]

I'm doing some electronic-submission integral calculus homework and for the life of me I can't figure out what I did wrong here.

A leaky bucket that weighs 6 lb and a rope of neglible weight are used to draw water from a well that is 422 ft deep. The bucket is filled with 41 lb of water and is pulled up at a rate of 3 ft/sec, but water leaks out of the bucket at a rate of 0.3 lb/sec. Find the work done in pulling the bucket to the top of the well (in ft-lb).

I worked it out by hand and tried to submit my answer but it wouldn't take.

Here's the WolframAlpha page where I checked my work. It got the same answer I did, so I must have set the problem up wrong.

Can anyone help me catch the problem? I've been working back through my own logic several times and can't find the bug.

[Vector]

Maybe because 41 =/= 47 =)

[Argembarger]

No, see, the bucket weighs 6 pounds. When it's full it's 47. At y = 410 the bucket no longer has water, but it's still a 6 pound bucket

[Vector]

No, see, the bucket weighs 6 pounds.

Well, that was stupid.  Sorry.  Looking at it...


I don't understand your (422 - y) bits.  Can you explain that?

EDIT: Never mind, still being dumb.  Hmm...

[Vector]

Are you sure you're supposed to submit this in foot-pounds?

[Argembarger]

I'm pretty sure it should be in foot-pounds. It doesn't say otherwise, anywhere. specifically says to do that.

(Edited due to your ninja edit)

[Vector]

Then it certainly looks to me like you got it right.  Someone better-versed in physics than I will have to help you, I suppose.

[Jim Groovester]

I don't see any immediate problems either.

Are you certain you're wrong?

[Argembarger]

No, but the system we submit our answers to says I am, and I'd have a hard time convincing the Profs that the system is wrong.

EDIT: To be honest I'm sort of expecting to suddenly realize the problem and have it be a big facepalm thing. But that's why I came here for a second opinion; I'd ask a peer but I'm doing this from home, and if you guys reassure me that I'm right I'll e-mail the Prof.

Just for the record, since Wolfram truncated the answer a bit (2,932,480 ft-lbs), my exact answer is 2,932,478.66666667 ft-lbs. But I've tried both of them and it accepts neither. (I have infinite chances, thankfully)

For decimals, the system typically has an acceptable error range. It's pretty small, but I should be well within it (assuming I set up the problem correctly)

[Jim Groovester]

Where's the 0 point in your integral? Is it at the top of the well or the bottom of the well? I think your integration limits might be wrong but I can't quite place it.

[Argembarger]

the low bound of 0 is set at the bottom of the well, making the distance at that point 422. (422-0)

...wait a minute...

Mm, nope, still can't find the problem. However, I am readying myself for a truly epic facepalm.

[Argembarger]

Here's the next problem in the set, by the way

Spoiler (click to show/hide)

I don't need any help with it, but I did lol pretty hard at the problem's story.

[Jim Groovester]

If you have infinite tries try replacing (422 - y) with just y in your integral.

You want the height to be zero when the bucket is full, not 422.

I think. Setting these problems up are always the most difficult part of them.

[dragnar]

Here's the next problem in the set, by the way

Spoiler (click to show/hide)

I don't need any help with it, but I did lol pretty hard at the problem's story.

Yeah, that's a pretty simple calculus problem... but that is by far one of the most amusing ones I've seen. Usually there's no explanation given for why you are draining the pool.

[RedWarrior0]

The answer is clear in that one: drink the Kool-Aid. All of it.

I suppose I'll finally give up on stubbornness with this one. It's sort of for the hell of it but fairly practical: finding a version of the quadratic formula, but for cubic functions.

Solve for x, with ax³ + bx² + cx + d = 0

So I try to rewrite the equation to make it easy to solve for x. My two initial tries (if you don't count the royal fuckup that led to me realizing "oh, hey, this graph can have turning points!":

Attempted method 1:

x(a(x+b/(2a))² + c - b²/(4a)) + d = 0

Problem: That x that got factored out has issues with going away

Attempted method 2:

ax³ + b(x + c/(2b)² + d - c²/(4b) = 0

Problem: that ax³ is a bitch.

Any suggestions?