Check my Calculus Homework :)
 

Heya, we just had a takehome quiz. And being only a 20 point quiz, I normally wouldn't worry too much about it, however, I am borderline on grades in that class. If anyone is just bored and wants to check these two problems really quick and let me know how they look, I would appreciate it! :)

1) If the work needeed to compress a spring 6 inches from its natural length is 54 inch/lb., find the work needed to compress it one food from its natural length.
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Before I could integrate this problem, I had to use the known work from the integration from 1 to 6 and solve for the force. After that I used that force into an integration from 1 to 12 using f(x) = 1.643x.

The answer I ended up with was 110.3245 in/lb.



2) A cylindrical tank, 10 ft. high and 10 ft. in diameter, contains half tank of oil weighing 50lb per cubit foot. Find the work required to pump out all the oil in the tank from a pipe that is one foot above the top of the tank.
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On this one, I used the following equasion:
Work = Integral from 0 to 10 of f(x) where f(x) = 50(pi)((5)^2)(11-x).

The answer I ended up with on this one was 137,500(pi) lb/ft.




Thanks for any help. Lemme know if I screwed anything up or if it looks okay :)

Wouldn't helping you with this before you've turned it in be considered cheating?

The professor said for us to try it on our own (as you can see, I did), and if we have problems we can ask some friends.

Im not asking for a correct answer (if mine is wrong), I am only looking for reassurance that this is correct, or a point in the correct direction if it is not. Like I said, I'm only bothering asking about this because of the borderline grade (A/B). With calculus being a 4 credit hour class, the A would be a great help ;)

You got question 1 wrong. According to your answer the work for a spring would be linear, for which it's not. The force for a spring is linear, not the work. And why would you integrate from 1 to 6 and not 0 to 6, and 1 to 12 and not 0 to 12? The spring force goes according to F(x) = - k * x, where x is the distant the spring is moved from it's natural length. See if this helps you at all.

I haven't tried your second question yet as it sounds a bit confusing to me.

And I think you may've also gotten question 2 wrong. However, the only problem I see with what you did was your limits of integration, so rethink those and give it another try. Also the units of the answer should just be lbs. and not lb/ft. I'm pretty sure I'm correct but I'm not 100%.

Thanks shred. I am not quite sure why I was going from 1-6 and 1-12.

So once I change the integration limits to 0 to 6 and solve for the force and then integrate using that from 0 to 12, that should give the correct result?

On problem two, the integration limits are the levels of the water. 0 to 10 should be correct ;)

I declare all of this to be meaningless. If perhaps everything were in a _real_ decimal system such as metric .. now that has meaning :D.

Try and it find out. Post your results.

Yes, I agree that the integration limits are the levels of the water, but you stated that the tank was only half-full, so wouldn't it just be half that?

Oh, just noticed this: shred_head, question #1 is in terms of work, which by definition requires a distance component (ie. these so-called "inches")

Then you wouldn't be pumping the oil out of the tank now would you ? ;)

Yes, I agree, I didn't state otherwise.

New answer for #1, using integrals starting at zero: 108

Yes, you still would, that was taken into account with the actual integrand.

/sigh

I completely missed that when reading over it. Thanks, my fault.

New answer for #2: 53,125 lbs

Nope, that is still incorrect. If the force for a spring is of the form F = - k * x, and work is the integral of force, so then use that to find the form for work on a spring. And then you can just use that and plug and chug and find the correct answer.

He's correct :) The integral is from 0 to 5 because the level of the oil is at 5 feet and going down to 0. The size of the container is taken into account when determining the height of the pipe, which is 1 foot above the container (11 - x)

That's correct.

I guess I'm making a mistake somewhere...
Here's what I am doing in problem 1....

54 = [integral from 0 to 6] (F / 6)(x) dx = (Fx^2) / 12 [0 to 6] = 3F - 0 = 3F

Since W = F * D.... 54 / 3 = Force = 18
Since F = -k*x, k = -18 / 12 = -3/2

So, W = [integral from 0 to 12] (3/2)(x) dx = 108 in/lb.


=/

I'm clueless as to why as your integrand is (F / 6), you totally lost me there. But that you did it wrong should be obvious because you can check yourself by doing your integral with the force, that you determined to be -1.5*x, and integrating it from 6 to 0. When you do this you should get the original work required to compress the spring 6 inches (54 inch/lb), but you don't, you come up with 27 inch/lb instead. So something is clearly wrong already.

Try it this way:

W = [integral from 6 to 0] F(x) dx
substituting in for F(x) by F(x) = - k * x
W = [integral from 6 to 0] (- k * x) dx

Now set W = 54 inch/lb and solve the integral and then you can determine k. Then set up your new integral from 12 to 0 with your determined value for k and that will give the work required to compress the spring 12 inches.

Hrm, what i did was...

F = -kx
F = -k*6
k = -F / 6

F(x) = (F/6)*x


I see what you're doing, but where is my mistake?

I don't think you can do what you did because starts some kind of circular logic thing, and it also clearly gets you to the wrong answer. Do what I said before, check why your method cannot be correct and then use my method, and see the difference in answers.

Okay, I got it.

Once I integrate my "F(x) = (F/6)*x", I need to take the result (18) and plug it back into my "k = F / 6", which gives me the spring constant 3, which is what I am getting using your method, F(x) = -kx

:)

Which is correct, now use that to find the new answer.

Got 216 in/lb. this time.

Correct or not, I gotta' crash for the night. Thanks for all the help everyone, especially shred :) I appreciate it!

You have the right answer...almost...in/lb is not a unit of work, but in*lb is. =P

:) thanks n0n

Hmmm...I may be really late, but these questions always bug me. I get something different from you Artsemis for #1. My reasoning is as follows :

if W = integral(f*dx) and f = -k*x, the W = int[-k*x*dx]. If you integrate that from 0 -> 6, you get W = -k*x^2/2, and apply limits of integration and all that, and you can back out k. Integrating the second one gives you something different from 216 for the W for a foot...but I'm curious about the answer when you get it back. Good luck.

For problems like these, units are really important and for your number 2 answer of "53,125 lbs" is definitely incorrect. The answer is 53,125 (pi) ft*lbs. Since ft*lbs is unit for work ( Work = Force * Distance; ft*lbs = (volume (ft^3)*density (lbs/ft^3)) * ft ) ).

W = -k*x^2/2 is correct. Now plug in 6 for the upper limit to get W = -k*36/2 = -k*18. Since W = 54 was given, we have 54 = -k*18, so k = -3.

Now all you have to do is integrate from 0 to 12 of -kx dx, which is 3x dx.

That help?


And for those that were wondering. I got full credit, but yes, my units were off. The correct answers were:

1) 216
2) 53,125

I saw the first one like this, but then I'm an engineer:

the work to compress any spring is 1/2K*(x^2)
where k is the spring constant.

So, you find the constant by backward calculating from your given, ie

54 = 1/2K (6^2)

therefore K = 54 (2) / 6^2

which equals 3.

Now plug that in for your new compression length, 12

W = 1/2 (3) (12^2)

= 216 in-lb

voila! No integrals needed!

1/2K*(x^2) -is- the integral ;) you just knew it offhand

Yah, they teach how to integrate it in math class. In engineering class they show that on the first day, then assume you can remember it the rest of the 4 years.