Reborn in Fire (hammerforge) wrote in calculus,
Reborn in Fire


I need a second pair of eyes. I have done something for my Calculus 3 class that is either clever or utter crap, and I am unsure of which. I am on the chapter for line integrals and have the following problem: A 160 lb man carries a 25 lb can of paint up a helical staircase that encircles a silo with a radius of 20 ft. Over the course of his climb, 9 lbs of paint leaks out? If the silo is 90 ft in height, and the man makes exactly 3 complete revolutions, how much work is done against gravity? From a physics perspective, this is quite simple. Gravity is a conservative field and so the only actual distance that matters is his height above the earth. And at 90', his height is not enough for the gravitational field to vary from its constant of 32'/s^2. Lacking any variable for time, we can express paint loss (and hence mass) as a function of height. !
  • height = 0g = gravity
  • M = Mass man
  • P = Mass paint
  • Delta P can be expressed as height/10. As integration of 1/10 will give us this value, delta P is mearly -1/10.
  • Work for a 160 lb man to carry a can of paint 90 ' vertically (532800 ft lbs/s^2),
  • Work for a 160 lb man to carry a 16 lb can of paint 90 ' vertically (506880 ftlbs/s^2),
This gives us the function of g(M+P-(1/10)32 dz to be integrated which will express all the values as a linier function of height. If this were my phsyics homework I would be done. Sadly this is not Physics but Calculus 3. I must express this as a Line Integral. For which I need to obtain i and j values that will go away when this problem is complete. 

Expressing the helix as a function of Height
  • 90' = 6z(pi) (for 3 full revolutions)
  • 2z(pi>/30) will give us a theta function
  • Radius = 20'
  • cos( 2z(pi>/30))i+sin(2z(pi>/30))j

Now that we have dummy values for i and j (but hopefully valid dummy values) we can attempt to express this as a line integral.

  • Integrate: F * T dz 0
  • Integrate <0i+0j+gk>*<20cos(2pi(z)/30)i+20sin(2pi(z)/30)j+(M+P-(1/10))k>
  • Integrate: g(M+P-(1/10))dz 0<z<90
  • Integrate 32(160+25-(1/10)dz 0<z<90

This gives a value of 532512 ftlbs/s^2
506880 ftlbs/s^2 < 532512 ftlbs/s^2 < 532800 ft lbs/s^2

While this is within the range that I would expect it to be, I cannot help but feel that this equation is a little bit too simple for a Calculus 3 line integral any opinions?

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