Need help with Calculus 2 Project.

  • Thread starter Techman07
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  • #1
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Looking for help on a certain project. I have posted the project in pdf format


http://home.ripway.com/2005-5/317800/proj1sum05.pdf

I have already solved the first problem (a), but problem b (finding the angle) doesn't make intuitive sense to me.If you all could be so kind, all ideas are greatly appreciated. Maybe if I understood the question better that would also help.
 

Answers and Replies

  • #2
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Well you are given the equation of the curve created by the cable. Thus you can find the tangent line to to curve at [itex]x=\pm b[/itex]. Thus you can create a triangle using the tangent line as the hypotenuse, a pole as a vertical leg, and the horizontal line tangent to the minimum of the curve. Try drawing a picture.
 
  • #3
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number one....

man i don't even think I did number one right now....
 
  • #4
2,210
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Maybe you'll see it better if its translated.

A curve with equation y = 2acosh(x/a) intersects the x axis at x = -b, b. Find the arclength of the curve. All you need to do is integrate arclength for x = -50 to 50.
 
  • #5
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its the letter a....

I don't mind integrating it, but I just can't figure out what a is.....
 
  • #6
Pyrrhus
Homework Helper
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Have you read the problem? it states a is a positive constant...

Considering real valued functions, this means

[tex] \{ a \epsilon \Re | a > 0 \} [/tex]
 
Last edited:
  • #7
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since the distance of the poles from another is 100m, then this means there are poles at [itex]x=\pm 50[/itex]. From then you can solve for [itex]a[/itex] since you know that [itex]S = 20[/itex] and you know what [itex]b[/itex] is.
 
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  • #8
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I end up with (a+20 = a cosh(50/a), I know that this in turn is the same as
[a + 20 = a (e^(50/a) + e^-50/a) all over 2). There are too many a's I still don't understand how to simplify, thank you for your help though.
 

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