# Need help with Calculus 2 Project.

1. May 31, 2005

### Techman07

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.

2. May 31, 2005

### Corneo

Well you are given the equation of the curve created by the cable. Thus you can find the tangent line to to curve at $x=\pm b$. 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. May 31, 2005

### Techman07

number one....

man i don't even think I did number one right now....

4. May 31, 2005

### whozum

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. May 31, 2005

### Techman07

its the letter a....

I don't mind integrating it, but I just can't figure out what a is.....

6. May 31, 2005

### Pyrrhus

Have you read the problem? it states a is a positive constant...

Considering real valued functions, this means

$$\{ a \epsilon \Re | a > 0 \}$$

Last edited: May 31, 2005
7. Jun 1, 2005

### Corneo

since the distance of the poles from another is 100m, then this means there are poles at $x=\pm 50$. From then you can solve for $a$ since you know that $S = 20$ and you know what $b$ is.

Last edited: Jun 1, 2005
8. Jun 1, 2005

### Techman07

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.