Calc 3 Project: Solve Arc Length Problem with y = 1/c cosh(cx + b) + a

In summary, a hanging cable has the shape y=1/c cosh(cx+b)+a for some constants a,b,c with c>0. If the length of the cable is known to be 100 units then determine a,b,c and then plot the graph. If you want to solve the equations for a,b,c then you will need to use Newton's method.
  • #1
SigurRos
25
0
Hey I got a project assigned for my Calc 3 class, and I was wondering what to do with the following:

A hanging cable has the shape

y = 1/c cosh(cx + b) + a

for some constants a,b,c with c>0. Suppose the ends are at P(0,10) and P2(30,5).

If the length of the cable is known to be 100 units then determine a,b,c and then plot the graph.

I know that dy/dx = sinh(cx + b), so the arc length formula would be:

100 = int(sqrt(1 + sinh(cx + b)^2)) from 0 to 30

but I'm having issues solving the equations in terms of a,b and c. I tried, and got an equation with lots of cosh's that myself and Maple could not solve or reduce.

Any advice?

Thanks alot!
 
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  • #2
SigurRos said:
Hey I got a project assigned for my Calc 3 class, and I was wondering what to do with the following:

A hanging cable has the shape

y = 1/c cosh(cx + b) + a

for some constants a,b,c with c>0. Suppose the ends are at P(0,10) and P2(30,5).

If the length of the cable is known to be 100 units then determine a,b,c and then plot the graph.

I know that dy/dx = sinh(cx + b), so the arc length formula would be:

100 = int(sqrt(1 + sinh(cx + b)^2)) from 0 to 30

but I'm having issues solving the equations in terms of a,b and c. I tried, and got an equation with lots of cosh's that myself and Maple could not solve or reduce.

Any advice?

Thanks alot!

you can simplify the
1 + sinh(cx + b)^2 into just (cosh(cx + b))^2.

And you were given 2 points which you can use to find equations relating a, b , and c.
 
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  • #3
d_leet said:
You forgot a 1/c in your derivative

Maple computed the derivative for me, and that is what it gave me.
Also, I did use those 2 points to construct 2 equations, but when I tried to solve for the system I got a complex equation with lots of cosh's that maple couldn't solve.
 
  • #4
SigurRos said:
Maple computed the derivative for me, and that is what it gave me.
Also, I did use those 2 points to construct 2 equations, but when I tried to solve for the system I got a complex equation with lots of cosh's that maple couldn't solve.

Ohh wow I feel dumb your deriavtive is right. I forgot to multiply by c..
 
  • #5
SigurRos said:
Maple computed the derivative for me, and that is what it gave me.
Also, I did use those 2 points to construct 2 equations, but when I tried to solve for the system I got a complex equation with lots of cosh's that maple couldn't solve.
Newton's method? :biggrin:
 
  • #6
SigurRos said:
Maple computed the derivative for me, and that is what it gave me.
Also, I did use those 2 points to construct 2 equations, but when I tried to solve for the system I got a complex equation with lots of cosh's that maple couldn't solve.

You are relying on Maple far too much. You should be able to compute the derivative by hand. You will need to guide Maple to find an approximate solution, with most of the work done by hand.

Hints:
1. Eliminate a.
2. Use the hyperbolic identities for cosh(u)-cosh(v) and sinh(u)-sinh(v).
3. Find tanh(15c+b).
 

FAQ: Calc 3 Project: Solve Arc Length Problem with y = 1/c cosh(cx + b) + a

1. What is the purpose of solving an arc length problem using the equation y = 1/c cosh(cx + b) + a?

The purpose of solving an arc length problem using this equation is to find the length of a curve, or the distance along the curve, between two points. This can be useful in many real-world applications, such as calculating the length of a road or the perimeter of a curved object.

2. How do you approach solving an arc length problem using this equation?

To solve an arc length problem using this equation, you will need to use the arc length formula, which is given by L = ∫√(1 + (dy/dx)^2)dx. This formula involves taking the integral of the square root of 1 plus the derivative of the equation with respect to x squared. Once you have the integral, you can use integration techniques to solve for the arc length.

3. What are the variables in the equation y = 1/c cosh(cx + b) + a and how do they affect the arc length?

The variables in this equation are c, b, and a. The variable c represents the curvature of the curve, b represents the horizontal shift of the curve, and a represents the vertical shift of the curve. These variables affect the arc length by changing the shape and position of the curve, which in turn affects the distance along the curve between two points.

4. Can this equation be used to solve arc length problems for any type of curve?

Yes, this equation can be used to solve arc length problems for any type of curve as long as it is continuous and differentiable. This means that the curve must have a defined arc length and its derivative must exist at all points along the curve.

5. Are there any limitations to using this equation to solve arc length problems?

One limitation of using this equation to solve arc length problems is that it can only be used for curves that can be expressed in the form y = 1/c cosh(cx + b) + a. This means that it cannot be used for more complex curves, such as those with multiple peaks or points of inflection. Additionally, it may be difficult to solve for the arc length using this equation if the integral is not easily integrable.

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