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# Catenary equation [Solved!]

### My question

Hello, in your article titled "Arc Length of a Curve using Integration", in example 3 regarding the Golden Gate Bridge cables. May you please elaborate how you "guessed and checked" the catenary equation of the cables. I am doing a math exploration internal assessment (for ib) about this topic and its mandatory to explain every equation I use, hence "guess and check" is not sufficient.

### Relevant page

11. Arc Length of a Curve using Integration

### What I've done so far

I plugged in the point (152,640) and tried solving for a, but there is no answer

X

Hello, in your article titled "<a href="http://www.intmath.com/applications-integration/11-arc-length-curve.php">Arc Length of a Curve using Integration</a>", in example 3 regarding the Golden Gate Bridge cables. May you please elaborate how you "guessed and checked" the catenary equation of the cables. I am doing a math exploration internal assessment (for ib) about this topic and its mandatory to explain every equation I use, hence "guess and check" is not sufficient.
Relevant page

<a href="/applications-integration/11-arc-length-curve.php">11. Arc Length of a Curve using Integration</a>

What I've done so far

I plugged in the point (152,640) and tried solving for a, but there is no answer

## Re: Catenary equation

Well, it's best if you figure out how to do it (rather than me just telling you :-)

Let's start with the equation for the simplest catenary: y=(e^[x]+e^[-x])/(2).

I said in the solution on that page "For convenience, we'll place the origin at the lowest point of the cable."

In the equation y=(e^[x]+e^[-x])/(2), when x=0, what is the value of y (this is the lowest point)?

X

Hello Shahad

Well, it's best if you figure out how to do it (rather than me just telling you :-)

Let's start with the equation for the simplest catenary: y=(e^[x]+e^[-x])/(2).

I said in the solution on that page "For convenience, we'll place the origin at the lowest point of the cable."

In the equation y=(e^[x]+e^[-x])/(2), when x=0, what is the value of y (this is the lowest point)?

## Re: Catenary equation

I am aware of that, so we also have point (0,0). so do I need to graph this simplest equation and then try transforming it? Or what I do exactly. I want to model the main cable of the Golden Gate Bridge, and it's actual vertex is 0,0. So I only need to do horizontal and vertical stretches and compressions right?

X

I am aware of that, so we also have point (0,0). so do I need to graph this simplest equation and then try transforming it? Or what I do exactly. I want to model the main cable of the Golden Gate Bridge, and it's actual vertex is 0,0. So I only need to do horizontal and vertical stretches and compressions right?

## Re: Catenary equation

I'm taking you step by step through the process I used to come up with the equation for the ropes.

When x=0, the equation y=(e^[x]+e^[-x])/(2) doesn't give us (0,0)!

X

I'm taking you step by step through the process I used to come up with the equation for the ropes.

When x=0, the equation y=(e^[x]+e^[-x])/(2) doesn't give us (0,0)!

## Re: Catenary equation

Seems like Shahad has disappeared. Anyone else like to have a go at finishing this?

X

Seems like Shahad has disappeared. Anyone else like to have a go at finishing this?

## Re: Catenary equation

I'm brave and I'll give it a go.

At x=0,

y=(e^[x]+e^[-x])/(2)=(e^+e^[-0])/(2)=(1+1)/2=1

So it passes through (0,1) at the lowest point.

X

I'm brave and I'll give it a go.

At x=0,

y=(e^[x]+e^[-x])/(2)=(e^+e^[-0])/(2)=(1+1)/2=1

So it passes through (0,1) at the lowest point.

## Re: Catenary equation

Yes, Stephen, you are correct.

This will also be the case if our function is, say,

y=(e^[3x]+e^[-3x])/(2),

or in fact, any number b in there, like

y=(e^[bx]+e^[-bx])/(2).

They all will pass through (0,1).

That's where the "-1" comes from in my equation (since I'm assuming the curve passes through (0,0), as stated in the solution).

So I knew it had to be of the form:

y=a((e^[bx]+e^[-bx])/(2)-1).

OK so far?

X

Yes, Stephen, you are correct.

This will also be the case if our function is, say,

y=(e^[3x]+e^[-3x])/(2),

or in fact, any number b in there, like

y=(e^[bx]+e^[-bx])/(2).

They all will pass through (0,1).

That's where the "-1" comes from in my equation (since I'm assuming the curve passes through (0,0), as stated in the solution).

So I knew it had to be of the form:

y=a((e^[bx]+e^[-bx])/(2)-1).

OK so far?

## Re: Catenary equation

Yes, it makes sense.

Now we have 2 equations in 2 unknowns, right?

The curve passes through (-640,152) and (640,152), so

When x=-640,

152=a((e^[-640b]+e^[640b])/(2)-1)

and

When x=640,

152=a((e^[640b]+e^[-640b])/(2)-1)

Putting them equal and cancelling off the a's gives:

(e^[-640b]+e^[640b])/(2)-1 = (e^[640b]+e^[-640b])/(2)-1

Adding 1 to both sides gives the obvious statement whcich can't be solved:

(e^[-640b]+e^[640b])/(2) = (e^[640b]+e^[-640b])/(2)

So I'm stuck.

X

Yes, it makes sense.

Now we have 2 equations in 2 unknowns, right?

The curve passes through (-640,152) and (640,152), so

When x=-640,

152=a((e^[-640b]+e^[640b])/(2)-1)

and

When x=640,

152=a((e^[640b]+e^[-640b])/(2)-1)

Putting them equal and cancelling off the a's gives:

(e^[-640b]+e^[640b])/(2)-1 = (e^[640b]+e^[-640b])/(2)-1

Adding 1 to both sides gives the obvious statement whcich can't be solved:

(e^[-640b]+e^[640b])/(2) = (e^[640b]+e^[-640b])/(2)

So I'm stuck.

## Re: Catenary equation

Good reply, Stephen - and so fast!

You're right - that approach does lead to a dead end.

We need to go back to this:

y=a((e^[bx]+e^[-bx])/(2)-1)

We need to choose an a and b that passes through the given points, AND has close to the correct shape for the Golden Gate Bridge.

In fact, there are an infinite number of catenaries that pass through the required points - we just need one that is close.

I can't remember now, but I probably used Desmos or Wolfram|Alpha to do my "Guessing and checking".

In this first image, you can see 3 curves.

The red one is y=100(\cosh (x/406.2)-1)

The blue one is y=1280(\cosh (x/1326)-1) (this is the one I ended up using)

As you can see, there is very little difference between those two.

The green one has a very small a value, for comparison:

y=0.1(\cosh (x/79.6)-1) Having chosen some (random, at first) value for a, I just tweaked the value for b until it passed through the required points (-640,152) and (640,152). Quick and easy to do using Desmos. (I could have solved it for b using Wolfram|Alpha, directly.)

When overlaying the 3 curves on an actual photo of the Golden Gate Bridge, we can see the first 2 are very close (basically indistinguishable from the actual shape), but the third is way off. [Bridge image source]

So Stephen and Shahad, sometimes when we have 2 unknowns, a bit of guess and check is necessary!

X

Good reply, Stephen - and so fast!

You're right - that approach does lead to a dead end.

We need to go back to this:

y=a((e^[bx]+e^[-bx])/(2)-1)

We need to choose an a and b that passes through the given points, AND has close to the correct shape for the Golden Gate Bridge.

In fact, there are an infinite number of catenaries that pass through the required points - we just need one that is close.

I can't remember now, but I probably used <a href="https://www.desmos.com/calculator">Desmos</a> or <a href="http://www.wolframalpha.com/">Wolfram|Alpha</a> to do my "Guessing and checking".

In this first image, you can see 3 curves.

The red one is y=100(\cosh (x/406.2)-1)

The blue one is y=1280(\cosh (x/1326)-1) (this is the one I ended up using)

As you can see, there is very little difference between those two.

The green one has a very small a value, for comparison:

y=0.1(\cosh (x/79.6)-1)

<img src="/forum/uploads/imf-3249-golden-gate.png" width="328" height="83" alt="Applications of Integration" />

Having chosen some (random, at first) value for a, I just tweaked the value for b until it passed through the required points (-640,152) and (640,152). Quick and easy to do using Desmos. (I could have <a href="http://bit.ly/2h1xZZZ">solved it for b using Wolfram|Alpha</a>, directly.)

When overlaying the 3 curves on an actual photo of the Golden Gate Bridge, we can see the first 2 are very close (basically indistinguishable from the actual shape), but the third is way off.

<img src="/forum/uploads/imf-3315-golden-gate.jpg" width="400" height="97" alt="Applications of Integration" />

[Bridge <a href="https://en.wikipedia.org/wiki/GoldenGateBridge">image source</a>]

So Stephen and Shahad, sometimes when we have 2 unknowns, a bit of guess and check is necessary!

## Re: Catenary equation

Hello, sorry for disappearing!
Thank you very much Murray and Stephen for solving this problem.

X

Hello, sorry for disappearing!
Thank you very much Murray and Stephen for solving this problem.

## Re: Catenary equation

Hello! I wanted to know why inside your general formula of the catenary equation you take bx as in:

y = a ((e^(bx) + e^(-bx))/2-1 )

and not x/a as in:

y = a ((e^(x/a) + e^(-x/a))/2-1 ) which is the accurate substitution of cosh (x/a)

Thank you

X

Hello! I wanted to know why inside your general formula of the catenary equation you take bx as in:

y = a ((e^(bx) + e^(-bx))/2-1 )

and not x/a as in:

y = a ((e^(x/a) + e^(-x/a))/2-1 ) which is the accurate substitution of cosh (x/a)

Thank you

## Re: Catenary equation

@Sofia: Fair enough question. I have added some more detail in the original answer to explain we are assuming it's a flattened catenary.

BTW, the formula for a (simple) catenary is y=a cosh (x/a) (with the a out front), which is equal to

y =a ((e^(x/a) + e^(-x/a))/2 )

I have -1 in my model since I'm placing the cable on either side of the y-axis..

X

@Sofia: Fair enough question. I have added some more detail in the <a href="/applications-integration/11-arc-length-curve.php">original answer</a> to explain we are assuming it's a flattened catenary.

BTW, the formula for a (simple) catenary is y=a cosh (x/a) (with the a out front), which is equal to

y =a ((e^(x/a) + e^(-x/a))/2 )

I have -1 in my model since I'm placing the cable on either side of the y-axis..

## Re: Catenary equation

Hi Murray.

Can you help me with my math exploration for IB about catenary equations?

X

Hi Murray.

Can you help me with my math exploration for IB about catenary equations?

## Re: Catenary equation

@belaldaqqah: Feel free to post a new question here on the IntMath Forum. Don't forget to show your working!

X

@belaldaq<span></span>qah: Feel free to post a new question here on the IntMath Forum. Don't forget to show your working!

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