How can an equation produce the iconic Gateway Arch using a catenary curve?

[turnip]

A catenary curve can be described by the following function:

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

where a is a non-zero constant
By selecting sutable axes, devise an equation that will produce the gateway arch.

Right next to the question is a graph of an arch that is 630 feet high (on the y axis) and 630 feet long (on the x axis).



I have no idea where to start. I was thinking that I could substitute the x and y values in of 630 to find the a value and thus the normal function (since its constant). But then why not just ask for that? Why go and say devise an equation, i don't follow.

Please, I would really appreciate someone getting me started on this!

 

[turnip]

here's what i could get:

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

630 x2 = a(e^630/a + e^-630/a)

1260/a = e^630/a + e^-630/a

ln 1260/a = 630/a -630/a = 630 - 630/a

then that would mean 0/a which is infinity (iam getting the idea that isn't the answer)

where am i going wrong?

 

[HallsofIvy]

A catenary curve can be described by the following function:

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

where a is a non-zero constant
By selecting sutable axes, devise an equation that will produce the gateway arch.

Right next to the question is a graph of an arch that is 630 feet high (on the y axis) and 630 feet long (on the x axis).



I have no idea where to start. I was thinking that I could substitute the x and y values in of 630 to find the a value and thus the normal function (since its constant). But then why not just ask for that? Why go and say devise an equation, i don't follow.

Please, I would really appreciate someone getting me started on this!

You aren't describing the graph sufficiently. What is the value of y when x= 0? You say it is "630 feet long (on the x axis)". Is the y value when x= 630 the same as when x= 0? And you say that it is "630 feet high (on the y axis). For what x is that? 630/2= 315?

 

[student01]

I have just been given this exact question to complete as part of an assignment.

Any hints or ideas how to devise a formula for the Gateway Arch.

Im a bit stuck.

Thanks.

 

[HallsofIvy]

I have just been given this exact question to complete as part of an assignment.

You said there was a picture- we do not have the picture.

Any hints or ideas how to devise a formula for the Gateway Arch.

Not unless you can answer my questions.

Im a bit stuck.

Thanks.

 

[uart]

Without the diagram we have to make a few assumptions but I'll bite.

Assume that the equation is valid over the full arch and that you can shift it up or down by an arbitrary amount. Initially I'll take "a" positive and make an upside-down arch, but you can flip it later by negating "a".

Clearly y() has even symmetry and it's min is at x=0.

The min is y(0) = a so to fit the requirements we want y(+/- 315) = a + 630.

So all you need to do is solve the following transcendental equation numerically for "a",

a + 630 = a/2 ( exp(-315/a) + exp(+315/a) )

Flip it upside-down and shift it up or down as required.