I am given a curve defined by:

"The a is a positive constant"

How do I handle this? Am I supposed to factor the a out of the original expression in order to get my and ?

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- August 9th 2010, 03:24 PMMechEngParametric Arc Length
I am given a curve defined by:

"The a is a positive constant"

How do I handle this? Am I supposed to factor the a out of the original expression in order to get my and ? - August 9th 2010, 03:36 PMPlato
What is the question?

Can you find - August 9th 2010, 03:36 PMArchie Meade
- August 9th 2010, 04:05 PMMechEng
Plato,

I am asked to find the length of curve, for 0__<__t__<__pi/2, defined by x and y above. If I supposed to pull a out of expression, I can find dx/dt and dy/dt... as follows:

- August 9th 2010, 04:12 PMMechEng
So, should my formula for arc length look like this?

Or is this supposed to say...

- August 9th 2010, 04:31 PMArchie Meade
- August 9th 2010, 04:46 PMMechEng
Thank you!

So I should wind up with:

- August 9th 2010, 05:01 PMArchie Meade
- August 9th 2010, 05:09 PMMechEng
the next one I am working on is:

So far I have:

I am getting stuck on the square root portion as follows:

Do you have a hint or a direction from this point? - August 9th 2010, 05:18 PMArchie Meade
- August 9th 2010, 05:43 PMMechEng
Thank you once again. I know exactly what I did wrong... for some reason, I treated e^t as e^tn when I took the derivative. That gave me te^t. Wrong either way, but at least I know why I was wrong.

Anyway... finishing this one out produces...

So...

- August 9th 2010, 06:25 PMMechEng
Roots are really causing me some grief tonight.

Same scenario... given the following:

I get...

I keep running myself into these corners. I can plug this into my calculator and come up with:

But I'm not exactly sure how to get there. In addition, this does not look particularly useful for the integration. So, I must have made some sort of error along the way. - August 9th 2010, 07:26 PMMechEng
Ok... Let me try that again...

I get...

This is where I ran into a dead-end last time...

Much more usable this time around...

Now, plugging this into my integral...

let

let

Wait... how did my terminals of integration become 1 to 1? - August 10th 2010, 03:15 AMArchie Meade
Some editing above....

If you continue from

then under the square root can be factored to

Now, the graph alternates from positive to negative, so be careful with your limits,

to avoid getting a result of zero, or subtracting any part of the area.

You are correct to look for the integral of the modulus.

However, you need to locate the x-axis crossing points of your graph

in order to evaluate the area between the resulting function and the x-axis

and if it is symmetrical.

gives the area between the curve and the x-axis.

You may also express

- August 10th 2010, 06:42 AMMechEng
Is this something specific to parametric equations? I don't think I have seen this sort of shift before. I have another problem that is giving me the same result. I start off with:

Plugging that into my surface are formula yields:

Now, to finish this up I would like to substitute:

If I understand this correctly... I want to change my terminals of integration to ensure that I do not try to "add" a negative portion of the curve to a positive portion. How does the logic work for this?

is one complete revolution. During this revolution the function crosses the x-axis twice. Is there some rule of thumb for ensuring that I am always dealing with a positive value?

I think a value of would accomplish what I want.