# arclength parameter

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• Apr 18th 2010, 07:46 PM
gralla55
arclength parameter
I'm to find the acrlength parameter for s = s(t) for the path

x = e^at cos bt i + e^at sin bt j + e^at k

I do know the formula for the arclength parameter, but I'm not quite sure how the unit vectors would fit in. The formula for arclength is

s(t) = (integral from a to t)||x'(b)||db

Do I just pretend they're not there and proceed normally? Thanks!
• Apr 18th 2010, 08:48 PM
dwsmith
Quote:

Originally Posted by gralla55
I'm to find the acrlength parameter for s = s(t) for the path

x = e^at cos bt i + e^at sin bt j + e^at k

I do know the formula for the arclength parameter, but I'm not quite sure how the unit vectors would fit in. The formula for arclength is

s(t) = (integral from a to t)||x'(b)||db

Do I just pretend they're not there and proceed normally? Thanks!

Why are you worried about unit vectors?

Just integrate the norm of the derivative of x(t)
• Apr 19th 2010, 01:58 PM
gralla55
I'm worried about the unit vectors because it's the only exercise in this chapter which has them, and there's no examples or even a correct answer given to it...

I figured I could just take the derivative normally, which would give:

x = e^at cos bt i + e^at sin bt j + e^at k

x'(t) = (ae^at cos bt - e^at sin bt) + (ae^at sin bt + e^at cos bt) + (aet)

is this correct?
• Apr 19th 2010, 02:01 PM
dwsmith
Quote:

Originally Posted by gralla55
I'm worried about the unit vectors because it's the only exercise in this chapter which has them, and there's no examples or even a correct answer given to it...

I figured I could just take the derivative normally, which would give:

x = e^at cos bt i + e^at sin bt j + e^at k

x'(t) = (ae^at cos bt - e^at sin bt) + (ae^at sin bt + e^at cos bt) + (aet)

is this correct?

You forgot to use the chain rule on the trig component of the derivative.

Then you need to take the norm of the derivative and integrate.

The derivative of cos(bt)=-bsin(bt)
• Apr 19th 2010, 02:45 PM
gralla55
Yes you're right, so the correct derivative should be:

x'(t) = (ae^at b cos bt - e^at b sin bt) + (ae^at b sin bt + e^at b cos bt) + (a e^at)

and arclength parameter:

(integral from a to t) (root)( (ae^ax b cos bx - e^ax b sin bx)^2 + (ae^ax b sin bx + e^ax b cos bx)^2 + (ae^ax)^2 )

Which isn't the easiest integral I've ever seen. I'm guessing I have to make some trig identities appear. Perhaps the first step would be:

integral from a to t) (root)( (ae^ax b (cos bx - sin bx) )^2 + (ae^ax b (sin bx + cos bx) )^2 + (ae^ax)^2 )

but then what? Thanks again!
• Apr 19th 2010, 02:47 PM
dwsmith
You will probably have some sine and cosine squared simplify down to 1 I am guessing.
• Apr 19th 2010, 03:19 PM
gralla55
That's what I figured. I guess I could rewrite it as:

(integral from a to t) (root)( (ae^ax b)^2 (cos bx - sin bx)^2 + (ae^ax b)^2 + (sin bx + cos bx)^2 + (ae^ax)^2 )

But I still don't see it...
• Apr 19th 2010, 03:20 PM
dwsmith
Quote:

Originally Posted by gralla55
That's what I figured. I guess I could rewrite it as:

(integral from a to t) (root)( (ae^ax b)^2 (cos bx - sin bx)^2 + (ae^ax b)^2 + (sin bx + cos bx)^2 + (ae^ax)^2 )

But I still don't see it...

Factor this out of the radical (ae^ax b)^2 since it is in all three terms.
• Apr 19th 2010, 03:28 PM
gralla55
The last one doesn't contain the b, but I could of course write:

(integral from a to t) (root)( (ae^ax b)^2 (cos bx - sin bx)^2 + (ae^ax b)^2 + (sin bx + cos bx)^2 + (ae^ax)^2 )

=

(integral from a to t) (root)( (ae^ax)^2 (b cos bx - b sin bx)^2 + (ae^ax)^2 + (b sin bx + b cos bx)^2 + (ae^ax)^2 )

=

(integral from a to t) (ae^ax)(root)((b cos bx - b sin bx)^2 + (b sin bx + b cos bx)^2)

See anything else? Thanks alot
• Apr 19th 2010, 03:31 PM
dwsmith
Just square it and see what you should get a sine and cosine squared and probably something that fits the double angle form of sin(2x)=2sinxcosx
• Apr 19th 2010, 03:42 PM
gralla55
(integral from a to t) (ae^ax)(root)((b cos bx - b sin bx)^2 + (b sin bx + b cos bx)^2)

=

(integral from a to t) (ae^ax)(root)((b cos bx)^2 - (2b cos bx sin bx) + (b sin bx)^2 + (b cos bx)^2 + (2b cos bx sin bx) + (b sin bx)^2 )

=

(integral from a to t) (ae^ax)(root)((b cos bx)^2 + (b sin bx)^2 + (b cos bx)^2 + (b sin bx)^2 )

=

(integral from a to t) (ae^ax)(root)((b^2 (cos^2 bx + sin^2 bx)^2 + b^2 (cos^2 bx + sin^2 bx)^2)

=

(integral from a to t) (ae^ax)(root)(2b^2)

=

(integral from a to t) (bae^ax)(root)2

Does this look right?
• Apr 19th 2010, 03:45 PM
dwsmith
Looks good. You can move the square root of 2 out of the integral since it is a constant.
• Apr 19th 2010, 04:00 PM
gralla55
So:

ab(root)2 (integral from a to t) e^ax dx

= b(root)2 * e^at - b(root)2 * e^a^2

?
• Apr 19th 2010, 04:05 PM
dwsmith
I am confused about something. Your differential should be dt not dx and if your differential is dt, you should integrate from [a,t]. Where did you get the bounds of integration and why do you have x's when there were none to start with?
• Apr 19th 2010, 05:18 PM
gralla55
I must have started writing x's instead of t's just out of habit... still, it should be correct if I just substitute the final x with a t:

ab(root)2 (integral from a to t) e^at dt

= b(root)2 * e^t^2 - b(root)2 * e^a^2

Now does it look right?
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