# Parametric Arc Length

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• August 9th 2010, 03:24 PM
MechEng
Parametric Arc Length
I am given a curve defined by:

$x=a\cos(t)+at\sin(t)$

$y=a\sin(t)-at\cos(t)$

"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 $\frac{dy}{dt}$ and $\frac{dx}{dt}$ ?
• August 9th 2010, 03:36 PM
Plato
What is the question?

Can you find $\dfrac{dx}{dt}~\&~\dfrac{dy}{dt}~?$
• August 9th 2010, 03:36 PM
Archie Meade
Quote:

Originally Posted by MechEng
I am given a curve defined by:

$x=a\cos(t)+at\sin(t)$

$y=a\sin(t)-at\cos(t)$

"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 $\frac{dy}{dt}$ and $\frac{dx}{dt}$ ?

$x=a(cost+tsint)$

$y=a(sint-tcost)$

$\displaystyle\huge\frac{dx}{dt}=a(....)$

$\displaystyle\huge\frac{dy}{dt}=a(,,,,)$
• August 9th 2010, 04:05 PM
MechEng
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:

$\frac{dx}{dt}=a(t \cos(t))$

$\frac{dy}{dt}=a(t \sin (t))$
• August 9th 2010, 04:12 PM
MechEng
So, should my formula for arc length look like this?

$\displaystyle s=\int_{0}^{\pi/2}{\sqrt{a(t\cos t)^2+a(t\sin t)^}}dt$

Or is this supposed to say...

$\displaystyle s=a\int_{0}^{\pi/2}{\sqrt{(t\cos t)^2+(t\sin t)^}}dt$
• August 9th 2010, 04:31 PM
Archie Meade
Quote:

Originally Posted by MechEng
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:

$\frac{dx}{dt}=a(t \cos(t))$

$\frac{dy}{dt}=a(t \sin (t))$

Then the arc length is

$\displaystyle\huge\int_{0}^{\frac{{\pi}}{2}}\sqrt{ \left(\frac{dy}{dt}\right)^2+\left(\frac{dx}{dt}\r ight)^2}\ dt$

$=\displaystyle\huge\int_{0}^{\frac{{\pi}}{2}}\sqrt {\left(atcost\right)^2+\left(atsint\right)^2}\ dt$

$=\displaystyle\huge\int_{0}^{\frac{{\pi}}{2}}\sqrt {a^2t^2\left(cos^2t\right)+a^2t^2\left(sin^2t\righ t)}\ dt$

$=\displaystyle\huge\int_{0}^{\frac{{\pi}}{2}}at\sq rt{\left(cos^2t\right)+\left(sin^2t\right)}\ dt$

$=\displaystyle\huge\ a\int_{0}^{\frac{{\pi}}{2}}t\ dt$
• August 9th 2010, 04:46 PM
MechEng
Thank you!

So I should wind up with:

$\displaystyle s=\frac{a\pi^2}{8}$
• August 9th 2010, 05:01 PM
Archie Meade
Quote:

Originally Posted by MechEng
Thank you!

So I should wind up with:

$\displaystyle s=\frac{a\pi^2}{8}$

Yes,

in cases like that, you can keep the constant with the other terms,
until you notice that there is a common factor within the square root.

$\displaystyle\huge\int{\sqrt{a^2t^2cos^2t+a^2t^2si n^2t}}\ dt=\int{\sqrt{(at)^2\left(cos^2t+sin^2t\right)}\ dt$

$=\displaystyle\huge\int{\sqrt{(at)^2}\sqrt{cos^2t+ sin^2t}\ dt=\int{at\sqrt{1}\ dt=a\int{t}\ dt$
• August 9th 2010, 05:09 PM
MechEng
the next one I am working on is:

$x=e^t \cos t$

$x=e^t \sin t$

So far I have:

$\displaystyle\frac{dx}{dt}=te^t \cos t - e^t \sin t$

$\displaystyle\frac{dx}{dt}=te^t \sin t - e^t \cos t$

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

$\displaystyle\sqrt{(te^t cos t - e^t sin t)^2+(te^t sin t + e^t cos t)^2}$

$\displaystyle\sqrt{e^{2t}(t cos t - sin t)^2+e^{2t}(t sin t + cos t)^2}$

$\displaystyle e^t \sqrt{(t cos t - sin t)^2+(t sin t + cos t)^2}$

Do you have a hint or a direction from this point?
• August 9th 2010, 05:18 PM
Archie Meade
Quote:

Originally Posted by MechEng
the next one I am working on is:

$x=e^t \cos t$

$y=e^t \sin t$

So far I have:

$\displaystyle\frac{dx}{dt}=e^t \cos t - e^t \sin t$

$\displaystyle\frac{dy}{dt}=e^t \sin t + e^t \cos t$

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

$\displaystyle\sqrt{(e^t cos t - e^t sin t)^2+(e^t sin t + e^t cos t)^2}$

$\displaystyle\sqrt{e^{2t}(cos t - sin t)^2+e^{2t}(sin t + cos t)^2}$

$\displaystyle e^t \sqrt{(cos t - sin t)^2+(sin t + cos t)^2}$

Do you have a hint or a direction from this point?

You had a stray "t" and a small typo on the declared function for "y".
It's been edited above.

Next, expand the expression under the square root and simplify..

$(cost-sint)(cost-sint)=cos^2t-2costsint+sin^2t$

$(cost+sint)(cost+sint)=cos^2t+2sintcost+sin^2t$

Bring those together...
• August 9th 2010, 05:43 PM
MechEng
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...

$s=\int_{0}^{\pi}{e^t\sqrt{2}}$

So...

$s=(e^{\pi}-1)\sqrt{2}$
• August 9th 2010, 06:25 PM
MechEng
Roots are really causing me some grief tonight.

Same scenario... given the following:

$0 \leq t \leq 2\pi$

$x= a \cos^3 t$

$y= a \sin^3 t$

$\frac{dx}{dt}= a 3 cos^2 t sin t$

$\frac{dy}{dt}= a 3 sin^2 t cos t$

I get...

$\sqrt{(a 3 cos^2 t sin t)^2 + (a 3 sin^2 t cos t)^2}$

$\sqrt{a3}\sqrt{(cos^2 t sin t)^2 + (sin^2 t cos t)^2}$

$\sqrt{a3}\sqrt{cos^4 t sin^2 t + sin^4 t cos^2 t}$

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

$\sqrt{a3}|sin^2 t cos^2 t|$

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 PM
MechEng
Ok... Let me try that again...

$0 \leq t \leq 2\pi$

$x= a \cos^3 t$

$y= a \sin^3 t$

$\frac{dx}{dt}= a 3 cos^2 t sin t$

$\frac{dy}{dt}= a 3 sin^2 t cos t$

I get...

$\sqrt{(a 3 cos^2 t sin t)^2 + (a 3 sin^2 t cos t)^2}$

$a\sqrt{(3 cos^2 t sin t)^2 + (3 sin^2 t cos t)^2}$

$a\sqrt{9 cos^4 t sin^2 t + 9 sin^4 t cos^2 t}$

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

$a3|sin^2 t cos^2 t|$

Much more usable this time around...

Now, plugging this into my integral...

$\displaystyle s= \int_{0}^{2\pi}{3a|\sin t \cos t|} dt$

$\displaystyle s= 3a\int_{0}^{2\pi}{|\sin t \cos t|} dt$

let $u= \cos t$

let $du= -\sin t$

$\displaystyle s= 3a\int_{\cos 0}^{\cos 2\pi}{|-u|} du$

$s= 3a \frac{u^2}{2}|_{1}^{1}$

Wait... how did my terminals of integration become 1 to 1?
• August 10th 2010, 03:15 AM
Archie Meade
Quote:

Originally Posted by MechEng
Ok... Let me try that again...

$0 \leq t \leq 2\pi$

$x= a \cos^3 t$

$y= a \sin^3 t$

$\frac{dx}{dt}= a 3 cos^2 t(- sin t)$

$\frac{dy}{dt}= a 3 sin^2 t cos t$

I get...

$\sqrt{(a 3 cos^2 t(- sin t))^2 + (a 3 sin^2 t cos t)^2}$

$\sqrt{(3a)^2( cos^2 t (-sin t))^2 + (3a)^2( sin^2 t cos t)^2}$

$\sqrt{(3a)^2(cos^4 t sin^2 t + sin^4 t cos^2 t)}$

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

$a3|\sqrt{sin^2 t cos^2 t}|$

Much more usable this time around...

Now, plugging this into my integral...

$\displaystyle s= \int_{0}^{2\pi}{3a|\sin t \cos t|} dt$

$\displaystyle s= 3a\int_{0}^{2\pi}{|\sin t \cos t|} dt$

let $u= \cos t$ ..........No! it's too early to do this! You need to examine the shape of this graph

let $du= -\sin t\ dt$

$\displaystyle s= 3a\int_{\cos 0}^{\cos 2\pi}{|-u|} du$

$s= 3a \frac{u^2}{2}|_{1}^{1}$

Wait... how did my terminals of integration become 1 to 1?

Some editing above....

If you continue from

$\displaystyle\huge\int_{0}^{2{\pi}}\sqrt{(3a)^2cos ^4tsin^2t+(3a)^2sin^4tcos^2t}\ dt$

then under the square root can be factored to

$\displaystyle\huge\int_{0}^{2{\pi}}\sqrt{(3a)^2\le ft(cos^2tsin^2t\right)\left(cos^2t+sin^2t\right)}\ dt$

$=\displaystyle\huge\int_{0}^{2{\pi}}\sqrt{(3a)^2}\ sqrt{(costsint)^2}(1)\ dt$

$=\displaystyle\huge\ 3a\int_{0}^{2{\pi}}costsint\ dt$

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.

$u=sint,\ du=costdt$

$t=0,\ u=0$

$t=\frac{{\pi}}{2},\ u=1$

$\displaystyle\huge\ (4)3a\int_{0}^1u\ du$

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

You may also express

$costsint=\frac{1}{2}sin2t$
• August 10th 2010, 06:42 AM
MechEng
Quote:

Originally Posted by Archie Meade

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.

$u=sint,\ du=costdt$

$t=0,\ u=0$

$t=\frac{{\pi}}{2},\ u=1$

What does this notation mean?

$\displaystyle\huge\ (4)3a\int_{0}^1u\ du$

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

You may also express

$costsint=\frac{1}{2}sin2t$

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:

$x=a(t-\sin t)$

$y=a(1-\cos t)$

$\frac{dx}{dt}=a(1-\cos t)$

$\frac{dy}{dt}=a(\sin t)$

$\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}=a\sqrt{ 2-2\cos t)}$

Plugging that into my surface are formula yields:

$\displaystyle s=2\pi\int_{0}^{2\pi}{a^2(1-\cos t)|\sqrt{2-2\cos t}|dt$

$s=2\sqrt{2}\pi a^2\int_{0}^{2\pi}{(1-\cos t)^{3/2}}dt$

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

$u=1-\cos t$

$du=\sin t dt$

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?

$2\pi$ is one complete revolution. During this revolution the function $\cos t$ 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 $\frac{\pi}{2}$ would accomplish what I want.
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