Calculating Arc Length of a Curve: Caculus III Formula and Reparameterization

[dswatson]

consider the curve r(t)=<5sin(t),4cos(t),3cos(t)> where 0<t<2*pi.
Find a formula for this curve's arc length function: s(t). Also, compute the total arc length.

Reparameterize this curve with respect to arc length (find r(s)) Don't forget to specify the range for the arc length parameter: ?a?<s<?b?.

I am completely lost. If someone could step me through at least the first part I would greatly appreciate it. Thanks in advance.

The general form for the length of an arc in polar coordinates is given by

Integral of

[ sqrt( r^2 - (dr/da)^2 ) ] da

Where "da" means d-theta

I have this in my notes but am still lost.

 

[Dick]

You aren't in polar coordinates. It's a parametric curve. Arc length is the integral of sqrt(x'(t)^2+y'(t)^2+z'(t)^2) dt. Try that.

 

[dswatson]

ok so i have gotten this far.
int[sqrt{(25+sin^2(t))}]
where do I go from here? It doesn't factor or that would make it easy.

 

[iamalexalright]

Notice that Dick said the arc length is the integral of sqrt(x'(t)^2+y'(t)^2+z'(t)^2) dt!

You only have sqrt(x(t)^2)!

Notice that x'(t) denotes the derivative of x(t) wrt. t. You also need to include y'(t)^2 and z'(t)^2. You should be able to simplify nicely if you do this correctly. Post what you get !

 

[dswatson]

I feel like I did do all of the derivatives...

I did x'=5cost, y'=-4sint, and z'=-3sint
then square them...
x'^2=25cos^2(t), y'^2=16sin^2(t), and z'^2=9sin^2(t)

the add them but convert x'^2 to a factor of sin^2(t) first.

25cos^2(t)=25(1+sin^2(t)) then distribute and I ended up with...

int[sqrt{25+25sin^2(t)+16sin^2(t)+9sin^2(t)}dt]

then add up the sine terms...

int[sqrt{25+50sin^2(t)}dt]

this is where I'm stuck...did I do something wrong?

 

[dswatson]

ok nevermind i see where i messed up...i shouldn't have made the cos a sin.
If i leave it as it was before...

int[sqrt{25cos^2(t)+16sin^2(t)+9sin^2(t)}dt]
then just combine the sin terms i get

int[sqrt{25cos^2(t)+25sin^2(t)}dt]

this gives me...


int[sqrt{25}
equals...
int[5]
equals
5t over interval 0 to 2*pi

so my answer is 10*pi right?

 

[iamalexalright]

Btw, I did misread your post #3 but I'm glad you figured out your problem !

That does look correct

 

[dswatson]

Btw, I did misread your post #3 but I'm glad you figured out your problem !

That does look correct

ok...so assuming that is is correct...where do I begin on the next step?

I am a little behind in calc and am trying to catch up...

I now have to "reparameterize this curve with respect to arc length (find "r(s)"). Don't forget to specify the range for the arc length parameter: [?a?]\leq[/?b?] "

this is a quoted question and have no clue how to do it...any suggestions?

 

[iamalexalright]

Well, you found the arc length function s(t). Why not try solving for t?

After doing this how would you reparameterize the curve? With a little thinking you'll get the new bounds !

 

[dswatson]

my problem is that i don't what you mean to solve for t...my equation for arc length was 5t evaluated over 0<t<2*pi. To solve for t, what is 5t equal to? this is where I'm completely confused...

 

[dswatson]

is this simply s(t)=5t plugged back into the original r(t)?

 

[iamalexalright]

yeah, so t = s/5

What range of values is allowed for s now?

 

[dswatson]

0<s<10*pi ?

 

[dswatson]

well nevermind that doesn't work because there are no "s"'s in the eqn

 

[dswatson]

so my final answer so far is...
r(s)=<5sin(s/5),4cos(s/5),3cos(s/5)>
so are my new parameters...0<s/5>2*pi so...0<s<10*pi

 

[iamalexalright]

looks good to me

 

[dswatson]

awesome! thank you so much for all of the help
I really appreciate it...