Arc Length for curve r(t)=(t,(t)sin(t),(t)cos(t))

[petrol.veem]

Let r be the curve r(t)=(t,(t)sin(t),(t)cos(t)). Find the length of r between the points (0,0,0) and (pi,0,-pi).

I think i know how to do this using the arc length formula (Integral (norm of r')) but the integral gets very messy very fast. any ideas?

 

[tkhunny]

You must be doing something fishy.

\(\displaystyle sin^{2}(t)\;+\;cos^{2}(t)\;=\;1\), right?

 

[petrol.veem]

r'(t)=(1, sin(t)+(t)cos(t), cos(t)-(t)sin(t))

so after some simplification the norm of r'(t)=sqrt(1+(sin^2(t)+cos^2(t))+t^2(cos^2(t)+sin^2(t))

further, r'(t)=sqrt(1+1+t^2)

so the arc length =integral(sqrt(2+t^2)).

correct?

how does one integrate that?

 

[galactus]

\(\displaystyle \L\\\int\sqrt{t^{2}+2}dt\)

You can use trig substitution.

Let \(\displaystyle \L\\t=\sqrt{2}tan{\theta}, \;\ dt=\sqrt{2}sec^{2}{\theta}d{\theta}\)

Make the subs and whittle it down.