Problem integrating - finding length of curve

**aplrt** · October 4th 2009, 01:54 AM

I have the parametrization of the curve:

x=e^t * cos(t)
y=e^t * sin(t)
z=t

and I am asked to find the length of the curve, between t=0 and t=2pi.

IF I haven't made any mistakes, I end up with:

INT (from 0 to 2pi) sqrt( 2*e^(2t) + 1 ) dt

Now, provided I haven't come to the wrong integral, how do I integrate this (i.e. sqrt (2*e^(2t) + 1) ) ?

I have forgotten all my integration skills, so thanks for any help.

**simplependulum** · October 4th 2009, 02:42 AM

$\int \sqrt{2e^{2t}+1}~dt$

Sub $x^2 = 2e^{2t} + 1 \implies 2xdx = 4e^{2t} dt = 2(x^2-1)dt$

$dt = \frac{ xdx}{x^2-1}$

The integral becomes

$\int \frac{x^2~dx}{x^2-1}$

$= \int ( 1 + \frac{1}{x^2-1})dx$

$= \sqrt{2e^{2t}+1} + \coth^{-1}(\sqrt{ 2e^{2t}+1}) + C$

**aplrt** · October 4th 2009, 05:03 AM

Thanks - nice choice of substitution (honestly wouldn't have thought of it myself ) ! I only have one slight concern. I don't remember much/anything about the hyperbolic functions, but I know the primitive function of the indefinite integral

1/ ( x^2 - a^2 )

is

(1/2a) ln (l x-a l / l x+a l )

which seems (for someone like me who dislikes the hyperbolic functions) simpler, and it is actually in this form that the answer has been given. So here's my, perhaps stupid, question: Is this the same primitive function you wrote ? I might be calculating that horrible equation wrong, but the two functions don't seem to give the same answer (in value) ?!

**Soroban** · October 4th 2009, 05:35 AM

Hello, aplrt!

Another approach . . . but it takes ten times longer!

I have the parametrization of the curve: . $\begin{array}{ccc}x &=& e^t\cos(t) \\<br /> y&=&e^t\sin(t) \\ z&=& t \end{array}$

and I am asked to find the length of the curve, between $t=0$ and $t=2\pi$ .

IF I haven't made any mistakes, I end up with: . $<br /> \int^{2\pi}_0 \sqrt{2e^{2t} + 1}\,dt$ . . . . Right!

Now how do I integrate this?

Let: . $\sqrt{2}\,e^t \:=\:\tan\theta$

Then: . $\sqrt{2e^{2t} + 1} \;\;=\;\; \sqrt{\left(\sqrt{2}e^t\right)^2 + 1} \;\;=\;\; \sqrt{\tan^2\theta + 1} \;\;=\;\;\sqrt{\sec^2\theta} \;\;=\;\;\sec\theta$

Also: . $\sqrt{2}\,e^t \:=\:\tan\theta \quad\Rightarrow\quad e^t \:=\:\frac{\tan\theta}{\sqrt{2}} \quad\Rightarrow\quad t \:=\:\ln\left(\frac{\tan\theta}{\sqrt{2}}\right) \;=\;\ln(\tan\theta) \;-\; \ln(\sqrt{2})$
. . Hence: . $dt \;=\;\frac{\sec^2\theta}{\tan\theta}\,dt$

Substitute: . $\int \sec\theta\cdot\frac{\sec^2\theta}{\tan\theta}\,dt \;=\;\int\frac{\sec^3\theta}{\tan\theta}\,dt$

We have: . $\frac{\sec^2\theta}{\tan\theta}\cdot\sec\theta \;=\;\frac{\tan^2\theta+1}{\tan\theta}\cdot\sec\th eta \;=\;\left(\tan\theta + \frac{1}{\tan\theta}\right)\,\sec\theta$

. . . . . . $= \; \sec\theta\tan\theta + \frac{\sec\theta}{\tan\theta} \;=\;\sec\theta\tan\theta + \frac{\frac{1}{\cos\theta}}{\frac{\sin\theta}{\cos \theta}}$

. . . . . . $=\; \sec\theta\tan\theta + \frac{1}{\sin\theta} \;=\;\sec\theta\tan\theta + \csc\theta$

So we have: . $\int(\sec\theta\tan\theta + \csc\theta)\,d\theta \;\;=\;\;\sec\theta + \ln|\csc\theta - \cot\theta| + C$

All you have to do is back-substitute and evaluate.

. . . I'll wait in the car . . .
.

**Krizalid** · October 4th 2009, 11:33 AM

solving the integral is easier than it seems:

put $u=\sqrt{2e^{2t}+1}$ so that $u^2=2e^{2t}+1\implies\frac u{u^2-1}\,du=\,dt$ and the integral becomes $\int_{u(0)}^{u(2\pi )}{\frac{u^{2}}{u^{2}-1}\,du}.$

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