Find the Arc Length of a Vector Function

**bkbowser** · February 11th 2013, 05:42 PM

the given vector and the bounds of integration are;
$r_{0}=\langle\sqrt{2},e^t,e^{-t}\rangle; 0\leq t \leq 1$

differentiation gives;
$r^\prime_0=\langle 0, e^t, -e^{-t}\rangle$

definition of the length of a curve (were $| r^\prime_0|$ refers to the magnitude of the vector $r^\prime_0$ );
$L=\int_{a}^{b} | r^\prime_0|dt$

substituting the magnitude of $r^\prime_0$ and the bounds of integration into the equation;
$L=\int_{0}^{1} [ 0+ (e^t)^2+ (-e^{-t})^2]^{\frac{1}{2}}dt$

So every problem, of this general form, so far has had a very simple integration where there is ultimately just a constant under the radical. So I think I'm looking for a way to combine the exponential functions into a constant and I can't come up with it so far.

If this is the wrong strategy please let me know.

**chiro** · February 11th 2013, 06:04 PM

Hey bkbowser.

One suggestion is to try making the elements of the vector simple polynomials and then integrate the function that way (or use trig functions).

**bkbowser** · February 11th 2013, 06:11 PM

I don't think we've discussed either of those methods in my program.

By trig functions do you mean the hyperbolic trig functions? We've been avoiding those sections in the text book and it's possible that this problem was assigned by mistake.

Is it possible to get more information on making the elements of the vector simple polynomials? Some keywords to look up on the internet perhaps?

**chiro** · February 11th 2013, 06:14 PM

If you just want practice for certain integrals, just either pick up a book or make your own up.

Personally I feel that once you get the hang of a few, you should just stop and move to the next thing.

Also if you are being tested exam wise on certain classes of integrals, focus on those since that is what you'll be marked on.

**HallsofIvy** · February 11th 2013, 06:39 PM

I recommend you go back and check the statement of the problem. The problem, as you give it, $\left(\sqrt{2}, e^t, e^{-t}\right)$ , is extremely difficult while $\left(\sqrt{2}t, e^t, e^{-t}\right)$ is a very easy problem.

**bkbowser** · February 12th 2013, 11:10 AM

Originally Posted by HallsofIvy

I recommend you go back and check the statement of the problem. The problem, as you give it, $\left(\sqrt{2}, e^t, e^{-t}\right)$ , is extremely difficult while $\left(\sqrt{2}t, e^t, e^{-t}\right)$ is a very easy problem.

You are correct. I have copied the problem wrong.

So it should be

the given vector and the bounds of integration are;
$r_{0}=\langlet\sqrt{2},e^t,e^{-t}\rangle; 0\leq t \leq 1$

differentiation gives;
$r^\prime_0=\langle \sqrt{2}, e^t, -e^{-t}\rangle$

definition of the length of a curve (were $| r^\prime_0|$ refers to the magnitude of the vector $r^\prime_0$ );
$L=\int_{a}^{b} | r^\prime_0|dt$

substituting the magnitude of $r^\prime_0$ and the bounds of integration into the equation;
$L=\int_{0}^{1} [ 2+ (e^t)^2+ (-e^{-t})^2]^{\frac{1}{2}}dt$

I'm still stuck here. I've been trying to get the radicand to either look something like $(something)^2$ or get it to be a constant. I can't seem to do either here.

**bkbowser** · February 12th 2013, 11:21 AM

the radicand simplifies like so;
$[2+(e^t)^2+(-e^{-t})^2]=2(cosh(t))^2$

is this correct?

**HallsofIvy** · February 12th 2013, 11:38 AM

No, it isn't. $cosh(2t)= \frac{e^{2t}+ e^{-2t}}{2}$ so $cosh^2(2t)= \left(\frac{e^{2t}+ e^{-2t}}{2}\right)^2= \frac{e^{4t}+ 2+ e^{-4t}}{4}$ so that $cosh^2(2t)= \frac{e^{4t}+ e^{-4t}{2}+ \frac{2}{4}= cos(4t)+ \frac{1}{2}$

**bkbowser** · February 12th 2013, 11:45 AM

Should I be trying to substitute in some hyperbolic trig function in the first place?

I ask because we're not covering anything involving hyperbolic trig functions in my colleges program, so there's a chance I might not have to know how to do this problem at the moment!

**Prove It** · February 12th 2013, 04:04 PM

Originally Posted by bkbowser

You are correct. I have copied the problem wrong.

So it should be

the given vector and the bounds of integration are;
$r_{0}=\langlet\sqrt{2},e^t,e^{-t}\rangle; 0\leq t \leq 1$

differentiation gives;
$r^\prime_0=\langle \sqrt{2}, e^t, -e^{-t}\rangle$

definition of the length of a curve (were $| r^\prime_0|$ refers to the magnitude of the vector $r^\prime_0$ );
$L=\int_{a}^{b} | r^\prime_0|dt$

substituting the magnitude of $r^\prime_0$ and the bounds of integration into the equation;
$L=\int_{0}^{1} [ 2+ (e^t)^2+ (-e^{-t})^2]^{\frac{1}{2}}dt$

I'm still stuck here. I've been trying to get the radicand to either look something like $(something)^2$ or get it to be a constant. I can't seem to do either here.

$\displaystyle \begin{align*} \int_0^1{\sqrt{2 + \left( e^t \right)^2 + \left( -e^{-t} \right)^2 }\,dt} &= \int_0^1{ \sqrt{e^{2t} + 2 + \frac{1}{e^{2t}} }\,dt } \\ &= \int_0^1{ \sqrt{ e^{2t} \left( 1 + \frac{2}{e^{2t}} + \frac{1}{e^{4t}} \right) } \, dt } \\ &= \int_0^1{ e^t \, \sqrt{ 1 + \frac{2}{e^{2t}} + \frac{1}{e^{4t}} } \, dt } \end{align*}$

Now let $\displaystyle \begin{align*} u = e^t \implies du = e^t\, dt \end{align*}$ and note that $\displaystyle \begin{align*} u(0) = 1 \end{align*}$ and $\displaystyle \begin{align*} u(1) = e \end{align*}$ , then the integral becomes

$\displaystyle \begin{align*} \int_0^1{ e^t \, \sqrt{ 1 + \frac{2}{e^{2t}} + \frac{1}{e^{4t}} } \, dt } &= \int_1^e{ \sqrt{ 1 + \frac{2}{u^2} + \frac{1}{u^4} }\, du } \\ &= \int_1^e { \sqrt{ \frac{ u^4 + 2u^2 + 1 }{u^4} } \, du } \\ &= \int_1^e{ \sqrt{ \frac{\left( u^2 + 1 \right) ^2 }{ u^4 } } \, du } \\ &= \int_1^e{ \frac{u^2 + 1}{u^2} \, du } \\ &= \int_1^e{ 1 + u^{-2} \, du } \\ &= \left[ u - \frac{u^{-3}}{3} \right]_1^e \\ &= \left[ u - \frac{1}{3u^3} \right]_1^e \\ &= \left( e - \frac{1}{3e^3} \right) - \left( 1 - \frac{1}{3} \right) \\ &= e - \frac{1}{3e^3} - \frac{2}{3} \end{align*}$

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