1. ## Arc Length

I think I'm missing something. This seems relatively easy but I am not getting the right answer.

Find the arc length.

$y=5 + 6x^{\frac{3}{2}}$ for $0\leq x\leq 1$

Using: $\int\limits^{b}_{a} \sqrt{1 + (f(x)')^2} dx$

I get:

$\int\limits^{1}_{0} \sqrt{(81x^{\frac{1}{4}}) +1}dx$

$\int\limits^{1}_{0} 9x^{\frac{1}{2} + 1}dx$

$= 6x^{\frac{3}{2}} + x$

Then when I evaluate at my limits, I get 7. Which is the wrong answer in Webassign. I also tried it using u-substitution and got like 54 for an answer.

CAn someone get me straight here?
Thanks!!

2. Originally Posted by mollymcf2009
I get:

$\int\limits^{1}_{0} \sqrt{(81x^{\frac{1}{4}}) +1}dx$
$y=5+6x^{3/2}$

$\Rightarrow\frac{dy}{dx}=\frac32\cdot6\cdot x^{3/2-1}=9x^{1/2}=9\sqrt x$

So

$\left(\frac{dy}{dx}\right)^2=(9\sqrt x)^2=81x.$

3. Thanks for clarifying, but I still get 7 after I evaluate my limits. Can you see if this is what you get when you evaluate it?

4. Originally Posted by mollymcf2009
Thanks for clarifying, but I still get 7 after I evaluate my limits. Can you see if this is what you get when you evaluate it?
$\int_a^b\sqrt{1+[f'(x)]^2}\,dx$

$=\int_0^1\sqrt{1+81x}\,dx$

$=\frac1{81}\int_0^181\sqrt{1+81x}\,dx$

$=\frac1{81}\left[\frac23(1+81x)^{3/2}\right]_0^1$

$=\frac1{81}\left(\frac2382\sqrt{82}-\frac23\right)$

$=\frac2{243}(82\sqrt{82}-1)\approx6.1032$

5. Originally Posted by Reckoner
$\int_a^b\sqrt{1+[f'(x)]^2}\,dx$

$=\int_0^1\sqrt{1+81x}\,dx$

$=\frac1{81}\int_0^181\sqrt{1+81x}\,dx$

$=\frac1{81}\left[\frac23(1+81x)^{3/2}\right]_0^1$

$=\frac1{81}\left(\frac2382\sqrt{82}-\frac23\right)$

$=\frac2{243}(82\sqrt{82}-1)\approx6.1032$

Thank you for that!! Sorry to bug you one more time, but can you give me a quick explanation as to how you know when to pull a number out like that? Can you look at my other post too and get me started on it?
Thank you SO much for your help!

6. Originally Posted by mollymcf2009
Thank you for that!! Sorry to bug you one more time, but can you give me a quick explanation as to how you know when to pull a number out like that?
You are essentially doing the chain rule in reverse. The derivative of the inner function, $1+81x,$ is just 81, so there needs to be an 81 outside of the radical in order to put the integrand into the form $f'(g(x))g'(x).$ And of course, if we put 81 into the integrand, we will have to divide by 81 somewhere to cancel it out.

If you have trouble "seeing" the patterns, you can go through the process more explicitly by using $u$-substitution (in this case, let $u=1+81x\Rightarrow du=81\,dx$). As you gain a little experience with basic integration, you will find that you no longer need to write out the substitution for the simpler integrals.

I will leave your other post for another member, as I need to be getting to bed. Take care.