# Thread: Finding the arc length

1. ## Finding the arc length

So the formula is

$
s = \sqrt{1 + (f'(x))^2}
$

My question is regarding the piecewise function.

$
x < 4, x = x^\frac{3}{2}
$

$
x = 4, x = 4
$

I currently know the area of this function which is 12.8 and has a height of 8 and a base of 4. Using pythagorem theorem we get

$
C^2 = 4^2 + 8^2
$

$
C = 4\sqrt{5}
$

$
C \approx 8.94
$

So with that in mind I now go about finding the derivative of my function which is really just finding the first part of the equation.

$
f'(x) = \frac{3}{2}x^\frac{1}{2}
$

Now putting back into the arc formula
$
s = \sqrt{ 1 + \frac{9}{4}x }
$

Removing the 1 and putting it into the fraction gives me.

$
s = \sqrt{\frac{13}{4}x}
$

getting rid of the square root we have
$
s=(\frac{13}{4}x)^\frac{1}{2}
$

If i have been correct thus far this is where things go a little bit hazzy and my answer doesn't look even remotely right.

$
\int_0^4 (\frac{13}{4}x)^\frac{1}{2}
$

$
\frac {2}{3}(\frac{13}{4}x)^\frac{3}{2}
$

Integrating I get approx 49.9 which is way above what I expected to get. Im expecting something close to why pythag gave me.

Maple suggests its roughly 9.6

2. This is all really hard to follow, how did you get this?

Originally Posted by gk99

Now putting back into the arc formula
$
s = \sqrt{ 1 + \frac{9}{4}x }
$

Removing the 1 and putting it into the fraction gives me.

$
s = \sqrt{\frac{13}{4}x}
$

3. I think my brain is finally fried. But I thought itd be a great idea to do something like this.

$
\frac{4}{4} + \frac{9}{4}x
$

which in the end got me
$
\frac{13}{4}x
$

Maybe I need to rethink this problem a bit more.

4. Originally Posted by gk99
So the formula is

$
s = \sqrt{1 + (f'(x))^2}
$

My question is regarding the piecewise function.

$
x < 4, x = x^\frac{3}{2}
$

$
x = 4, x = 4
$
So this function just has a "displaced point"? That will change neither area under the curve nor arclength.

I currently know the area of this function which is 12.8 and has a height of 8 and a base of 4. Using pythagorem theorem we get
I have no idea what you are talking about. A "function" doesn't have an area so you must be talking about the area below a graph of the function. But over what x range? From 0 to 4, this function rises from 0 to 8 so I guess you are talking about the area under $y= x^{3/2}$ from x= 0 to x= 4. Yes, that is 12.8.

$
C^2 = 4^2 + 8^2
$

$
C = 4\sqrt{5}
$

$
C \approx 8.94
$

So with that in mind I now go about finding the derivative of my function which is really just finding the first part of the equation.
So all of that was to find the length of the straight line from (0, 0) to (4, 8)? Why? Just to get an estimate before you calculate the arclength?

$
f'(x) = \frac{3}{2}x^\frac{1}{2}
$

Now putting back into the arc formula
$
s = \sqrt{ 1 + \frac{9}{4}x }
$

Removing the 1 and putting it into the fraction gives me.

$
s = \sqrt{\frac{13}{4}x}
$
No, $1+ \frac{9}{4}x$ is NOT equal to $\frac{13}{4}x$.

In general, a+ bx is NOT equal to (a+b)x

getting rid of the square root we have
$
s=(\frac{13}{4}x)^\frac{1}{2}
$
Well, you didn't "get rid of the square root, you just wrote it as a 1/2 power.

If i have been correct thus far this is where things go a little bit hazzy and my answer doesn't look even remotely right.

$
\int_0^4 (\frac{13}{4}x)^\frac{1}{2}
$

$
\frac {2}{3}(\frac{13}{4}x)^\frac{3}{2}
$

Integrating I get approx 49.9 which is way above what I expected to get. Im expecting something close to why pythag gave me.

Maple suggests its roughly 9.6
You want to integrate $\int_0^4 \sqrt{1+ \frac{9}{4}x} dx$.

Do that by substituting $u= 1+ \frac{9}{4}x$.

5. Originally Posted by gk99

$
\frac{4}{4} + \frac{9}{4}x
$
This is a big no no. These are not like terms, they can't be added.