\(\displaystyle

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

\)

My question is regarding the piecewise function.

\(\displaystyle

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

\)

\(\displaystyle

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

\(\displaystyle

C^2 = 4^2 + 8^2

\)

\(\displaystyle

C = 4\sqrt{5}

\)

\(\displaystyle

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.

\(\displaystyle

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

\)

Now putting back into the arc formula

\(\displaystyle

s = \sqrt{ 1 + \frac{9}{4}x }

\)

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

\(\displaystyle

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

\)

getting rid of the square root we have

\(\displaystyle

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.

\(\displaystyle

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

\)

\(\displaystyle

\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