Find the arc length of x=3y^{3/2}+1 from y=0 to y=4.
Now we can proceed with the derivative of the function:
x = 3y^{3/2} + 1
x' = \cfrac{9}{2} y^{1/2}
Great, with the derivative already made, we can substitute in the arc length formula:
\displaystyle L = \int_{0}^{4} \sqrt{1 + \left[\cfrac{9y^{1/2}}{2} \right]^{2}} \ dy
We square the fraction and add:
\displaystyle = \int_{0}^{4} \sqrt{1 + \cfrac{81}{4} y}\ dy
\displaystyle = \int_{0}^{4} \sqrt{\cfrac{4 + 81y}{4}}\ dy
We will apply square root to the denominator 4 and we will extract it from the integral:
\displaystyle =\int_{0}^{4} \cfrac{\sqrt{4 + 81y}}{2} dy= \cfrac{1}{2} \int_{0}^{4} \sqrt{4 + 81y} \ dy
To solve that integral, what we will do is replace 4+81y with the letter u and then we will derive it:
u = 4 + 81y
du = 81dy\quad \Rightarrow \quad \cfrac{du}{81} = dy
Now we proceed to substitute in the integral:
\displaystyle =\cfrac{1}{2} \int_{0}^{4} \sqrt{u} \cfrac{du}{81}
\displaystyle =\cfrac{1}{162} \int_{0}^{4} \sqrt{u} \ du
We proceed to perform the integral, to make it easier we will take the square root as an exponent to the one medium:
\displaystyle =\cfrac{1}{162} \int_{0}^{4} u^{1/2} \ du
= \left. \cfrac{1}{162} \cdot \cfrac{2u^{3/2}}{3}\right]_{0}^{4}
We perform the corresponding multiplications and divisions:
= \left. \cfrac{u^{3/2}}{243} \right]_{0}^{4}
Now you have to replace the u back, remember that u = 4 + 81y:
L = \left. \cfrac{\left(4 + 81y \right)^{3/2}}{243} \right]_{0}^{4}
We evaluate:
L = \left[ \left( \cfrac{\left( 328\right)^{3/2}}{243} \right) - \left( \cfrac{8}{243} \right)\right]
=\cfrac{8\left(82 \right)^{3/2} - 8}{243}
In the fraction we can factorize the 8 and in the same way we can take out the denominator:
\cfrac{8}{243} \left( 82^{3/2} - 1\right) \ \text{u}
The \text{u} means units.
So finally we have our result for the arc length of this example!
L = \cfrac{8}{243} \left( 82^{3/2} - 1\right) \ \text{u}\approx 24.41
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