Question

Find arc length of curve (sketch also) x=y^4 /8 + 1/4 y^2 , y=1 to y = 4

Answer 1

Find arc length of curve (sketch also) $x=\frac{y^4}{8} + \frac{1}{4y^2}$ , $y=1$ to $y = 4$ .

Arc length of curve:

$L=\int_{1}^{4} \sqrt{1+(\frac{dx}{dy})^2}dy=\int_{1}^{4} \sqrt{1+(\frac{y^3}{2}-\frac{1}{2y^3})^2}dy=\int_{1}^{4} \sqrt{(\frac{y^3}{2}+\frac{1}{2y^3})^2}dy=\int_{1}^{4} (\frac{y^3}{2}+\frac{1}{2y^3})dy=$

$(\frac{y^4}{8}-\frac{1}{4y^2})|_1^4=\frac{256}{8}-\frac{1}{4\cdot 16}-(\frac18-\frac14)=32\frac{7}{64}$

Answer: $L=32\frac{7}{64}.$