Find the exact length of the curve. Y=y48+14y2, 1 less than or equal to Y less than or equal 2.

Given: Y=y48+14y2,1≤y≤2 As, the length of the curve, L is given by: L=∫ab1+(dydx)2dx According to the given question, we have: a=1 , b=2 Since, Y=y48+14y2 Therefore, dYdy=ddy(y48+14y2) dYdy=438−24y3 dYdy=y32=12y3 Therefore, the length of the given curve is: L=∫121+(y32−12y3)2dy L=∫121+14(y6−1y3)2dy L=∫12(4Y6+(y6−1)24y6)12dy L=∫12(4y6+y12−2y6+14y6)12dy L=∫12(y12+2y6+14y6)12dy

Best solutions to - "Find the exact length of the curve. Y=y48+14y2,..."

ZIHLOLEp3

Answered question

2021-11-30

Find the exact length of the curve.

Y = \frac{y^{4}}{8} + \frac{1}{4} y^{2},

1 less than or equal to Y less than or equal 2.

Answer & Explanation

Stephanie Mann

Beginner2021-12-01Added 25 answers

Given:

Y = \frac{y^{4}}{8} + \frac{1}{4 y^{2}}, 1 \leq y \leq 2

As, the length of the curve, L is given by:

L = \int_{a}^{b} \sqrt{1 + {(\frac{d y}{d x})}^{2}} d x

According to the given question, we have:
a=1 , b=2
Since,

Y = \frac{y^{4}}{8} + \frac{1}{4 y^{2}}

Therefore,

\frac{d Y}{d y} = \frac{d}{d y} (\frac{y^{4}}{8} + \frac{1}{4 y^{2}})

\frac{d Y}{d y} = \frac{4^{3}}{8} - \frac{2}{4 y^{3}}

\frac{d Y}{d y} = \frac{y^{3}}{2} = \frac{1}{2 y^{3}}

Therefore, the length of the given curve is:

L = \int_{1}^{2} \sqrt{1 + {(\frac{y^{3}}{2} - \frac{1}{2 y^{3}})}^{2}} d y

L = \int_{1}^{2} \sqrt{1 + \frac{1}{4} {(\frac{y^{6} - 1}{y^{3}})}^{2}} d y

L = \int_{1}^{2} {(\frac{4 Y 6 + {(y^{6} - 1)}^{2}}{4 y^{6}})}^{\frac{1}{2}} d y

L = \int_{1}^{2} {(\frac{4 y^{6} + y^{12} - 2 y^{6} + 1}{4 y^{6}})}^{\frac{1}{2}} d y

L = \int_{1}^{2} {(\frac{y^{12} + 2 y^{6} + 1}{4 y^{6}})}^{\frac{1}{2}} d y

This is helpful 0 Flag

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Integral Calculus

Didn't find what you were looking for?