Question

Find the arch length of the curve given by: `x= (1/4)y^3 + (1/3y)` ; `1 le y le`2 ?

Answer 1

the answer `L=int_1^2sqrt[1+(3/4*y^2-1/(3y^2))^2]*dy=23/12 (unite)`

Explanation:

we can determine the length of arc by using

`L=int_a^bsqrt[1+(f'(y))^2]*dy`

`f(y)=(1/4)y^3 + (1/(3y))`

`f'(y)=3/4*y^2-1/(3y^2)`

`L=int_1^2sqrt[1+(3/4*y^2-1/(3y^2))^2]*dy`

`intsqrt[1+(3/4*y^2-1/(3y^2))^2]*dy=y^3/4-1/(3*y)=(3*y^4-4)/(12*y)`

`L=int_1^2sqrt[1+(3/4*y^2-1/(3y^2))^2]*dy=23/12 (unite)`