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

1 Answer
May 2, 2018

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)#