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`x=t , y=t^5/10 +1/(6t^3) , 1

The formula of arc length of a parametric equation on the interval `alt=tlt=b` is:


`L = int_a^b sqrt((dx/dt)^2+(dy/dt)^2) dt`


The given parametric equation is:


`x=t`


`y=t^5/10 + 1/(6t^3)`


The derivative of x and y are:


`dx/dt = 1`


`dy/dt =(5t^4)/10 + (-3)/(6t^4) =t^4/2-1/(3t^4)`


So the integral needed to compute the arc length of the given parametric equation on the interval `1 lt=tlt=2` is:


`L=int_1^2 sqrt(1^2 + (t^4/2-1/(3t^4))^2) dt`


The simplified form of the integral is:


`L= int_1^2 sqrt(1 + t^8/4-1/2+1/(4t^8))dt`


`L=int_1^2 sqrt(t^8/4+1/2+1/(4t^8))dt`


`L=int _1^2 sqrt((t^16+2t^8+1)/(4t^8))dt`


`L=int_1^2 sqrt( ((t^8+1)^2)/(4t^8))dt`


`L=int_1^2 (t^8+1)/(2t^4)dt`


Evaluating this yields:


`L=int_1^2 (t^4/2 + 1/(2t^4))dt`


`L= int_1^2 (t^4/2+ t^(-4)/2)dt`


`L= t^5/(2*5) + t^(-3)/(2*(-3))`  `|_1^2`


`L= t^5/10-1/(6t^3)`  `|_1^2`


`L= (2^5/10 - 1/(6*2^3))-(1^5/10-1/(6*1^3))`


`L=(32/10-1/48) - (1/10 - 1/6)`


`L=(768/240-5/240)-(3/30-5/30)`


`L=763/240+2/30`


`L=763/240+16/240`


`L=779/240`


Therefore, the arc length of the curve is `779/240` units.

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