`y = x^4/8 + 1/(4x^2) , [1, 3]` Find the arc length of the graph of the function over the indicated interval.

Expert Answers
gsarora17 eNotes educator| Certified Educator

Arc length (L) of the function y=f(x) on the interval [a,b] is given by the formula,

`L=int_a^b sqrt(1+(dy/dx)^2)dx` , if y=f(x) a `<=`  x `<=`  b,

Now `y=x^4/8+1/(4x^2)`

Now we need to differentiate the above function with respect to x,

`dy/dx=1/8(4)x^(4-1)+1/4(-2)x^(-2-1)`

`dy/dx=1/2x^3-1/2x^(-3)`

`dy/dx=x^3/2-1/(2x^3)`

`dy/dx=(x^6-1)/(2x^3)`  

Now arc length L=`int_1^3 sqrt(1+((x^6-1)/(2x^3))^2)dx`

`=int_1^3sqrt(1+(x^12-2x^6+1)/(4x^6))dx`

`=int_1^3sqrt((4x^6+x^12-2x^6+1)/(4x^6))dx`

`=int_1^3sqrt((x^12+2x^6+1)/(4x^6))dx`

`=int_1^3sqrt(((x^6+1)/(2x^3))^2)dx`

`=int_1^3(x^6+1)/(2x^3)dx`

`=int_1^3(x^6/(2x^3)+1/(2x^3))dx`

`=int_1^3(x^3/2+1/2x^(-3))dx`

`=[1/2x^4/4+1/2(x^(-3+1)/(-3+1))]_1^3`

`=[x^4/8-1/(4x^2)]_1^3`

`=[3^4/8-1/(4(3)^2)]-[1^4/8-1/(4(1)^2)]`

`=[81/8-1/36]-[1/8-1/4]`

`=[(729-2)/72]-[(1-2)/8]`

`=[727/72]-[-1/8]`

`=727/72+1/8`

`=(727+9)/72`

`=736/72`

`=92/9`

So, the Arc length=`92/9`