Find the Length of the curve y=(x^4/8)+ (1/(4x^2)) on [1,3] For the derivative I got (x^3/2)-(1/(2x^3)) The square of that is x^6/4 + 1/4x^6 - 1/2...

You need to find the length of the given curve, on the interval [1,3], using the formula `int_1^3 ds` , where `ds = sqrt(1+((dy)/(dx))^2).`

You need to differentiate the equation of the curve `y = x^4/8 + 1/(4x^2)` , with respect to x, such that:

`(dy)/(dx) = (4x^3)/8 + (1/4)*(-2)*(x^(-3))`

`(dy)/(dx) = (x^3)/2 - 1/(2x^3)`

`(dy)/(dx) = (x^6 - 1)/(2x^3)`

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

`1 + ((dy)/(dx))^2 = 1 + x^6/4 - 2*((x^3)/2)* (1/(2x^3)) + 1/(4x^6)`

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You need to find the length of the given curve, on the interval [1,3], using the formula `int_1^3 ds` , where `ds = sqrt(1+((dy)/(dx))^2).`

You need to differentiate the equation of the curve `y = x^4/8 + 1/(4x^2)` , with respect to x, such that:

`(dy)/(dx) = (4x^3)/8 + (1/4)*(-2)*(x^(-3))`

`(dy)/(dx) = (x^3)/2 - 1/(2x^3)`

`(dy)/(dx) = (x^6 - 1)/(2x^3)`

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

`1 + ((dy)/(dx))^2 = 1 + x^6/4 - 2*((x^3)/2)* (1/(2x^3)) + 1/(4x^6)`

`1 + ((dy)/(dx))^2 = 1 - 1/2 + x^6/4 + 1/(4x^6)`

`1 + ((dy)/(dx))^2 = 1/2 + x^6/4 + 1/(4x^6)`

Notice that if you raise to square the binomial `(x^3)/2 - 1/(2x^3)` yields ` x^6/4 + 1/(4x^6) - 1/2.` Now, in the last result, you have obtained `1/2 + x^6/4 + 1/(4x^6),` hence, you may recognize the formula of special binomial product `1/2 + x^6/4 + 1/(4x^6) = ((x^3)/2 + 1/(2x^3))^2`

You may take now the square root, such that:

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

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

You may evaluate the definite integral, such that:

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

`int_1^3 ((x^3)/2+ 1/(2x^3))dx = (x^4/8- 1/(4x^2))|_1^3`

`int_1^3 ((x^3)/2+ 1/(2x^3))dx = 92/9`

Hence, evaluating the length of the curve `y = x^4/8 + 1/(4x^2)` , on interval `[1,3],` yields `l = 92/9` .