Question

How do you find the arc length of the curve `f(x)=x^3/6+1/(2x)` over the interval [1,3]?

Answer 1

The arc length is `14/3` units.

Explanation:

The arc length of a curve on the interval `[a, b]` is given by evaluating `int_a^b sqrt(1 + (dy/dx)^2)dx`.

The derivative of `f'(x)`, given by the power rule, is

`f'(x) = 1/2x^2 - 1/(2x^2) = (x^4 - 1)/(2x^2)`

Substitute this into the above formula.

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

Expand.

`int_1^3 sqrt(1 + (x^8 - 2x^4 + 1)/(4x^4))dx`

Put on a common denominator.

`int_1^3sqrt((x^8 + 2x^4 + 1)/(4x^4)dx`

Factor the numerator as the perfect square trinomial, and recognize the denominator can be written of the form `(ax)^2`.

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

Eliminate the square root using `(a^2)^(1/2) = a`

`int_1^3 (x^4+ 1)/(2x^2)dx`

Factor out a `1/2` and put it in front of the integral.

`1/2int_1^3 (x^4 + 1)/x^2dx`

Separate into different fractions.

`1/2int_1^3 x^4/x^2 + 1/x^2dx`

Simplify using `a^n/a^m = a^(n - m)` and `1/a^n = a^-n`.

`1/2int_1^3 x^2 + x^-2dx`

Integrate using `intx^ndx = x^(n + 1)/(n + 1)`, with `n in RR, n!= -1`.

`1/2[1/3x^3 - 1/x]_1^3`

Evaluate using the second fundamental theorem of calculus, which states that for `int_a^b F(x) = f(b) - f(a)`, if `f(x)` is continuous on `[a, b]` and where `f'(x) = F(x)`.

`1/2(1/3(3)^3 - 1/3 - (1/3(1)^3 - 1/1))`

Combine fractions and simplify.

`1/2(9 - 1/3 - 1/3 + 1)`

`1/2(10 -2/3)`

`5 - 1/3`

`14/3`

Hopefully this helps!