Question

How do you find the arc length of the curve `y=(5sqrt7)/3x^(3/2)-9` over the interval [0,5]?

Answer 1

`s=int_0^5sqrt(1+175/4x)color(white).dx=1/525(879^(3/2)-8)`

Explanation:

The arc length of a function `y` on the interval `[a,b]` is given by:

`s=int_a^bsqrt(1+(dy/dx)^2)color(white).dx`

Here:

`y=(5sqrt7)/3x^(3/2)-9`

`dy/dx=(5sqrt7)/3(3/2x^(1/2))=(5sqrt7)/2x^(1/2)`

Then the arc length desired is:

`s=int_0^5sqrt(1+((5sqrt7)/2x^(1/2))^2)color(white).dx`

`s=int_0^5sqrt(1+175/4x)color(white).dx`

Let `u=1+175/4x`, which implies that `du=175/4dx`. Moreover, note the change of bounds of the integral under this substitution: `x=0=>u=1` and `x=5=>u=879/4`. The integral is then:

`s=4/175int_1^(879//4)u^(1/2)color(white).du`

Integrating using the power rule for integration:

`s=4/175(2/3u^(3/2))|_1^(879//4)`

`s=8/525((879/4)^(3/2)-1^(3/2))`

`s=8/525(879^(3/2)/8-1)`

`s=1/525(879^(3/2)-8)`