Question

How do you find the length of the curve for `y=x^(3/2)` for (0,6)?

Answer 1

See below.

Explanation:

We need `int_a^b sqrt(1+(dy/dx)^2) dx`.

`int_0^6 sqrt(1+(9x)/4) dx = 4/9 int_0^6 sqrt(1+(9x)/4) 9/4dx`

`= 4/9 [2/3(1+(9x)/4)^(3/2)]_0^6`

`= 8/27[(4+9x)^(3/2)/8]_0^6`

`= 1/27(58sqrt58 - 8)`

Answer 2

16.064 (to 3 places of decimals)

Explanation:

Denoting some short segment of the curve by `ds`, and abusing terminology a little, note that it is possible to approximate `ds` using the Pythagorean relationship

`ds^2 = dx^2 + dy^2`

Extracting `dx^2` as a factor from the right hand side

`ds^2 = (1 + dy^2/dx^2) dx^2`

Taking square roots of both sides

`ds = sqrt((1 + dy^2/dx^2)) dx`

from which

`s = int sqrt((1 + (dy/dx)^2)) dx`

Now noting

`y = x^(3/2)`

implies

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

so

`s = int sqrt((1 + (3/2 x^(1/2))^2)) dx`

that is

`s = int (sqrt(1 + 9/4 x)) dx`

so the required solution is

`int_0^6 (sqrt(1 + 9/4 x)) dx`

`= 1/2 int_0^6 (sqrt(4 + 9x)) dx` (after rearrangement)

Gritting teeth and going for integration by substitution with

`u = 4+9x`

so that

`(du)/dx= 9`

from which

`1/9 int du = int dx`

and noting required limits under the substitution

`u(0) = 4 and u(6) = 58`

so

`s = (1/2)(1/9) int_4^58 (sqrt(u)) du`

that is

`s = (1/2)(1/9)(2/3) [u^(3/2)]_4^58`

`= (1/27)(58^(3/2) - 4^(3/2))`

`= 16.064` (3 places of decimals)