Question

Calculate the length of the arc of `y = 2x-x^2` for #x in [0,2]?

Answer 1

Arc Length `~~ 2.970` to 3dp

Explanation:

Arc Length is given by:

`L = int_alpha^beta \ sqrt(1+(dy/dx)^2) \ dx`

We have:

`y = 2x-x^2`

Differentiating wrt `x` we have:

`dy/dx = 2-2x`

So the Required Arc Length, is given by:

`L = int_0^2 \ sqrt( 1 + (2-2x)^2 ) \ dx`
`\ \ = int_0^2 \ sqrt( 1 + 4-8x+4x^2 ) \ dx`
`\ \ = int_0^2 \ sqrt( 4x^2-8x+5 ) \ dx`

Now, let us estimate this integral numerically by using the Trapezium rules with `10` sub-intervals working to 6dp. The values of the function are tabulated as follows;

`A = 0.2/2 * { 2.236068 + 2.236068 +`
`\ \ \ \ \ \ \ \ \ 2*(1.886796 + 1.56205 + 1.280625 + 1.077033 +`
`\ \ \ \ \ \ \ \ \ 1 + 1.077033 + 1.280625 + 1.56205 + 1.886796) }`
`\ \ \ = 0.1 * { 4.472136 + 2*(12.613008) }`
`\ \ \ = 0.1 * { 4.472136 + 25.226016 }`
`\ \ \ = 0.1 * 29.698152`
`\ \ \ = 2.969815`

So we estimate that:

`L = 2.970` to 3dp

Actual Value

As it happens this integral can be evaluated and we find that

`L = 1/2(sinh^(-1)(2)+2sqrt(5)) = 2.957885715089195`