Question

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

Answer 1

Arc Length `= 1/4sinh^-1 6 + 3/2sqrt(37)`

Explanation:

The Arc Length `l` is given by the integration formula
`l=int_a^b sqrt(1+(dy/dx)^2)dx`

With `y=x^2 => dy/dx=2x`. And so:

`l = int_0^3 sqrt(1+(2x)^2)dx`
`:. l = int_0^3 sqrt(1+4x^2)dx`

I will quote the result, but if you want to see how to perform the integration, please use this link

`l = [sinh^-1(2x)/4 + (xsqrt(4x^2+1))/2]_0^3`
`:. l = (sinh^-1 6/4 + (3sqrt(36+1))/2) - (sinh^-1 0/4 + 0)`
`:. l = (sinh^-1 6/4 + (3sqrt(37))/2) - (0 + 0)`
`:. l = 1/4sinh^-1 6 + 3/2sqrt(37)`