Question

What is the general equation for the arclength of a line?

Answer 1

If we wish to find the arc length of `y = mx + b` on `[a, b]`, then `(b - a)sqrt(1 + m^2)` will give the correct arc length.

Explanation:

The general equation of a line is `y = mx + b`.

Recall the formula for arc length is `A = int_a^b sqrt(1 + (dy/dx)^2)dx`.

The derivative of the linear function is `y' = m`.

`A = int_a^b sqrt(1 + m^2)dx`

`m` is simply a constant, we can use the power rule to integrate.

`A = [sqrt(1+ m^2)x]_a^b`

`A = bsqrt(1 + m^2) - asqrt(1 + m^2)`

`A = (b - a)sqrt(1 + m^2)`

Now let's verify to see if our formula is correct. Let `y = 2x + 1` and the arc length we wish to find being on the x-interrval `[2, 6]`.

`A = (6 - 2)sqrt(1 + 2^2) = 4sqrt(5)`

If we were to use pythagoras, by connecting a horizontal line to a vertical line, we would get the following"

`y(2) = 5`
`y(6) = 13`
`Delta y = 13 - 5 = 8`

`Delta x = 4`

Thus `A^2 = Delta^2y + Delta^2x = 8^2 +4^2`

`A = sqrt(80) = sqrt(16 * 5) = 4sqrt(5)`

As obtained using our formula.

Hopefully this helps!

Answer 2

`S = (b - a)sqrt(1 + m^2)`

Explanation:

For the arc length of a linear function given its slope `m` and an interval `[a, b]`, using the arc length formula:

`S = int_a^b sqrt(1 + (color(red)(dy/dx))^2)dx`

Let `y = mx + b`

`=> color(red)(dy/dx = m)`

`S = int_a^b sqrt(1 + m^2) dx`

This may look scary because of all of the variables, but `m` is technically just a constant: the slope of the line.

The antiderivative is `sqrt(1 - m^2) * x`, and substituting the limits of integration:

`S = sqrt(1 - m^2) * b - sqrt(1 - m^2) * a`

`S = (b - a)sqrt(1 - m^2)`