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

Question

Perman7z · Accepted Answer

The general equation of a line is y=mx+b
Recall the formula for arc length is

A = \int_{a}^{b} \sqrt{1 + (\frac{d y}{d x})^{2}} d x

The derivative of the linear function is y'=m

A = \int_{a}^{b} \sqrt{1 + m^{2}} d x

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

A = [\sqrt{1 + m^{2}} x]_{a}^{b}

A = b \sqrt{1 + m^{2}} - a \sqrt{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}} = 4 \sqrt{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

△ y = 13 - 5 = 8

△ x = 4

Thus

A^{2} = △^{2} y + △^{2} x = 8^{2} + 4^{2}

A = \sqrt{80} = \sqrt{16 * 5} = 4 \sqrt{5}

As obtained using our formula.

fythynwyrk0 · Accepted Answer

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 + (\frac{d y}{d x})^{2}} d x

Let y=mx+b

\Rightarrow \frac{d y}{d x} = m

S = \int_{a}^{b} \sqrt{1 + m^{2}} d x

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}}

What is the general equation for the arclength of a