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

Jul 12, 2018

If we wish to find the arc length of $y = m x + b$ on $\left[a , b\right]$, then $\left(b - a\right) \sqrt{1 + {m}^{2}}$ will give the correct arc length.

#### Explanation:

The general equation of a line is $y = m x + b$.

Recall the formula for arc length is $A = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}$.

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

$A = {\int}_{a}^{b} \sqrt{1 + {m}^{2}} \mathrm{dx}$

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

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

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

$A = \left(b - a\right) \sqrt{1 + {m}^{2}}$

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

$A = \left(6 - 2\right) \sqrt{1 + {2}^{2}} = 4 \sqrt{5}$

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

$y \left(2\right) = 5$
$y \left(6\right) = 13$
$\Delta y = 13 - 5 = 8$

$\Delta x = 4$

Thus ${A}^{2} = {\Delta}^{2} y + {\Delta}^{2} x = {8}^{2} + {4}^{2}$

$A = \sqrt{80} = \sqrt{16 \cdot 5} = 4 \sqrt{5}$

As obtained using our formula.

Hopefully this helps!

Jul 12, 2018

$S = \left(b - a\right) \sqrt{1 + {m}^{2}}$

#### Explanation:

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

$S = {\int}_{a}^{b} \sqrt{1 + {\left(\textcolor{red}{\frac{\mathrm{dy}}{\mathrm{dx}}}\right)}^{2}} \mathrm{dx}$

Let $y = m x + b$

$\implies \textcolor{red}{\frac{\mathrm{dy}}{\mathrm{dx}} = m}$

$S = {\int}_{a}^{b} \sqrt{1 + {m}^{2}} \mathrm{dx}$

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 $\sqrt{1 - {m}^{2}} \cdot x$, and substituting the limits of integration:

$S = \sqrt{1 - {m}^{2}} \cdot b - \sqrt{1 - {m}^{2}} \cdot a$

$S = \left(b - a\right) \sqrt{1 - {m}^{2}}$