Mylee Underwood

2022-07-04

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

Perman7z

Expert

The general equation of a line is y=mx+b
Recall the formula for arc length is $A={\int }_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}{\right)}^{2}}dx$
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.
$A=\left[\sqrt{1+{m}^{2}}x{\right]}_{a}^{b}$
$A=b\sqrt{1+{m}^{2}}-a\sqrt{1+{m}^{2}}$
$A=\left(b-a\right)\sqrt{1+{m}^{2}}$
Now let's verify to see if our formula is correct. Let y=2x+1 and the arc length we wish to find being on the x-interrval [2,6]
$A=\left(6-2\right)\sqrt{1+{2}^{2}}=4\sqrt{5}$
If we were to use pythagoras, by connecting a horizontal line to a vertical line, we would get the following"
y(2)=5
y(6)=13
$\mathrm{△}y=13-5=8$
$\mathrm{△}x=4$
Thus ${A}^{2}={\mathrm{△}}^{2}y+{\mathrm{△}}^{2}x={8}^{2}+{4}^{2}$
$A=\sqrt{80}=\sqrt{16\ast 5}=4\sqrt{5}$
As obtained using our formula.

fythynwyrk0

Expert

For the arc length of a linear function given its slope m and an interval [a,b], using the arc length formula:
$S={\int }_{a}^{b}\sqrt{1+\left(\frac{dy}{dx}{\right)}^{2}}dx$
Let y=mx+b
$⇒\frac{dy}{dx}=m$
$S={\int }_{a}^{b}\sqrt{1+{m}^{2}}dx$
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 $\sqrt{1-{m}^{2}}\ast x$, and substituting the limits of integration:
$S=\sqrt{1-{m}^{2}}\ast b-\sqrt{1-{m}^{2}}\ast a$
$S=\left(b-a\right)\sqrt{1-{m}^{2}}$

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