Mylee Underwood

Answered

2022-07-04

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

Answer & Explanation

Perman7z

Expert

2022-07-05Added 13 answers

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

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

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.

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

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

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

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

$\mathrm{\u25b3}y=13-5=8$

$\mathrm{\u25b3}x=4$

Thus ${A}^{2}={\mathrm{\u25b3}}^{2}y+{\mathrm{\u25b3}}^{2}x={8}^{2}+{4}^{2}$

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

As obtained using our formula.

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

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.

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

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

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

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

$\mathrm{\u25b3}y=13-5=8$

$\mathrm{\u25b3}x=4$

Thus ${A}^{2}={\mathrm{\u25b3}}^{2}y+{\mathrm{\u25b3}}^{2}x={8}^{2}+{4}^{2}$

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

As obtained using our formula.

fythynwyrk0

Expert

2022-07-06Added 7 answers

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

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

Let y=mx+b

$\Rightarrow \frac{dy}{dx}=m$

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

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:

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

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

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

Let y=mx+b

$\Rightarrow \frac{dy}{dx}=m$

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

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:

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

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

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