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

2 Answers
Jul 12, 2018

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!

Jul 12, 2018

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)