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

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!

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)#