Arc Length

 

Or, Integrating the Hypotenuses of Infinitesimally Small Triangles

Say each Pi is actually the Hypotenuse of a triangle of sides Δx and Δy. The length of each Pi would be sqrt((x1-x2)^2+(y1-y2)^2) or sqrt((Δx)^2+(Δy)^2) as shown above. The Δx can be factored out to obtain the following:

From here we can clearly integrate to find the Length, L, of the curve. 

This can also be done in terms of y. Just switch the variables.

Arc Length of Parametric Plots

What to do when we have 3 variables in 2 dimensions....

x(t)=6cos(t)+t, y(t)=6sin(2t+3), t{-3,3}

We can take advantage of a simple substitution to make everything work.

So let's solve for the arc length of the parametric plot above... And while were at it, let's take a look at the tangent of the curve at any point, t.

1. The tangent at any point t, would thus be dy/dt/dx/dt:

          (1-6sin(t)) / 12cos(2t+3)

   And then just plug in the t-value to get the slope/tangent line.

2. Let's rewrite these derivatives more clearly and plug them into the integral.