We now need to discuss some calculus topics in terms of
We will start with finding tangent lines to polar
curves. In this case we are going to
assume that the equation is in the form . With the equation in this form we can
actually use the equation for the derivative we derived when we looked at tangent lines with parametric equations. To do this however requires us to come up
with a set of parametric equations to represent the curve. This is actually pretty easy to do.
From our work in the previous section we have the following
set of conversion equations for going from polar coordinates to Cartesian coordinates.
Now, we’ll use the fact that we’re assuming that the
equation is in the form . Substituting this into these equations gives
the following set of parametric equations (with θ as the parameter) for the curve.
Now, we will need the following derivatives.
The derivative is then,
Derivative with Polar
Note that rather than trying to remember this formula it
would probably be easier to remember how we derived it and just remember the
formula for parametric equations.
Let’s work a quick example with this.
Example 1 Determine
the equation of the tangent line to at .
We’ll first need the following derivative.
The formula for the derivative becomes,
The slope of the tangent line is,
Now, at we have . We’ll need to get the corresponding x-y
coordinates so we can get the tangent line.
The tangent line is then,
For the sake of completeness here is a graph of the curve
and the tangent line.