# Implicit use of the Implicit Function Theorem when finding tangent lines to polar curves.

Recently I found myself having to teach students how to find the slope of a tangent line to a curve in $\mathbb R^2$ given in polar coordinates by the equation $r = f(\theta)$. The students' calculus book instructs them that surely the slope of the tangent line must be given by $\frac{dy}{dx}$ and uses the chain rule to calculate $$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$$ Then, using the fact that $x = r\cos\theta$ and $y = r\sin\theta$ they are therefore able to find a formula for $\frac{dy}{dx}$ in terms of $r$ and $\theta$. There is however a problem with this line of reasoning, namely that there is no compelling reason for $y$ to be defined even locally as a function of $x$. Knowledge of the implicit function theorem allows one to formulate the hypothesis necessary to make the above use of the chain rule correct. However it would be nice to justify the use of the chain rule here using only methods available to a first year calculus student. My question is therefore

Is there a nice way (intuitive or rigorous) of explaining when and why $y$ is a differentiable function of $x$ in this case, using only methods available to a first year calculus student?

Yes, it is possible to look through the proof of the implicit function theorem and simplify it in this case, but I am especially interested in hearing the thoughts of experienced teachers, so I hope I am justified in asking the question.

• The potential problems don't begin with the chain rule. The first potential problem is that there might not be such a thing as "the slope of the tangent line", either because the curve doesn't have a (unique) tangent at that point or because the tangent is vertical. Asking for the slope seems to imply that there is a slope, and that implies that $y$ is defined locally as a function of $x$. The problem in applying the chain rule isn't that $y$ might not be defined locally as a function of $x$, but that the slope might be defined without $y$ and $x$ being differentiable with respect to $\theta$. Commented Mar 11, 2012 at 8:02
• Yes, but I think it is reasonable (depending upon the class) to give the students a test they can perform on $f(\theta)$ to determine whether or not the slope exists, and then proceed from there. Commented Mar 11, 2012 at 17:04

The first is to solve for $\theta$ as a function of $x$. If you write $x-f(\theta)\cos\theta=0$, you use the implicit function theorem to do it. But if you write it $x=f(\theta)\cos\theta:=g(\theta)$, it is just the inverse function theorem for real functions of one real variable, which is included in he syllabus of many first year calculus courses. If it is not included, you can justify it as follows: if $g'(\theta_0)\ne0$, then $g$ is strictly monotone on a neighborhood around $\theta_0$, there is an inverse, the inverse is also strictly monotone and is differentiable.
The second is writing the curve in parametric form \begin{align*} x&=f(\theta)\cos\theta,\\ y&=f(\theta)\sin\theta. \end{align*} Then $(x',y')$ is a vector in the direction of the tangent (if it is not $(0,0)$). From it you can compute the slope of the tangent, and find when that tangent is vertical.
• @joriki No, but under certain conditions that one can easily check, there is a tangent line, and the tangent line has a slope. If $x'(\theta_0)\ne0$, then the curve has a tangent at the point $(x(\theta_0),y(\theta_0))$, and the tangent has a slope. If $x'(\theta_0)=0$ and $y'(\theta_0)\ne0$, then the curve has a vertical tangent line at the point $(x(\theta_0),y(\theta_0))$. If $x'(\theta_0)=y'(\theta_0)=0$, then we have a singularity: there may exist no tangent. Commented Mar 11, 2012 at 9:06