Polar coordinates and Cartesian slopes

Understand the idea

Polar coordinates describe a point by a signed radius and an angle. To differentiate a polar curve as a curve in the Cartesian plane, convert its coordinate functions to $\displaystyle x=r(\theta)\cos\theta$ and $\displaystyle y=r(\theta)\sin\theta$ , then use the parametric derivative ratio.

A negative radius points in the direction opposite the stated angle, so the same geometric point can have several coordinate descriptions. Keep the relevant angle interval when sketching or finding intersections. The radial derivative $\displaystyle dr/d\theta$ measures changing radius, not the slope of the plotted curve.

Apply the product rule to both coordinate functions because the radius may vary with angle. Horizontal and vertical tangent tests require checking both coordinate derivatives. At the pole, or when both derivatives vanish, inspect the curve's limiting behavior rather than dividing automatically.

\frac{dy}{dx}=\frac{r'(\theta)\sin\theta+r(\theta)\cos\theta}{r'(\theta)\cos\theta-r(\theta)\sin\theta}

A method you can apply

Locate the point using the signed radius and angle.
Differentiate $\displaystyle r\cos\theta$ and $\displaystyle r\sin\theta$ with the product rule.
Form and evaluate the derivative ratio where its denominator is nonzero.

Worked example

For $\displaystyle r=1+\cos\theta$ , find the tangent line at $\displaystyle \theta=\pi/2$ .

At $\displaystyle \pi/2$ , the radius is $\displaystyle 1$ , giving the point $\displaystyle (0,1)$ .
The radial derivative is $\displaystyle r'=-\sin\theta=-1$ at that angle.
The coordinate derivatives are $\displaystyle dx/d\theta=-1$ and $\displaystyle dy/d\theta=-1$ , so the Cartesian slope is $\displaystyle 1$ .

The tangent line is

\displaystyle y-1=x

.

Try it yourself

For the circle $\displaystyle r=2$ , find $\displaystyle dy/dx$ at $\displaystyle \theta=\pi/4$ .

Show solution and reasoning

Here $\displaystyle r'=0$ , so the slope is $\displaystyle (2\cos\theta)/(-2\sin\theta)=-\cot\theta$ . At $\displaystyle \pi/4$ , the slope is $\displaystyle -1$ .