
# 10.2: Slopes in Polar Coordinates

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When we describe a curve using polar coordinates, it is still a curve in the$$x-y$$ plane. We would like to be able to compute slopes and areas for these curves using polar coordinates.

We have seen that $$x=r\cos\theta$$ and $$y=r\sin\theta$$ describe the relationship between polar and rectangular coordinates. If in turn we are interested in a curve given by $$r=f(\theta)$$, then we can write $$x=f(\theta)\cos\theta$$ and $$y=f(\theta)\sin\theta$$, describing $$x$$ and $$y$$ in terms of $$\theta$$ alone. The first of these equations describes $$\theta$$ implicitly in terms of $$x$$, so using the chain rule we may compute

${dy\over dx}={dy\over d\theta}{d\theta\over dx}.$

Since $$d\theta/dx=1/(dx/d\theta)$$, we can instead compute

${dy\over dx}={dy/d\theta\over dx/d\theta}= {f(\theta)\cos\theta + f'(\theta)\sin\theta\over -f(\theta)\sin\theta + f'(\theta)\cos\theta}.$

Example $$\PageIndex{1}$$:

Find the points at which the curve given by $$r=1+\cos\theta$$ has a vertical or horizontal tangent line. Since this function has period $$2\pi$$, we may restrict our attention to the interval $$[0,2\pi)$$ or $$(-\pi,\pi]$$, as convenience dictates. First, we compute the slope:

${dy\over dx}={(1+\cos\theta)\cos\theta-\sin\theta\sin\theta\over -(1+\cos\theta)\sin\theta-\sin\theta\cos\theta}= {\cos\theta+\cos^2\theta-\sin^2\theta\over -\sin\theta-2\sin\theta\cos\theta}.$

This fraction is zero when the numerator is zero (and the denominator is not zero). The numerator is $$2\cos^2\theta+\cos\theta-1$$ so by the quadratic formula $$\cos\theta={-1\pm\sqrt{1+4\cdot2}\over 4} = -1 \quad\hbox{or}\quad {1\over 2}.$$ This means $$\theta$$ is $$\pi$$ or $$\pm \pi/3$$. However, when $$\theta=\pi$$, the denominator is also $$0$$, so we cannot conclude that the tangent line is horizontal.

Setting the denominator to zero we get \eqalign{ -\theta-2\sin\theta\cos\theta &= 0\cr \sin\theta(1+2\cos\theta)&=0,\cr} so either $$\sin\theta=0$$ or $$\cos\theta=-1/2$$. The first is true when $$\theta$$ is $$0$$ or $$\pi$$, the second when $$\theta) is \(2\pi/3$$ or $$4\pi/3$$. However, as above, when $$\theta=\pi$$, the numerator is also $$0$$, so we cannot conclude that the tangent line is vertical. Figure 10.2.1 shows points corresponding to $$\theta$$ equal to $$0$$, $$\pm\pi/3$$, $$2\pi/3$$ and $$4\pi/3$$ on the graph of the function. Note that when $$\theta=\pi$$ the curve hits the origin and does not have a tangent line.

Figure 10.2.1. Points of vertical and horizontal tangency for $$r=1+\cos\theta$$.

We know that the second derivative $$f''(x)$$ is useful in describing functions, namely, in describing concavity. We can compute $$f''(x)$$ in terms of polar coordinates as well. We already know how to write $$dy/dx=y'$$ in terms of $$\theta$$, then

${d\over dx}{dy\over dx}= {dy'\over dx}={dy'\over d\theta}{d\theta\over dx}={dy'/d\theta\over dx/d\theta}.$

Example $$\PageIndex{2}$$:

We find the second derivative for the cardioid $$r=1+\cos\theta$$:

\eqalign{ {d\over d\theta}{\cos\theta+\cos^2\theta-\sin^2\theta\over -\sin\theta-2\sin\theta\cos\theta}\cdot{1\over dx/d\theta} &=\cdots= {3(1+\cos\theta)\over (\sin\theta+2\sin\theta\cos\theta)^2} \cdot{1\over-(\sin\theta+2\sin\theta\cos\theta)}\cr &={-3(1+\cos\theta)\over(\sin\theta+2\sin\theta\cos\theta)^3}.\cr}

The ellipsis here represents rather a substantial amount of algebra. We know from above that the cardioid has horizontal tangents at $$\pm \pi/3$$; substituting these values into the second derivative we get $$y''(\pi/3)=-\sqrt{3}/2$$ and $$y''(-\pi/3)=\sqrt{3}/2$$, indicating concave down and concave up respectively. This agrees with the graph of the function.

### Contributors

• Integrated by Justin Marshall.