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 10.2.1 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 $\ds \ds 2\cos^2\theta+\cos\theta-1$ 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 $$\eqalign{ -\sin \theta-2\sin\theta\cos\theta &= 0\cr \sin\theta(1+2\cos\theta)&=0,\cr} $$ 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. $\square$
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 10.2.2 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 $\ds y''(\pi/3)=-\sqrt{3}/2$ and $\ds y''(-\pi/3)=\sqrt{3}/2$, indicating concave down and concave up respectively. This agrees with the graph of the function. $\square$
Exercises 10.2
Compute $y'=dy/dx$ and $\ds y''=d^2y/dx^2$.
Ex 10.2.1 $r=\theta$ (answer)
Ex 10.2.2 $r=1+\sin\theta$ (answer)
Ex 10.2.3 $r=\cos\theta$ (answer)
Ex 10.2.4 $r=\sin\theta$ (answer)
Ex 10.2.5 $r=\sec\theta$ (answer)
Ex 10.2.6 $r=\sin(2\theta)$ (answer)
Sketch the curves over the interval $[0,2\pi]$ unless otherwise stated. Use the first and second derivative to identify horizontal and vertical tangents and local maximum and minimum points. You can check your work with Sage.
Ex 10.2.7 $r=\sin\theta+\cos\theta$
Ex 10.2.8 $r=2+2\sin\theta$
Ex 10.2.9 $\ds r={3\over2}+\sin\theta$
Ex 10.2.10 $r= 2+\cos\theta$
Ex 10.2.11 $\ds r={1\over2}+\cos\theta$
Ex 10.2.12 $\ds r=\cos(\theta/2), 0\le\theta\le4\pi$
Ex 10.2.13 $r=\sin(\theta/3), 0\le\theta\le6\pi$
Ex 10.2.14 $\ds r=\sin^2\theta$
Ex 10.2.15 $\ds r=1+\cos^2(2\theta)$
Ex 10.2.16 $\ds r=\sin^2(3\theta)$
Ex 10.2.17 $\ds r=\tan\theta$
Ex 10.2.18 $\ds r=\sec(\theta/2), 0\le\theta\le4\pi$
Ex 10.2.19 $\ds r=1+\sec\theta$
Ex 10.2.20 $\ds r={1\over 1-\cos\theta}$
Ex 10.2.21 $\ds r={1\over 1+\sin\theta}$
Ex 10.2.22 $\ds r=\cot(2\theta)$
Ex 10.2.23 $\ds r=\pi/\theta, 0\le\theta\le\infty$
Ex 10.2.24 $\ds r=1+\pi/\theta, 0\le\theta\le\infty$
Ex 10.2.25 $\ds r=\sqrt{\pi/\theta}, 0\le\theta\le\infty$