Derivatives of polar functions

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We differentiate polar functions.

The previous section discussed a special class of parametric functions called polar functions. We know that $\begin{align*} \d x &= x'(t) \d t\\ \d y &= y'(t) \d t, \end{align*}$ and so we can compute the derivative of $y$ with respect to $x$ using differentials: $\frac {\d y}{\d x} = \frac {y'(t) \d t}{x'(t) \d t} = \frac {y'(t)}{x'(t)}$ provided that $x'(t) \ne 0$ . With polar functions we have $\begin{align*} x(\theta) &= r(\theta) \cdot \cos(\theta) \\ y(\theta) &= r(\theta) \cdot \sin(\theta), \end{align*}$ so $\begin{align*} \dd[y]{x} &= \frac{y'(\theta)}{x'(\theta)} \\ &= \frac{r'(\theta) \sin(\theta)+r(\theta) \cos(\theta)}{r'(\theta)\cos(\theta)-r(\theta)\sin(\theta)} \end{align*}$ provided that $x'(\theta )\ne 0$ .

Consider the limaçon $r(\theta ) =1+2\sin (\theta )$ on the interval $[0,2\pi ]$ :

Find the equation of the tangent line to the curve at $\theta =\pi /4$ .

We start by computing the derivative in rectangular coordinates. Recall that $\begin{align*} x(\theta) &= \left(1+2\sin(\theta)\right)\cdot \cos(\theta) = \cos(\theta)+2\sin(\theta)\cos(\theta)\\ y(\theta) &= \left(1+2\sin(\theta)\right)\cdot \sin(\theta) = \sin(\theta)+2\sin^2(\theta) \end{align*}$ Consequently, $\begin{align*} x'(\theta) &= \answer[given]{-\sin(\theta) + 2\cos^2(\theta)-2\sin^2(\theta)} \\ y'(\theta) &= \answer[given]{\cos(\theta) + 4\sin(\theta)\cos(\theta)} \end{align*}$ and therefore $\dd [y]{x} = \answer [given]{\frac {\cos (\theta ) + 4\sin (\theta )\cos (\theta )}{-\sin (\theta ) + 2\cos ^2(\theta )-2\sin ^2(\theta )}}.$ When $\theta =\pi /4$ , $\begin{align*} x(\theta) &= \answer[given]{1+\sqrt{2}/2}\\ y(\theta) &=\answer[given]{1+\sqrt{2}/2}\\ \dd[y]{x} &=\answer[given]{-2\sqrt{2}-1} \end{align*}$ Thus the equation of the line (in polarrectangular coordinates) tangent to the limaçon at $\theta =\pi /4$ is $y=(-2\sqrt {2}-1)\big (x-(1+\sqrt {2}/2)\big )+1+\sqrt {2}/2 \approx -3.83 x+8.24.$ $\graph {r=1+2\sin (\theta ), y=(-2\sqrt {2}-1)(x-(1+\sqrt {2}/2))+1+\sqrt {2}/2}$

Consider the limaçon given by $r(\theta ) =1+2\sin (\theta )$ on the interval $[0,2\pi ]$ . Find Cartesian coordinates for the points where the tangent line to this limaçon is horizontal.

To find these points where the tangent lines is horizontal, we seek points where $\dd [y]{x} = \answer [given]{0}$ . From earlier, we discovered $\dd [y]{x} =\frac {\cos (\theta ) + 4\sin (\theta )\cos (\theta )}{-\sin (\theta ) + 2\cos ^2(\theta )-2\sin ^2(\theta )}$ so now we must find where the numerator is $0$ . Since the numerator is equal to $\cos (\theta )(1+ 4\sin (\theta ))$ and since a product is zero when botheither of the terms is zero, the numerator is zero when

$\cos (\theta )=0$ or
$1+4\sin (\theta )=0$ .

Let’s examine the case of $\cos (\theta ) = 0$ first, and restrict to the situation where $\theta$ is between $0$ and $2\pi$ . If $\theta$ is in the interval $[0,\pi ]$ , then $\cos (\theta )=0$ when $\theta =\answer [given]{\pi /2}$ . If $\theta$ is in the interval $[\pi ,2\pi ]$ , then $\cos (\theta )=0$ when $\theta =\answer [given]{3\pi /2}$ .

The corresponding points in rectangular coordinates are

$\begin{align*} x(\pi/2) &= \left(1+2\sin(\pi/2)\right)\cdot\cos(\pi/2)\\ &= 0,\\ y(\pi/2) &= \left(1+2\sin(\pi/2)\right)\cdot\sin(\pi/2)\\ &= 3, \end{align*}$ meaning $(x,y) = (0,3)$ , and $\begin{align*} x(3\pi/2) &= \left(1+2\sin(3\pi/2)\right)\cdot\cos(3\pi/2)\\ &= 0,\\ y(3\pi/2) &= \left(1+2\sin(3\pi/2)\right)\cdot\sin(3\pi/2)\\ &= 1, \end{align*}$ meaning $(x,y) = (0,1)$ .

But the numerator was the product of $\cos (\theta )$ and another term, namely $1+4\sin (\theta )$ So the numerator is also zero when

$\begin{align*} 1+4\sin(\theta)&=0\\ 4\sin(\theta)&=-1\\ \sin(\theta)&=\frac{-1}{4}. \end{align*}$ At this point we have a choice to make: We can either apply the arcsin function and solve for $\theta$ , or we can work with $\sin (\theta )$ directly. Perhaps the easier route is to work with $\sin (\theta )$ directly. To explore $\sin (\theta )$ , we may choose an algebraic method employing the Pythagorean identity, or a geometric method looking at the unit circle with the Pythagorean theorem. Let’s take the geometric route to compute $\cos (\theta )$ .

Hence the derivative is zerois oneis undefined at $\begin{align*} x(\theta) &= \left(1+2\sin(\theta)\right)\cdot\cos(\theta)\\ &= \left(1+2(-1/4)\right)\cdot \sqrt{15}/4\\ &= \sqrt{15}/8,\\ y(\theta) &= \left(1+2\sin(\theta)\right)\cdot\sin(\theta)\\ &= \left(1+2(-1/4)\right)\cdot (-1/4)\\ &= -1/8, \end{align*}$ meaning $(x,y) = (\sqrt {15}/8, -1/8)$ , and $\begin{align*} x(\theta) &= \left(1+2\sin(\theta)\right)\cdot\cos(\theta)\\ &= \left(1+2(-1/4)\right)\cdot (-\sqrt{15}/4)\\ &= -\sqrt{15}/8,\\ y(\theta) &= \left(1+2\sin(\theta)\right)\cdot\sin(\theta)\\ &= \left(1+2(-1/4)\right)\cdot (-1/4)\\ &= -1/8, \end{align*}$ meaning $(x,y) = (-\sqrt {15}/8, -1/8)$ .

In summary, the graph of $r(\theta ) = 1+2\sin (\theta )$ on the interval $[0,2\pi ]$ has horizontal tangent lines at the points

$(x,y) = (0,\answer [given]{3})$ ,
$(x,y) = (0,1)$ ,
$(x,y) = (\sqrt {15}/8, -1/8)$ , and
$(x,y) = (-\sqrt {15}/8, -1/8)$ .

You can confirm this by looking at the graph below: $\graph {r=1+2\sin (\theta ),y=3,y=1,y=-1/8}$

Again consider our friend the limaçon given by $r(\theta ) =1+2\sin (\theta )$ on the interval $[0,2\pi ]$ . Find Cartesian coordinates for the points where the tangent line to this limaçon is vertical.

To find the vertical tangent lines, we seek points where the denominator of $\dd [y]{x}=\frac {\cos (\theta ) + 4\sin (\theta )\cos (\theta )}{-\sin (\theta ) + 2\cos ^2(\theta )-2\sin ^2(\theta )}$ is equal to zero. To start, we will convert $\cos ^2(\theta )$ to $1-\answer [given]{\sin ^2(\theta )}$ , and express the denominator as a quadratic expression in $\sin (\theta )$ : $\begin{align*} -\sin(\theta) + 2\cos^2(\theta)-2\sin^2(\theta) &=0\\ -\sin(\theta) + 2(1-\sin^2(\theta))-2\sin^2(\theta) &=0\\ \answer[given]{4} \sin^2(\theta) + \answer[given]{1}\sin(\theta) + \answer[given]{-2} &= 0. \end{align*}$ Since this is a quadratic equation in the variable $\sin (\theta )$ , we invoke the quadratic formula to see $\begin{align*} \sin(\theta) &= \frac{-1\pm\sqrt{\answer[given]{33}}}{8}. \end{align*}$ Now we can find cosine using either geometrically via the unit circle or algebraically with the Pythagorean identity. This time, let’s take the more algebraic route with the Pythagorean identity. When $\sin (\theta ) =\frac {-1+\sqrt {33}}{8}$ , then $\begin{align*} \cos^2(\theta) + \sin^2(\theta) &= 1\\ \cos^2(\theta) + \left(\frac{-1+\sqrt{33}}{8}\right) &= 1. \end{align*}$ So $\cos (\theta ) = \pm \frac {\sqrt {(\answer [given]{15}+\sqrt {33})/2}}{4}$ and when $\sin (\theta ) =\frac {-1-\sqrt {33}}{8}$ , $\begin{align*} \cos^2(\theta) + \sin^2(\theta) &= 1\\ \cos^2(\theta) + \left(\frac{-1-\sqrt{33}}{8}\right) &= 1. \end{align*}$ So $\cos (\theta ) = \pm \frac {\sqrt {(\answer [given]{15}-\sqrt {33})/2}}{4}.$ So now we know that the graph of $r(\theta ) =1+2\sin (\theta )$ on $[0,2\pi ]$ has vertical tangent lines at the four points

Recalling that $\begin{align*} x(\theta) &= \left(1+2\sin(\theta)\right)\cdot\cos(\theta)\\ y(\theta) &= \left(1+2\sin(\theta)\right)\cdot\sin(\theta) \end{align*}$ we see that the graph of $r(\theta ) =1+2\sin (\theta )$ on $[0,2\pi ]$ has vertical tangent lines at

You can confirm this visually by looking at the graph below: $\graph {r=1+2\sin (\theta ), x=\left (1+\frac {-1+\sqrt {33}}{4}\right )\frac {\sqrt {(15+\sqrt {33})/2}}{4}, x=\left (1+\frac {-1+\sqrt {33}}{4}\right )\frac {-\sqrt {(15+\sqrt {33})/2}}{4}, x=\left (1+\frac {-1-\sqrt {33}}{4}\right )\frac {\sqrt {(15-\sqrt {33})/2}}{4}, x=\left (1+\frac {-1-\sqrt {33}}{4}\right )\frac {-\sqrt {(15-\sqrt {33})/2}}{4}}$

When the graph of the polar function $r(\theta )$ intersects the origin (sometimes called the “pole”), then $r(\alpha )=0$ for some angle $\alpha$ .

When $r(\alpha ) = 0$ , what is the formula for $\dd [y]{x}$ ?

$\dd [y]{x}= \sin (\alpha )$ . $\dd [y]{x}= \cos (\alpha )$ . $\dd [y]{x}= \tan (\alpha )$ .

The answer to the question above leads us to an interesting point. It tells us the slope of the tangent line at the pole. When a polar graph touches the pole at $\theta =\alpha$ , the equation of the tangent line in polar coordinates at the pole is $\theta =\alpha$ .

Find the equations of the lines tangent to the graph of $r(\theta )=1+2\sin (\theta )$ at the pole.

We need to know when $r(\theta )=0$ . $\begin{align*} 1+2\sin(\theta) &= 0\\ \sin(\theta) &= -1/2\\ \theta &= \frac{7\pi}{6},\ \frac{\answer[given]{11}\pi}6. \end{align*}$ Thus the equations of the tangent lines, in polarrectangular coordinates, are $\begin{align*} \theta &=7\pi/6 \theta \\ \theta &= \answer[given]{11}\pi/6. \end{align*}$ In rectangular form, the tangent lines are $y=\tan (7\pi /6)x$ and $y=\answer [given]{\tan (11\pi /6)}x$ . You can confirm this by looking at the graph below: $\graph {r=1+2\sin (\theta ),y=\tan (7\pi /6)x,y=\tan (11\pi /6)x}$