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 are 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 $\sin (\theta )$ : $\begin{align*} -\sin(\theta) + 2\cos^2(\theta)-2\sin^2(\theta) & = -\sin(\theta) + 2(1-\sin^2(\theta))-2\sin^2(\theta) \\ &= \answer[given]{-4} \sin^2(\theta) + \answer[given]{-1}\sin(\theta) + \answer[given]{2}. \end{align*}$ We then set this equal to zero and note that the equation is quadratic equation in the variable $\sin (\theta )$ . The quadratic formula gives that $\begin{align*} \sin(\theta) &= \frac{-1\pm\sqrt{\answer[given]{33}}}{8}. \end{align*}$ Now we can find cosine 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: $\graph {r=1+2\sin (\theta ),y=\tan (7\pi /6)x,y=\tan (11\pi /6)x}$