Derivatives of Polar Functions

• The position of points on the plane can be described in different coordinate systems. Besides the Cartesian coordinate system, the polar coordinate system is also widespread. In this system, the position of any point $$M$$ is described by two numbers (see Figure $$1$$):

• the length of the radius vector (r) drawn from the origin (O) (pole) to the point (M);
• the polar angle $$\theta$$ formed by segment $$OM$$ and the positive direction of the $$x$$-axis. The angle $$\theta$$ is measured counterclockwise.

The equation $$r = f\left( \theta \right)$$, which expresses the dependence of the length of the radius vector $$r$$ on the polar angle $$\theta$$ describes a curve in the plane and is called the polar equation of the curve.

For example, the Archimedean spiral (Figure $$2$$) is described by the polar equation

$r = a\theta ,$

where $$a$$ is a parameter determining the density of spiral turns.

The separation distance between successive turnings in the Archimedian spiral is constant and equal to $$2\pi a.$$

The general formulas for converting the polar coordinates $$\left( {r,\theta } \right)$$ to Cartesian ones $$\left( {x,y} \right)$$ are as follows:

${x = r\cos \theta ,\;\;}\kern-0.3pt{y = r\sin \theta .}$

If a curve is given by a polar equation $$r = f\left( \theta \right),$$ then in Cartesian coordinates it is described by the system of equations

\left\{ \begin{aligned} x &= f\left( \theta \right)\cos\theta \\ y &= f\left( \theta \right)\sin\theta \end{aligned} \right.,

As you can see, these equations are the parametric equations of the polar curve where $$\theta$$ is a parameter. Then the derivative $$\large{\frac{{dy}}{{dx}}}\normalsize$$ of a polar function $$r = f\left( \theta \right)$$ is defined by the formula for the derivative of a parametric function:

${\frac{{dy}}{{dx}} = {y’_x} = \frac{{{y’_\theta }}}{{{x’_\theta }}} } = {\frac{{{{\left( {f\left( \theta \right)\sin \theta } \right)}^\prime }}}{{{{\left( {f\left( \theta \right)\cos\theta } \right)}^\prime }}} } = {\frac{{f’\left( \theta \right)\sin \theta + f\left( \theta \right)\cos\theta }}{{f’\left( \theta \right)\cos\theta – f\left( \theta \right)\sin \theta }}.}$

Consider examples of calculating derivatives for some polar curves.

• Solved Problems

Click a problem to see the solution.

Example 1

Find the derivative $$\large\frac{{dy}}{{dx}}\normalsize$$ of the Archimedean spiral.

Example 2

Find the derivative $$\large\frac{{dy}}{{dx}}\normalsize$$ of the cardioid given by the equation $r = f\left( \theta \right) = a\left( {1 + \cos \theta } \right).$

Example 3

Find the angle of intersection $$\color{darkgreen}{\alpha}$$ of two cardioids $${r_1} = 1 + \cos \theta$$ and $${r_2} = 1 – \cos \theta.$$

Example 4

Find the derivative $$\large{\frac{{dy}}{{dx}}}\normalsize$$ of the polar function $$r = 1 + \sin \theta$$ at $$\theta = \large{\frac{\pi }{4}}\normalsize.$$

Example 5

Find the derivative $$\large{\frac{{dy}}{{dx}}}\normalsize$$ of the polar function $$r = 3 + 2\cos \theta$$ at $$\theta = \large{\frac{\pi }{6}}\normalsize.$$

Example 6

Find the derivative $$\large\frac{{dy}}{{dx}}\normalsize$$ of the logarithmic spiral defined by the equation $r = f\left(\theta\right)=ae^{b\theta},$ where $$a$$, $$b$$ are real numbers.

Example 7

Find the angle $$\alpha$$ between the polar curves $${r_1} = {e^\theta }$$ and $${r_2} = {e^{-\theta}}$$ at the point of intersection.

Example 8

Find the derivative $$\large{\frac{{dy}}{{dx}}}\normalsize$$ of the hyperbolic spiral $$r = \large{\frac{a}{\theta }}\normalsize.$$

Example 9

Find the derivative $$\large\frac{{dy}}{{dx}}\normalsize$$ of the circle and calculate its values for the polar angles $$\theta = {\large\frac{\pi }{4}\normalsize},{\large\frac{{3\pi }}{4}\normalsize}.$$

Example 10

Find the derivative $$\large\frac{{dy}}{{dx}}\normalsize$$ of the lemniscate of Bernoulli given by the equation ${r^2} = \cos 2\theta .$

Example 11

Find the derivative $$\large\frac{{dy}}{{dx}}\normalsize$$ of Galileo’s spiral.

Example 12

Find the derivative $$\large\frac{{dy}}{{dx}}\normalsize$$ of Fermat’s spiral given by the polar equation $r = \sqrt \theta.$

Example 13

Find the derivative $$\large{\frac{{dy}}{{dx}}}\normalsize$$ of the polar curve $$r = \large{\frac{a}{{\sqrt \theta }}}\normalsize$$ (the lituus curve).

Example 14

Compute the derivative $$\large{\frac{{dy}}{{dx}}}\normalsize$$ of the polar curve $$r = {\sin ^2}\theta$$ at $$\theta = \large{\frac{\pi }{3}}\normalsize.$$

Example 15

Find the derivative $$\large{\frac{{dy}}{{dx}}}\normalsize$$ of the polar function $$r = \cot \theta$$ at $$\theta = \large{\frac{\pi }{6}}\normalsize.$$

Example 16

Find the derivative $$\large{\frac{{dy}}{{dx}}}\normalsize$$ of the polar function $$r = \sec \left( {2\theta } \right)$$ at $${\theta = \large{\frac{\pi }{4}}\normalsize}.$$

Example 17

The golden spiral is defined by the polar equation $$r = {\varphi ^{\large{\frac{{2\theta }}{\pi }}\normalsize}},$$ where $$\varphi = \large{\frac{{1 + \sqrt 5 }}{2}}\normalsize \approx 1.618$$ is the golden ratio. Find the angle $$\theta$$ at which the tangent line to the curve is horizontal.

Example 1.

Find the derivative $$\large\frac{{dy}}{{dx}}\normalsize$$ of the Archimedean spiral.

Solution.

The equation of the Archimedean spiral (see Figure $$2$$ above) in polar coordinates can be written as

$r = f\left( \theta \right) = a\theta .$

The derivative $$\large\frac{{dy}}{{dx}}\normalsize$$ is given by

${\frac{{dy}}{{dx}} = {y’_x} = \frac{{{y’_\theta }}}{{{x’_\theta }}} } = {\frac{{f’\left( \theta \right)\sin \theta + f\left( \theta \right)\cos\theta }}{{f’\left( \theta \right)\cos\theta – f\left( \theta \right)\sin \theta }}.}$

Substituting the function $$f\left( \theta \right),$$ we have

${\frac{{dy}}{{dx}} }={ \frac{{{{\left( {a\theta } \right)}^\prime }\sin \theta + a\theta \cos\theta }}{{{{\left( {a\theta } \right)}^\prime }\cos\theta – a\theta \sin \theta }} } = {\frac{{a\sin \theta + a\theta \cos \theta }}{{a\cos \theta – a\theta \sin \theta }} } = {\frac{{\sin \theta + \theta \cos \theta }}{{\cos \theta – \theta \sin \theta }}.}$

Divide the numerator and denominator by $${\cos \theta }$$ (assuming that $$\theta \ne \large\frac{\pi }{2}\normalsize + \pi n,$$ $$n \in \mathbb{Z}$$). Then we get the following expression for the derivative:

${\frac{{dy}}{{dx}} }={ \frac{{\tan\theta + \theta }}{{1 – \theta \tan \theta }}.}$

The last formula can be further simplified using the trigonometric identity

${\tan \left( {\alpha + \beta } \right) }={ \frac{{\tan \alpha + \tan \beta }}{{1 – \tan \alpha \cdot \tan \beta }}.}$

For this we represent the angle $$\theta$$ as

$\theta = \tan \left( {\arctan\theta } \right).$

As a result, we obtain

${\frac{{dy}}{{dx}} }={ \frac{{\tan\theta + \theta }}{{1 – \theta \tan \theta }} } = {\frac{{\tan\theta + \tan \left( {\arctan\theta } \right)}}{{1 – \tan \theta \cdot \tan \left( {\arctan\theta } \right)}} } = {\tan \left( {\theta + \arctan\theta } \right).}$

It is interesting that the derivative of the Archimedean spiral does not depend on the radius $$r$$, but is defined only by the angle $$\theta$$. This reflects the self-similarity property of the Archimedean spiral. This feature is also typical for many other plane curves.

Example 2.

Find the derivative $$\large\frac{{dy}}{{dx}}\normalsize$$ of the cardioid given by the equation $r = f\left( \theta \right) = a\left( {1 + \cos \theta } \right).$

Solution.

First we calculate the derivative of the polar function:

${f’\left( \theta \right) = {\left( {a\left( {1 + \cos \theta } \right)} \right)^\prime } } = { – a\sin \theta .}$

Then the derivative $$\large\frac{{dy}}{{dx}}\normalsize$$ of the curve is given by

${\frac{{dy}}{{dx}} = {y’_x} = \frac{{{y’_\theta }}}{{{x’_\theta }}} } = {\frac{{f’\left( \theta \right)\sin \theta + f\left( \theta \right)\cos\theta }}{{f’\left( \theta \right)\cos\theta – f\left( \theta \right)\sin\theta }} } = {\frac{{ – {{\sin }^2}\theta + \cos \theta + {{\cos }^2}\theta }}{{ – \sin \theta \cos \theta – \sin \theta – \sin \theta \cos \theta }}}.$

Using the double angle formulas

${\cos 2\theta = {\cos ^2}\theta – {\sin ^2}\theta ,\;\;}\kern-0.3pt {\sin 2\theta = 2\sin \theta \cos \theta ,}$

we get

$\frac{{dy}}{{dx}} = – \frac{{\cos 2\theta + \cos \theta }}{{\sin 2\theta + \sin \theta }}.$

We then transform the expression for the derivative using the trigonometric identities

${{\cos \alpha + \cos \beta }={ 2\cos \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha – \beta }}{2},\;\;}}\kern-0.3pt {{\sin \alpha + \sin \beta }={ 2\sin \frac{{\alpha + \beta }}{2}\cos \frac{{\alpha – \beta }}{2}.}}$

As a result, we have

$\require{cancel} {\frac{{dy}}{{dx}} = – \frac{{\cos 2\theta + \cos \theta }}{{\sin 2\theta + \sin \theta }} } = { – \frac{{2\cos \frac{{2\theta + \theta }}{2}\cos \frac{{2\theta – \theta }}{2}}}{{2\sin \frac{{2\theta + \theta }}{2}\cos \frac{{2\theta – \theta }}{2}}} } = { – \frac{{\cos \frac{{3\theta }}{2}\cancel{\cos \frac{\theta }{2}}}}{{\sin \frac{{3\theta }}{2}\cancel{\cos \frac{\theta }{2}}}} } = { – \cot \frac{{3\theta }}{2}.}$

The derivative $$\large\frac{{dy}}{{dx}}\normalsize$$ is defined under conditions

${\left\{ \begin{array}{l} \cos\frac{\theta }{2} \ne 0\\ \sin \frac{{3\theta }}{2} \ne 0 \end{array} \right.,\;\;}\Rightarrow {\left\{ \begin{array}{l} \frac{\theta }{2} \ne \frac{\pi }{2} + \pi n\\ \frac{{3\theta }}{2} \ne \pi k \end{array} \right.,\;\;}\Rightarrow {\left\{ \begin{array}{l} \theta \ne \pi + 2\pi n,\;n \in \mathbb{Z}\\ \theta \ne \frac{2}{3}\pi k,\;k \in \mathbb{Z} \end{array} \right..}$

In the interval $$\left( {-\pi,\pi} \right)$$, these restrictions correspond to the values $$\theta = – \pi , – \frac{{2\pi }}{3},0,\frac{{2\pi }}{3},\pi .$$ The derivative of the cardioid does not exist at the indicated points.

The cardioid curve (Figure $$3$$) resembles the image of the heart (the name “cardioid” comes from the Greek word for “heart”) and has a number of remarkable properties.

Beautiful mathematical objects and structures of the same type often arise in various fields, which at first glance are not related to each other. Such examples make us think again about the amazing unity of the world and nature. So, a cardioid unexpectedly appears in the famous Mandelbrot fractal set, occupying its central part (Figure $$4$$).

Example 3.

Find the angle of intersection $$\color{darkgreen}{\alpha}$$ of two cardioids $${r_1} = 1 + \cos \theta$$ and $${r_2} = 1 – \cos \theta.$$

Solution.

First we determine the point of intersection of these curves:

${{r_1} = {r_2},}\;\; \Rightarrow {1 + \cos \theta = 1 – \cos \theta ,} \Rightarrow {2\cos \theta = 0,}\;\; \Rightarrow {\theta = \frac{\pi }{2}.}$

Compute the derivatives $$\large{\frac{{dy}}{{dx}}}\normalsize$$ of both curves at $${\theta = \large{\frac{\pi }{2}}}\normalsize.$$

For the $$1$$st cardioid $${r_1} = 1 + \cos \theta:$$

${{\left( {\frac{{dy}}{{dx}}} \right)_1} = \frac{{f^\prime\left( \theta \right)\sin \theta + f\left( \theta \right)\cos \theta }}{{f^\prime\left( \theta \right)\cos \theta – f\left( \theta \right)\sin \theta }} }={ \frac{{\left( { – \sin \theta } \right)\sin \theta + \left( {1 + \cos \theta } \right)\cos \theta }}{{\left( { – \sin \theta } \right)\cos \theta – \left( {1 + \cos \theta } \right)\sin \theta }} }={ \frac{{ – {{\sin }^2}\theta + \cos \theta + {{\cos }^2}\theta }}{{ – \sin \theta \cos \theta – \sin \theta – \sin \theta \cos \theta }} }={ – \frac{{\cos 2\theta + \cos \theta }}{{\sin 2\theta + \sin \theta }}.}$

When $${\theta = \large{\frac{\pi }{2}}\normalsize},$$ the derivative is equal to

${\frac{{dy}}{{dx}}{\left( {\theta = \frac{\pi }{2}} \right)_1} }={ – \frac{{\cos \pi + \cos \frac{\pi }{2}}}{{\sin \pi + \sin \frac{\pi }{2}}} }={ – \frac{{\left( { – 1} \right) + 0}}{{0 + 1}} }={ 1.}$

Similarly we find the derivative for the $$2$$nd curve $${r_2} = 1 – \cos \theta:$$

${{\left( {\frac{{dy}}{{dx}}} \right)_2} = \frac{{f^\prime\left( \theta \right)\sin \theta + f\left( \theta \right)\cos \theta }}{{f^\prime\left( \theta \right)\cos \theta – f\left( \theta \right)\sin \theta }} }={ \frac{{\sin \theta \sin \theta + \left( {1 – \cos \theta } \right)\cos \theta }}{{\sin \theta \cos \theta – \left( {1 – \cos \theta } \right)\sin \theta }} }={ \frac{{{{\sin }^2}\theta + \cos \theta – {{\cos }^2}\theta }}{{\sin \theta \cos \theta – \sin \theta + \sin \theta \cos \theta }} }={ – \frac{{\cos 2\theta – \cos \theta }}{{\sin 2\theta – \sin \theta }}.}$

Respectively when $${\theta = \large{\frac{\pi }{2}}\normalsize},$$ we have

${\frac{{dy}}{{dx}}{\left( {\theta = \frac{\pi }{2}} \right)_2} = – \frac{{\cos \pi – \cos \frac{\pi }{2}}}{{\sin \pi – \sin \frac{\pi }{2}}} }={ – \frac{{\left( { – 1} \right) – 0}}{{0 – 1}} }={ – 1.}$

As the slope $$k$$ of the tangent line is equal to the value of the derivative, we get $${k_1} = 1,$$ $${k_2} = -1.$$ Hence, the tangent of the angle $$\alpha$$ between the curves is given by

${\tan \alpha = \frac{{{k_2} – {k_1}}}{{1 + {k_1}{k_2}}} }={ \frac{{ – 1 – 1}}{{1 + 1 \cdot \left( { – 1} \right)}} }={ \infty .}$

This means that $${\alpha = \large{\frac{\pi }{2}}\normalsize}.$$

Example 4.

Find the derivative $$\large{\frac{{dy}}{{dx}}}\normalsize$$ of the polar function $$r = 1 + \sin \theta$$ at $$\theta = \large{\frac{\pi }{4}}\normalsize.$$

Solution.

We calculate the derivative $$\large{\frac{{dy}}{{dx}}}\normalsize$$ using the formula

$\frac{{dy}}{{dx}} = \frac{{f^\prime\left( \theta \right)\sin \theta + f\left( \theta \right)\cos \theta }}{{f^\prime\left( \theta \right)\cos \theta – f\left( \theta \right)\sin \theta }}.$

Given that $$f\left( \theta \right) = 1 + \sin \theta ,$$ we get

${f^\prime\left( \theta \right) = \left( {1 + \sin \theta } \right)^\prime }={ \cos \theta .}$

Hence

${\frac{{dy}}{{dx}} \text{ = }}\kern0pt{\frac{{\cos \theta \sin \theta + \left( {1 + \sin \theta } \right)\cos \theta }}{{\cos \theta \cos \theta – \left( {1 + \sin \theta } \right)\sin \theta }} }={ \frac{{\sin \theta \cos \theta + \cos \theta + \sin \theta \cos \theta }}{{{{\cos }^2}\theta – \sin \theta – {{\sin }^2}\theta }} }={ \frac{{2\sin \theta \cos \theta + \cos \theta }}{{{{\cos }^2}\theta – {{\sin }^2}\theta – \sin \theta }} }={ \frac{{\sin 2\theta + \cos \theta }}{{\cos 2\theta – \sin \theta }}.}$

At the point $$\theta = \large{\frac{\pi }{4}}\normalsize,$$ we have

$\require{cancel}{\frac{{dy}}{{dx}}\left( {\theta = \frac{\pi }{4}} \right) }={ \frac{{\sin \frac{\pi }{2} + \cos \frac{\pi }{4}}}{{\cos \frac{\pi }{2} – \sin \frac{\pi }{4}}} }={ \frac{{1 + \frac{{\sqrt 2 }}{2}}}{{0 – \frac{{\sqrt 2 }}{2}}} }={ \frac{{\frac{{2 + \sqrt 2 }}{2}}}{{ – \frac{{\sqrt 2 }}{2}}} }={ – \frac{{2 + \sqrt 2 }}{{\sqrt 2 }} }={ – \frac{{\left( {2 + \sqrt 2 } \right)\sqrt 2 }}{{\sqrt 2 \cdot \sqrt 2 }} }={ – \frac{{2\sqrt 2 + 2}}{2} }={ – \frac{{\cancel{2}\left( {\sqrt 2 + 1} \right)}}{\cancel{2}} }={ – \sqrt 2 – 1.}$

Example 5.

Find the derivative $$\large{\frac{{dy}}{{dx}}}\normalsize$$ of the polar function $$r = 3 + 2\cos \theta$$ at $$\theta = \large{\frac{\pi }{6}}\normalsize.$$

Solution.

The derivative $$\large{\frac{{dy}}{{dx}}}\normalsize$$ for a polar function $$r = f\left( \theta \right)$$ is expressed in the form

$\frac{{dy}}{{dx}} = \frac{{f^\prime\left( \theta \right)\sin \theta + f\left( \theta \right)\cos \theta }}{{f^\prime\left( \theta \right)\cos \theta – f\left( \theta \right)\sin \theta }}.$

As $$r = 3 + 2\cos \theta,$$ we have

${f^\prime\left( \theta \right) = \left( {3 + 2\cos \theta } \right)^\prime }={ – 2\sin \theta .}$

The derivative is given by

${\frac{{dy}}{{dx}} \text{ = }}\kern0pt{\frac{{\left( { – 2\sin \theta } \right)\sin \theta + \left( {3 + 2\cos \theta } \right)\cos \theta }}{{\left( { – 2\sin \theta } \right)\cos \theta + \left( {3 + 2\cos \theta } \right)\sin \theta }} }={ \frac{{2\left( {{{\cos }^2}\theta – {{\sin }^2}\theta } \right) + 3\cos \theta }}{{ – 4\sin \theta \cos \theta – 3\sin \theta }} }={ – \frac{{2\cos \left( {2\theta } \right) + 3\cos \theta }}{{2\sin \left( {2\theta } \right) + 3\sin \theta }}.}$

Substitute $$\theta = \large{\frac{\pi }{6}}\normalsize:$$

${\frac{{dy}}{{dx}}\left( {\theta = \frac{\pi }{6}} \right) = – \frac{{2\cos \frac{\pi }{3} + 3\cos \frac{\pi }{6}}}{{2\sin \frac{\pi }{3} + 3\sin \frac{\pi }{6}}} }={ – \frac{{2 \cdot \frac{1}{2} + 3 \cdot \frac{{\sqrt 3 }}{2}}}{{2 \cdot \frac{{\sqrt 3 }}{2} + 3 \cdot \frac{1}{2}}} }={ – \frac{{3\sqrt 3 + 2}}{{2\sqrt 3 + 3}}.}$

Page 1
Problems 1-5
Page 2
Problems 6-17