Calculus

Differentiation of Functions

Derivatives of Polar Functions

Page 1
Problem 1
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Problems 2-7

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.
Position of a point in polar coordinate system

Figure 1.

Archimedean spiral

Fig.2 Archimedean Spiral

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 it can be seen, we have actually got the parametric equations of the curve where the angle \(\theta\) plays the role of the parameter \(t\). In this case, the derivative of the polar curve can be found 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 on problem description to see 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 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 4

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 5

Find the derivative \(\large\frac{{dy}}{{dx}}\normalsize\) of the lemniscate of Bernoulli given by the equation

\[{r^2} = \cos 2\theta .\]

 Example 6

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

 Example 7

Find the derivative \(\large\frac{{dy}}{{dx}}\normalsize\) of Fermat’s spiral given by the polar equation

\[r = \sqrt \theta.\]

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.

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Problem 1
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Problems 2-7