There are many figures and shapes in nature that are better described using polar curves, like a snail's shell, the arrangement of the seeds in a sunflower, the wind patterns that lead to a tornado, and many more. We can also use polar curves to describe artificial phenomena as well!
The signature of modern times is the existence of telecommunications. Internet, television, and many more features make our lives easier. In order for telecommunications to work, scientists and engineers are tasked with sending satellites to the Earth's orbit. Since orbits are better described using polar coordinates, it is important to know how to find the derivatives of polar functions.
Definition of the Derivatives of Polar Functions
A polar function is a function defined in polar coordinates. Just like when working with Cartesian coordinates (also known as rectangular coordinates) you write functions as
\[ y = f(x),\]
in polar coordinates, you write functions as
\[ r = f(\theta).\]
Consider the example of an Archimedean spiral described by
\[ r = 3\theta,\]
its graph is given as follows.
Figure 1. Graph of an Archimedean spiral
You might be tempted to find its derivative the way you are used to using rectangular coordinates, so you would get
\[ \frac{\mathrm{d}r}{\mathrm{d}\theta} = 3.\]
The above expression is the derivative of \( r \) with respect to \( \theta,\) which is telling you how the distance from the origin to a point on the polar curve changes as \( \theta\) is changing. Since the distance between two consecutive segments of an Archimedean spiral is constant, you can expect that this derivative is a constant as well.
Figure 2. The distance between two consecutive segments of an Archimedean spiral is constant
While the derivative of \( r \) with respect to \( \theta \) represents the change of \( r \) with respect to a change of \( \theta\), it does not represent the slope to a line tangent to the polar curve.
If you wanted to find a line tangent to a polar curve, you will need to use a special formula that involves derivatives using polar coordinates.
Figure 3. A line tangent to a point on the Archimedean spiral
Equations of the Derivatives of Polar Functions
The derivative of a function written as
\[ y = f(x) \]
can help you find a line tangent to the curve \( f(x)\), so you can use this idea for finding the line tangent to a polar curve. In this case, you need to find the derivative
\[ \frac{\mathrm{d}y}{\mathrm{d}x}\]
in terms of polar coordinates \( r\) and \( \theta\). In order to do so, you can use the following formula.
Let \( r= f(\theta) \) be a polar function, then its derivative is given by
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{ \frac{\mathrm{d}r}{\mathrm{d}\theta} \sin{\theta} + r \, \cos{\theta}}{\frac{\mathrm{d}r}{\mathrm{d}\theta} \cos{\theta} - r \, \sin{\theta}}.\]
Since \( r=f(\theta)\), you might also find this formula written in terms of \( f\) and using prime notation, that is
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{f'(\theta) \cdot \sin{\theta} + f(\theta) \cdot \cos{\theta}}{f'(\theta) \cdot \cos{\theta}-f(\theta) \cdot \sin{\theta}}.\]
You can evaluate this derivative for any value of \( \theta \), which will give you the slope of a line tangent to the curve at the point defined by \( f(\theta) \).
Please note that in the above formula it is possible to have \( 0\) in the denominator. While you are usually told to avoid this scenario, in this context this means that the line tangent to the curve is vertical.
Steps for Finding the Derivatives of Polar Functions
The formula used to find the derivative of a polar function might look intimidating, so let's break it down into steps.
- Find the derivative \( f'(\theta) \) using any relevant differentiation rules.
- Use the formula for the derivative of a polar function.
- Substitute \( f(\theta) \) and \( f'(\theta) \) into the formula for the derivative of a polar function.
- Simplify the resulting expression.
Steps are always better understood with examples. Here is one!
Find the derivative of the Archimedean spiral described by
\[ f(\theta) = 3 \theta.\]
Solution:
Here you can follow the steps introduced in this section to find the derivative of the Archimedean spiral shown at the start of the article.
1. Find the derivative \( f'(\theta) \) using any relevant differentiation rules.
Remember that the derivative involved in this step is the usual derivative of \( f(\theta) \) with respect to \( \theta\), that is
\[ f'(\theta) = \frac{\mathrm{d}f}{\mathrm{d}\theta},\]
so you can find it with help of the Power Rule, giving you
\[ f'(\theta) = 3. \]
2. Use the formula for the derivative of a polar function.
Now you can use the formula for the derivative of a polar function,
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{f'(\theta) \cdot \sin{\theta} + f(\theta) \cdot \cos{\theta}}{f'(\theta) \cdot \cos{\theta}-f(\theta) \cdot \sin{\theta}}.\]
3. Substitute \( f(\theta) \) and \( f'(\theta) \) into the formula for the derivative of a polar function.
You were given \( f(\theta) = 3\theta \) and you found that \( f'(\theta)=3\), so substitute these values into the formula for the derivative of a polar function, that is
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(3)\sin{\theta}+(3\theta)\cos{\theta}}{(3)\cos{\theta}-(3\theta)\sin{\theta}}.\]
4. Simplify the resulting expression.
Finally, simplify the derivative by factoring out \( 3 \) in both the numerator and denominator, that is
\[ \begin{align} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{3\left(\sin{\theta}+\theta\cos{\theta}\right)}{3\left(\cos{\theta}-\theta\sin{\theta}\right)} \\ &= \frac{\cancel{3}\left(\sin{\theta}+\theta\cos{\theta}\right)}{\cancel{3}\left(\cos{\theta}-\theta\sin{\theta}\right)} \\ &= \frac{\sin{\theta}+\theta\cos{\theta}}{\cos{\theta}-\theta\sin{\theta}} . \end{align}\]
The resulting derivative can be evaluated at any value of \( \theta\), which will give you the slope of the line tangent to the spiral. For example, you can use \( \theta = \frac{\pi}{2}\) and obtain
\[\begin{align} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\sin{\frac{\pi}{2}} +\frac{\pi}{2}\cos{\frac{\pi}{2}} }{\cos{\frac{\pi}{2}}-\frac{\pi}{2}\sin{\frac{\pi}{2}}} \\ &= \frac{1+\frac{\pi}{2}(0)}{0-\frac{\pi}{2}(1)} \\ &= \frac{1}{-\frac{\pi}{2}} \\ &= -\frac{2}{\pi}. \end{align}\]
Since \( r= f(\theta), \) you can find the point on the curve when \( \theta=\frac{\pi}{2} \) by substituting this value of \( \theta \) in \( f,\) that is
\[ \begin{align} r &= f \left( \frac{\pi}{2} \right) \\ &= 3\cdot\frac{\pi}{2} \\&= \frac{3\pi}{2}, \end{align} \]
which gives you the distance from the origin to the point.
Figure 4. Graph of a line tangent to the Archimedean spiral when \( \theta = \frac{\pi}{2} \)
Note that, since you found the slope to be negative, the line is a decreasing function.
Types of Derivatives of Polar Functions
As mentioned earlier, when finding the derivatives of polar functions you can find either
\[ f'(\theta)=\frac{\mathrm{d}r}{\mathrm{d}\theta}\]
in a direct way, or you can also find
\[ \frac{\mathrm{d}y}{\mathrm{d}x}\]
using the formula introduced in the previous chapter to relate it to the line tangent to the polar curve. Note that to use the formula for
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{f'(\theta) \cdot \sin{\theta} + f(\theta) \cdot \cos{\theta}}{f'(\theta) \cdot \cos{\theta}-f(\theta) \cdot \sin{\theta}},\]
you actually need to find \( f'(\theta) \).
Usually, the derivative of a polar function refers to the lines tangent to a polar curve, so you should use the above formula to compute them.
Derivatives of Polar Functions in Physics
A common way to describe the motion of things is by means of polar coordinates. In this case, the coordinates \( r \) and \( \theta \) become functions of time, which is usually represented using the letter \( t\). This means that, given any instant of time, you can find the coordinates of an object in terms of its polar coordinates, that is
\[ r = r(t), \]
and
\[ \theta = \theta (t). \]
This way you can find the derivatives of \( r \) and \( \theta \) with respect to time, which receive special names.
The \( r \) coordinate is usually known as the radial coordinate, and its derivative is known as the radial speed.
The radial speed of a particle in motion is defined as \[ v = \frac{\mathrm{d}r}{\mathrm{d}t}.\] The radial speed of an object can be seen as how fast such an object is changing its distance relative to the center of a coordinate system.
And what about the angular coordinate \( \theta \)?
The angular speed of a particle in motion is defined as \[ \omega = \frac{\mathrm{d}\theta}{\mathrm{d}t}.\] The angular speed of an object can be seen as how fast such object is changing its orientation with respect to a fixed axis of a coordinate system, which is usually the \(x-\) axis.
You might come up with the terms radial velocity and angular velocity. These quantities are the corresponding vector quantity for each speed.
The motion of a particle described in polar coordinates is given by
\[ r(t) = t^2-2t\]
and
\[ \theta(t) = \sin{(2\pi\,t)}.\]
Find its radial and angular speeds.
Solution:
You can obtain the radial speed of the particle by finding the derivative of \( r(t)\), which can be done with the help of the Power Rule, that is
\[ \begin{align} v &= \frac{\mathrm{d}r}{\mathrm{d}t} \\ &= 2t-2. \end{align} \]
For the angular speed, you instead find the derivative of \( \theta(t)\), where you have to use the fact that the derivative of the sine function is the cosine function. Do not forget to use the Chain Rule as well! This will give you
\[ \begin{align} \omega &= \frac{\mathrm{d}\theta}{\mathrm{d}t} \\ &= 2\pi\cos{(2\pi\,t)}. \end{align}\]
Simple, isn't it?
Examples of Derivatives of Polar Functions
Here you can practice finding the derivatives of polar functions with some examples.
Consider the limaçon described by
\[ f( \theta ) = 2+3\sin{\theta}. \]
Find the derivative of this polar curve.
Solution:
1. Find the derivative \( f'(\theta) \) using any relevant differentiation rules.
Since the derivative of the sine function is the cosine function, you can write
\[ f'(\theta) = 3\cos{\theta}. \]
2. Use the formula for the derivative of a polar function.
Next, you need to use the formula
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{f'(\theta) \cdot \sin{\theta} + f(\theta) \cdot \cos{\theta}}{f'(\theta) \cdot \cos{\theta}-f(\theta) \cdot \sin{\theta}}.\]
3. Substitute \( f(\theta) \) and \( f'(\theta) \) into the formula for the derivative of a polar function.
This step is rather straightforward, doing so will give you\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(3\cos{\theta})\sin{\theta} + (2+3\sin{\theta})\cos{\theta}}{(3\cos{\theta}) \cos{\theta}-(2+3\sin{\theta})\sin{\theta}}.\]
4. Simplify the resulting expression.
This step usually involves plenty of algebra.
\[ \begin{align} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{(3\cos{\theta})\sin{\theta} + (2+3\sin{\theta})\cos{\theta}}{(3\cos{\theta}) \cos{\theta}-(2+3\sin{\theta})\sin{\theta}} \\ &= \frac{3(\cos{\theta})(\sin{\theta})+2\cos{\theta}+3(\sin{\theta})(\cos{\theta})}{3\cos^2{\theta}-2\sin{\theta}-3\sin^2{\theta}} \\ &= \frac{6(\sin{\theta})(\cos{\theta})+2\cos{\theta}}{3\cos^2{\theta}-3\sin^2{\theta}-2\sin{\theta}}. \end{align}\]
This is as good as it gets. Usually these derivatives will look this way.
Maybe the derivative of a rose curve looks nicer.
Consider the rose curve described by
\[ f(\theta) = 3\cos{(2\theta)}.\]
Find the derivative of this polar curve.
Solution:
1. Find the derivative \( f'(\theta) \) using any relevant differentiation rules.
First find the derivative of \( f\) using the fact that the derivative of the cosine function is the negative of the sine function, that is
\[ f'(\theta) = -6\sin{(2\theta)}.\]
2. Use the formula for the derivative of a polar function.
As usual, use the formula
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{f'(\theta) \cdot \sin{\theta} + f(\theta) \cdot \cos{\theta}}{f'(\theta) \cdot \cos{\theta}-f(\theta) \cdot \sin{\theta}}\]
to find the derivative of the polar curve.
3. Substitute \( f(\theta) \) and \( f'(\theta) \) into the formula for the derivative of a polar function.
Before substituting anything, you should try using some trigonometric identities to rewrite \( f(\theta)\) and its derivative. You can use double-angle identities and find that
\[\begin{align} f(\theta) &= 3\cos{(2\theta)} \\ &= 3(\cos^2{\theta}-\sin^2{\theta}) \\ &= 3\cos^2{\theta}-3\sin^2{\theta}\end{align}\]
and
\[\begin{align} f'(\theta) &= -6\sin{(2\theta)} \\ &= -6(2\sin{\theta}\,\cos{\theta}) \\ &= -12\sin{\theta}\,\cos{\theta}. \end{align}\]
Now you can substitute the above expressions into the formula for the derivative of a polar curve, so
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(-12\sin{\theta}\,\cos{\theta})(\sin{\theta})+(3\cos^2{\theta}-3\sin^2{\theta})(\cos{\theta})}{(-12\sin{\theta}\,\cos{\theta})(\cos{\theta})-(3\cos^2{\theta}-3\sin^2{\theta})(\sin{\theta})}. \]
4. Simplify the resulting expression.
Finally, simplify the above expression as much as you can, so
\[ \begin{align} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{(-12\sin{\theta}\,\cos{\theta})(\sin{\theta})+(3\cos^2{\theta}-3\sin^2{\theta})(\cos{\theta})}{(-12\sin{\theta}\,\cos{\theta})(\cos{\theta})-(3\cos^2{\theta}-3\sin^2{\theta})(\sin{\theta})} \\ &= \frac{-12\sin^2{\theta}\,\cos{\theta}+3\cos^3{\theta}-3\sin^2{\theta}\,\cos{\theta}} {-12\sin{\theta}\,\cos^2{\theta}-3\sin{\theta}\,\cos^2{\theta}+3\sin^3{\theta}} \\ &= \frac{-15\sin^2{\theta}\,\cos{\theta}+3\cos^3{\theta}}{-15\sin{\theta}\,\cos^2{\theta}+3\sin^3{\theta}}. \end{align}\]
From here, you can factor \(-3\) out of both the numerator and denominator, giving you
\[ \begin{align} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{\cancel{(-3)}(5\sin^2{\theta}\,\cos{\theta}-\cos^3{\theta})}{\cancel{(-3)}(5\sin{\theta}\,\cos^2{\theta} -\sin^3{\theta}) } \\ &= \frac{5\sin^2{\theta}\,\cos{\theta}-\cos^3{\theta}}{5\sin{\theta}\,\cos^2{\theta} -\sin^3{\theta} }. \end{align}\]
Remember that if you find that the slope of a line is infinity, it means that it is a vertical line.
Show that the line tangent to
\[ f(\theta) = 3\]
at the point when \( \theta = 0 \) is vertical.
Solution:
The given polar curve is a curve with constant \(r\), which is a circumference. The radius of this circumference is equal to \(3\).
Figure 5. Graph of the polar curve \( f(\theta) = 3\)
In order to find the line tangent to the curve at the requested point, you first need to find its derivative.
1. Find the derivative \( f'(\theta) \) using any relevant differentiation rules.
Since the given function is a constant function, its derivative is equal to zero, that is
\[ f'(\theta) = 0. \]
2. Use the formula for the derivative of a polar function.
Next, you will need to use the formula
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{f'(\theta) \cdot \sin{\theta} + f(\theta) \cdot \cos{\theta}}{f'(\theta) \cdot \cos{\theta}-f(\theta) \cdot \sin{\theta}}\]
to find the derivative of the given polar curve.
3. Substitute \( f(\theta) \) and \( f'(\theta) \) into the formula for the derivative of a polar function.
This time the function and its derivative are constants, so the substitution is straightforward. Doing so will yield you
\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(0)(\sin{\theta}) + (3)(\cos{\theta})}{(0)(\cos{\theta})-(3)(\sin{\theta})}.\]
4. Simplify the resulting expression.
You will find that simplifying expressions before evaluating them is very useful. In this case, the above expression simplifies as
\[ \begin{align} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{3\cos{\theta} }{-3\sin{\theta}} \\ &= -\cot{\theta}. \end{align}\]
Now that you found the derivative of the polar curve, you can substitute \( \theta=0,\) however you will find an issue here
\[ \cot{0} = \infty. \]
Do not worry! Since the above expression gives you the slope of the tangent line, this actually proves what you were asked, as the slope of a vertical line is infinity.
Figure 6. The line tangent to the polar curve at \( \theta=0\) is vertical
Derivatives of Polar Functions - Key takeaways
- Functions in polar coordinates are usually written as\[ r =f(\theta), \] where \( r \) is the distance from the origin to a point of the polar curve.
- You can find the derivative \( f'(\theta) \) the same way you find the derivative of any other function, that is,\[ f'(\theta) = \frac{\mathrm{d}f}{\mathrm{d}\theta}.\]However this derivative does not describe the slope of a line tangent to a point of the polar function.
- Let \( r= f(\theta) \) be a polar function, then its derivative, also known as the derivative of the polar curve, is given by\[ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{f'(\theta) \cdot \sin{\theta} + f(\theta) \cdot \cos{\theta}}{f'(\theta) \cdot \cos{\theta}-f(\theta) \cdot \sin{\theta}}.\]
- You can use the derivative \( \frac{\mathrm{d}y}{\mathrm{d}x} \) to find the slope of a line tangent to a point of a polar function.
- Note that the usual derivative \( f'(\theta)\) is involved in the formula for the derivative of a polar curve.
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