7.7: Polar Coordinates

Suppose that you wanted to write the equation of a spiral, like the one in Figure [fig:spiral]. The curve is clearly not the graph of a function \(y=f(x)\) in Cartesian coordinates, as it violates the vertical line test. However, this spiral is simple to express using polar coordinates.¹⁶ Recall that any point \(P\) distinct from the origin (denoted by \(O\)) in the \(xy\)-plane is a distance \(r>0\) from the origin, and the ray \(\overrightarrow{OP}\) makes an angle \(\theta\) with the positive \(x\)-axis, as in Figure [fig:polar]. Imagine \(\overrightarrow{OP}\) swings around the “pole” at the origin.

The pair \((r,\theta)\) contains the polar coordinates of \(P\), and the positive \(x\)-axis is called the polar axis of this coordinate system. For the angle \(\theta\) measured in radians, \((r,\theta) = (r,\theta + 2\pi k)\) for all integers \(k\). Thus, the polar coordinates of a point are not unique. By convention \(r\) can be negative, by defining \((-r,\theta) = (r,\theta + \pi)\) for any angle \(\theta\): the ray \(\overrightarrow{OP}\) is drawn in the opposite direction from the angle \(\theta\), as in Figure [fig:negpolar]. When \(r=0\), the point \((r,\theta) = (0,\theta)\) is the origin \(O\), regardless of the value of \(\theta\).

Example \(\PageIndex{1}\): spiral

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Solution

Express the spiral in Figure [fig:spiral] in polar coordinates, such that the distance between any two points separated by \(2\pi\) radians is always \(1\).

Solution: The goal is to find some equation involving \(r\) and \(\theta\) (measured in radians) that describes the spiral. The distance between any two points separated by \(2\pi\) radians is always \(1\), for example:

\[\begin{aligned} \theta ~=~ 0 \quad&\Rightarrow\quad r ~=~ 1\\ \theta ~=~ 2\pi \quad&\Rightarrow\quad r ~=~ 2\\ \theta ~=~ 4\pi \quad&\Rightarrow\quad r ~=~ 3\\ &\ldots\\ \theta ~=~ 2\pi\,k \quad&\Rightarrow\quad r ~=~ 1+k\end{aligned} \nonumber \]

for all integers \(k \ge 0\). In fact \(r=1+k\) when \(\theta = 2\pi k\) for all real \(k \ge 0\), by the assumption about the distance. So solving for \(k\) in terms of \(\theta\) yields the polar equation \(r = 1 + \frac{\theta}{2\pi}\) for all \(\theta \ge 0\).

You might be familiar with graphing paper, for plotting points or functions given in Cartesian coordinates. Such paper consists of a rectangular grid, where the horizontal and vertical lines represent where \(x\) and \(y\), respectively, are constants, at regular intervals. Similar graphing paper exists for polar coordinates, as in Figure [fig:polargraphpaper].

This polar grid is radial, not rectangular. The concentric circles around the origin \(O\) are where \(r\) is constant (e.g. \(r=1\), \(r=2\)), while the lines through the origin are where \(\theta\) is constant, at regular intervals for each. The angle \(\theta\) shown here is in degrees, though radians are often preferred for their “unitless” nature. In general, polar coordinates are useful in describing plane curves that exhibit symmetry about the origin (though there are other situations), which arise in many physical applications.

Figure [fig:polarconvert] shows how to convert between polar coordinates and Cartesian coordinates. Namely, for a point with polar coordinates \((r,\theta)\) and Cartesian coordinates \((x,y)\):

\[\label{eqn:polartorect} x ~=~ r\,\cos\,\theta \qquad y ~=~ r\,\sin\,\theta \]

\[\label{eqn:recttopolar} r ~=~ \pm\;\sqrt{x^2 ~+~ y^2} \qquad \tan\;\theta ~=~ \frac{y}{x} ~~\text{if $x \ne 0$} \]

In formula ([eqn:recttopolar]), if \(x = 0\) then \(\theta = \pi/2\) or \(\theta = 3\pi/2\). If \(x \ne 0\) and \(y \ne 0\) then the two possible solutions for \(\theta\) in the equation \(\tan \theta = \frac{y}{x}\) are in opposite quadrants (for \(0 \le \theta < 2\pi\)). If the angle \(\theta\) is in the same quadrant as the point \((x,y)\), then \(r = \sqrt{x^2 + y^2}\) (i.e. \(r\) is positive); otherwise \(r = -\sqrt{x^2 + y^2}\) (i.e. \(r\) is negative).

Example \(\PageIndex{1}\): polarcircle

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Solution

Write the equation of the unit circle \(x^2 + y^2 = 1\) in polar coordinates.

Solution: By formula ([eqn:recttopolar]), \(r^2 = x^2 + y^2 = 1\), so in polar coordinates the equation is simply \(r = 1\). In general the circle \(x^2+y^2=a^2\) of radius \(a>0\) has the simpler expression \(r=a\) in polar coordinates.

Write the equation \(x^2 + (y-4)^2 = 16\) in polar coordinates.

Solution: This is the equation of a circle of radius \(4\) centered at the point \((0,4)\). Expand the equation:

\[\begin{aligned} x^2 ~+~ (y-4)^2 ~&=~ 16\\ x^2 ~+~ y^2 ~-~ 8y ~+~ 16 ~&=~ 16\\ x^2 ~+~ y^2 ~&=~ 8y\\ r^2 ~&=~ 8\,r\sin\;\theta \quad\text{(by formulas (\ref{eqn:polartorect}) and (\ref{eqn:recttopolar}))}\\ r ~&=~ 8\,\sin\;\theta \end{aligned} \nonumber \]

Is it valid to cancel \(r\) from both sides in the last step? Yes. The point \((0,0)\) is on the circle, so canceling \(r\) does not eliminate \(r=0\) as a potential solution of the equation (e.g. \(\theta=0\) would make \(r = 8 \sin\,\theta = 0\)). Thus, the polar equation is \(r = 8 \sin\,\theta\).

Notice that this polar equation is actually less intuitive than its Cartesian equivalent. From the equation \(x^2 + (y-4)^2 = 16\) it is easy to identify the curve as a circle and read off its radius and center; these properties are not so obvious from the polar equation. Since the circle’s center is not the origin, there is no symmetry about the origin, which is when polar coordinates are often better suited.

Suppose that the polar coordinates \((r,\theta)\) for a plane curve are related by a function: \(r=r(\theta)\). Then by formula ([eqn:polartorect]), \(x=r(\theta)\,\cos\,\theta\) and \(y=r(\theta)\,\sin\,\theta\) are now parametric equations for the curve in the parameter \(\theta\). Thus, by the Product Rule and formulas ([eqn:paramderiv1]) and ([eqn:paramderiv2]) from Section 7.6, with \(\dx=x'(\theta)\,\dtheta\) and \(\dy=y'(\theta)\,\dtheta\):

Example \(\PageIndex{1}\): polaronepluscos

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Solution

Sketch the graph of \(r = 1 + \cos\,\theta\).

Solution: First sketch the graph treating \((r,\theta)\) as Cartesian coordinates, for \(0 \le \theta \le 2\pi\) as in Figure [fig:polargraph](a). Then use that graph to trace out a rough graph in polar coordinates, as in Figure [fig:polargraph](b).¹⁷

To find the maxima, minima, and inflection points it is still necessary to find \(\dydx\) and \(\frac{d^2y}{\dx^2}\). It is left as an exercise to use formula ([eqn:polarderiv]) and double-angle identities to show that

\[\dydx ~=~ -\frac{\cos\,\theta + \cos\,2\theta}{\sin\,\theta + \sin\,2\theta} \qquad\text{and}\qquad \frac{d^2y}{\dx^2} ~=~ -\frac{3\,(1 + \cos\,\theta)}{(\sin\,\theta + \sin\,2\theta)^3} \nonumber \]

and that for \(\theta\) in \(\ival{0}{2\pi}\), \(\dydx = 0\) only when \(\theta = \frac{\pi}{3}\) and \(\frac{5\pi}{3}\), while \(\dydx\) is undefined for \(\theta=0\), \(\frac{2\pi}{3}\), \(\pi\), \(\frac{4\pi}{3}\), and \(2\pi\). Since \(\frac{d^2y}{\dx^2} \Bigr|_{\theta=\frac{\pi}{3}} < 0\) and \(\frac{d^2y}{\dx^2} \Bigr|_{\theta=\frac{5\pi}{3}} > 0\) then the curve has a local maximum when \(\theta=\frac{\pi}{3}\) and a local minimum when \(\theta=\frac{5\pi}{3}\). It can also be shown that \(\frac{d^2y}{\dx^2}\) changes sign around \(\theta=0\), \(\frac{2\pi}{3}\), \(\pi\), \(\frac{4\pi}{3}\), so that the inflection points occur at those values of \(\theta\), as shown (with the correct concavity) in Figure [fig:polargraph](b).

In some cases polar coordinates can simplify evaluation of a definite integral or finding an area. To determine the polar form of a definite integral, suppose that \(r\) is a function of \(\theta\): \(r=f(\theta)\). A polar region swept out by \(r=f(\theta)\) between \(\theta=\alpha\) and \(\theta=\beta\) would look like the shaded region in Figure [fig:polararea] with area \(A\):

\[\begin{aligned} \dA ~&=~ \tfrac{1}{2}\,(r)\,(r+\dr)\,\sin\,\dtheta\

\[4pt] &=~ \tfrac{1}{2}\,r^2 \dtheta ~+~ \tfrac{1}{2}\,r\,(\dr)\,\dtheta\

\[4pt] &=~ \tfrac{1}{2}\,r^2 \dtheta ~+~ \tfrac{1}{2}\,r\,(f'(\theta)\,\dtheta)\,\dtheta\

\[4pt] &=~ \tfrac{1}{2}\,r^2 \dtheta ~+~ 0 ~=~ \tfrac{1}{2}\,r^2 \dtheta\end{aligned} \nonumber \]

since \((\dtheta)^2=0\). The area \(A\) of the region is the sum of these infinitesimal areas \(\dA\):

Use polar coordinates to show that the area of a circle of radius \(R\) is \(\pi R^2\).

Solution: Let the origin be the center of the circle. Then \(r=R\) is the polar equation of the circle, with \(0 \le \theta \le 2\pi\) sweeping out exactly one full circle. The area \(A\) inside the circle is then

\[A ~=~ \int_{0}^{2\pi} \tfrac{1}{2}\,r^2 \dtheta ~=~ \int_{0}^{2\pi} \tfrac{1}{2}\,R^2 \dtheta ~=~ \tfrac{1}{2}\,R^2 \theta~\Biggr|_0^{2\pi} ~=~ \tfrac{1}{2}\,R^2\,(2\pi - 0) ~=~ \pi R^2 ~~.\quad\checkmark \nonumber \]

Note the simplicity of this integral compared to the trigonometric substitution required when using Cartesian coordinates, as in Section 6.3. Notice also that using a larger interval—say \(\ival{0}{4\pi}\)—for \(\theta\) would result in an incorrect area (\(4\pi R^2\)) even though the region inside the curve is the same.

Find the area inside the curve \(r = 1 + \cos\,\theta\).

Solution: Choose \(0 \le \theta \le 2\pi\) as in Example

Example \(\PageIndex{1}\): polaronepluscos

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Solution

, so that the area \(A\) is:

\[\begin{aligned} A ~&=~ \int_{0}^{2\pi} \tfrac{1}{2}\,r^2 \dtheta ~=~ \int_{0}^{2\pi} \tfrac{1}{2}\,(1 + \cos\,\theta)^2 \dtheta ~=~ \int_{0}^{2\pi} \left(\frac{1}{2} \;+\; \cos\,\theta \;+\; \frac{\cos^2 \theta}{2} \right)~\dtheta\

\[4pt] &=~ \int_{0}^{2\pi} \left(\frac{1}{2} \;+\; \cos\,\theta \;+\; \frac{1 + \cos\,2\theta}{4} \right)~\dtheta ~=~ \int_{0}^{2\pi} \left(\frac{3}{4} \;+\; \cos\,\theta \;+\; \frac{\cos\,2\theta}{4} \right)~\dtheta\

\[4pt] &=~ \frac{3}{4}\,\theta ~+~ \sin\,\theta ~+~ \frac{1}{8}\,\sin\,2\theta~\Biggr|_0^{2\pi} ~=~ \frac{3\pi}{2}\end{aligned} \nonumber \]

[sec7dot7]

For Exercises 1-8 write the given equation in polar coordinates.

4

\((x-3)^2 + y^2 = 9\)

\(y = x\)

\(x^2 - y^2 = 1\)

\(3x^2 + 4y^2 - 6x = 9\)

4

\(y = -x\)

\(y = x + 1\)

\(y = x^2\)

\(y = x^3\)

Write the polar equation \(r^2 = 4\,\cos\,2\theta\) in Cartesian coordinates.

2

Find the tangent line to \(r=\cos\,2\theta\) at \(\left(\frac{1}{2},\frac{\pi}{6}\right)\).

Find the tangent line to \(r=8\,\sin^2\theta\) at \(\left(2,\frac{5\pi}{6}\right)\).

Recall the curve \(r = 1 + \cos\,\theta\) from Example

Example \(\PageIndex{1}\): polaronepluscos

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Solution

.

1. Verify that for this curve,

   \[\dydx ~=~ -\frac{\cos\,\theta + \cos\,2\theta}{\sin\,\theta + \sin\,2\theta} \qquad\text{and}\qquad \frac{d^2y}{\dx^2} ~=~ -\frac{3\,(1 + \cos\,\theta)}{(\sin\,\theta + \sin\,2\theta)^3} ~. \nonumber \]

2. Verify that for \(\theta\) in \(\ival{0}{2\pi}\), \(\dydx = 0\) only when \(\theta = \frac{\pi}{3}\) and \(\frac{5\pi}{3}\), and is undefined for \(\theta=0\), \(\frac{2\pi}{3}\), \(\pi\), \(\frac{4\pi}{3}\), and \(2\pi\).
3. Verify that \(\frac{d^2y}{\dx^2} \Biggr|_{\theta=\frac{\pi}{3}} < 0~\) and \(~\frac{d^2y}{\dx^2} \Biggr|_{\theta=\frac{5\pi}{3}} > 0\).
4. Verify that \(\frac{d^2y}{\dx^2}\) changes sign around \(\theta=0\), \(\frac{2\pi}{3}\), \(\pi\), and \(\frac{4\pi}{3}\).

For Exercises 13-15, sketch the graph of the given curve and indicate all local maxima and minima. [[1.]]

3

\(r = 1 + \sin\,\theta\)

\(r = 1 - \cos\,\theta\)

\(r = \sin\,2\theta\)

2

Find the area inside \(r = 1 + \sin\,\theta\).

Find the area inside \(r = \sin\,2\theta\).

[[1.]]

Sketch a rough graph of the meridian voltage component \(E_{\theta}\) for a half-wavelength linear antenna:

\[E_{\theta} ~=~ r(\theta) ~=~ \frac{\cos\,(\frac{\pi}{2} \cos\,\theta)}{\sin\,\theta} \nonumber \]

Show that the distance \(d\) between two points \((r_1 , \theta_1)\) and \((r_2 , \theta_2)\) in polar coordinates is

\[d ~=~ \sqrt{r_1^2 ~+~ r_2^2 ~-~ 2r_1r_2\,\cos\,(\theta_1 - \theta_2)} ~~. \nonumber \]

For a point \(P=(r,\theta)\) on the curve \(r=r(\theta)\), let \(\alpha\) be the angle that the tangent line through \(P\) makes with the positive \(x\)-axis. Let \(\psi=\alpha-\theta\) be the angle between the tangent line and the line through \(P\) and the origin. Show that \(\tan\,\psi = \frac{r(\theta)}{r'(\theta)}\). (Hint: Use formulas ([eqn:polarderiv]) and ([eqn:tangentangle]).)

For the parabola with focus at \((0,0)\), vertex at \((0,-p)\), and directrix \(y=-2p\) (with \(p > 0\)), show that the polar equation is

\[r ~=~ \frac{2p}{1 \;-\; \sin\,\theta} ~~. \nonumber \]

For the ellipse with foci \((0,0)\) and \((2c,0)\), vertexes \((c \pm a,0)\), and eccentricity \(e=\frac{c}{a}\) (with \(0 < c < a\)), show that the polar equation is

\[r ~=~ \frac{a\,(1 \;-\; e^2)}{1 \;-\; e\,\cos\,\theta} ~~. \nonumber \]

For the hyperbola with foci \((0,0)\) and \((2c,0)\), vertexes \((c \pm a,0)\), and eccentricity \(e=\frac{c}{a}\) (with \(0 < a < c\)), show that the polar equation is

\[r ~=~ \frac{a\,(e^2 \;-\; 1)}{1 \;+\; e\,\cos\,\theta} ~~. \nonumber \]

[[1.]]

The bipolar coordinates¹⁹ \((\xi,\eta)\) of a point \(P=(x,y)\) relative to two poles \(F_1=(-a,0)\) and \(F_2=(a,0)\) are given by \(\xi = \ln\,\frac{d_1}{d_2}\) and \(\eta =\angle F_1PF_2\), where \(d_1=F_1P\) and \(d_2=F_2P\), as in the figure on the right . Then \(-\infty < \xi < \infty\) except at \(F_1\) (which corresponds to \(\xi=-\infty\)) and \(F_2\) (\(\xi=\infty\)), while \(0 \le \eta \le \pi\) for points on or above the \(x\)-axis, and \(\pi < \eta < 2\pi\) below the \(x\)-axis.

1. Show that

   \[x ~=~ \frac{a\,\sinh\,\xi}{\cosh\,\xi \;-\; \cos\,\eta} \qquad\text{and}\qquad y ~=~ \frac{a\,\sin\,\eta}{\cosh\,\xi \;-\; \cos\,\eta} ~~.\qquad\qquad \nonumber \]

   (Hint: Use the distance formula, the Law of Cosines, and the Law of Sines.)

2. Show that for constants \(\tau \ne 0\) and \(\sigma \ne 0\) or \(\pi\), the curves \(\xi = \tau\) and \(\eta = \sigma\) are the respective circles

   \[(x \;-\; a\,\coth\,\tau)^2 ~+~ y^2 ~=~ \frac{a^2}{\sinh^2 \tau} \qquad\text{and}\qquad x^2 ~+~ (y \;-\; a\,\cot\,\sigma)^2 ~=~ \frac{a^2}{\sin^2 \sigma} ~~. \nonumber \]