1.8: Optional — Polar Coordinates

So far we have always written vectors in two dimensions in terms of the basis vectors \(\hat{\pmb{\imath}}\) and \(\hat{\pmb{\jmath}}\text{.}\) This is not always convenient. For example, when working in polar coordinates it is often convenient to use basis vectors \(\hat{\textbf{r}} (\theta)\text{,}\) \(\hat{\boldsymbol{\theta}}(\theta)\) which depend on the value of the current polar coordinate \(\theta\) — though one usually just writes \(\hat{\textbf{r}} \text{,}\) \(\hat{\boldsymbol{\theta}}\text{,}\) suppressing the dependence on \(\theta\) from the notation. When one is at the point with polar coordinates \((r,\theta)\text{,}\) these basis vectors are defined by

Equation 1.8.1

\[\begin{align*} \hat{\textbf{r}} (\theta) &= \cos\theta\,\hat{\pmb{\imath}} + \sin\theta\,\hat{\pmb{\jmath}}\\ \hat{\boldsymbol{\theta}}(\theta) &= -\sin\theta\,\hat{\pmb{\imath}} + \cos\theta\,\hat{\pmb{\jmath}} \end{align*}\]

Note that this basis has two very nice properties.

1. \(|\hat{\textbf{r}}(\theta)| = |\hat{\boldsymbol{\theta}}(\theta)| = 1\text{,}\) \(\hat{\textbf{r}}(\theta) \perp \hat{\boldsymbol{\theta}}(\theta)\) (orthonormality)
2. \(\dfrac{d\hat{\textbf{r}}}{d\theta}(\theta)= \hat{\boldsymbol{\theta}}(\theta)\text{,}\) \(\dfrac{d\hat{\boldsymbol{\theta}}}{d\theta}(\theta) = -\hat{\textbf{r}}(\theta)\)

That \(\dfrac{d\hat{\textbf{r}}}{d\theta}(\theta)\) is some scalar multiple of \(\hat{\boldsymbol{\theta}}(\theta)\) follows just from the fact that \(|\hat{\textbf{r}}(\theta)| = 1\text{.}\)

\[\begin{align*} |\hat{\textbf{r}}(\theta)| = 1 &\implies \hat{\textbf{r}} (\theta)\cdot\hat{\textbf{r}} (\theta)=1\\ &\implies \hat{\textbf{r}}(\theta)\cdot \dfrac{d\hat{\textbf{r}}}{d\theta}(\theta) =\frac{1}{2}\dfrac{d}{dt}\big( \hat{\textbf{r}} (\theta)\cdot\hat{\textbf{r}}(\theta)\big) =0\\ &\implies \dfrac{d\hat{\textbf{r}}}{d\theta}(\theta)\perp \hat{\textbf{r}} (\theta) \implies \dfrac{d\hat{\textbf{r}}}{d\theta}(\theta)\parallel \hat{\boldsymbol{\theta}}(\theta) \end{align*}\]

Similarly, that \(\dfrac{d\hat{\boldsymbol{\theta}}}{d\theta}(\theta)\) is some scalar multiple of \(\hat{\textbf{r}}(\theta)\) follows just from the fact that \(|\hat{\boldsymbol{\theta}}(\theta)| = 1\text{.}\)

Lemma 1.8.2

If we parametrize a curve by giving its polar coordinates¹ \(\big(r(t)\,,\,\theta(t)\big)\text{,}\) then

1. the position vector of the point at time \(t\) is

   \[\begin{gather*} \vecs{r} (t) = r(t)\ \hat{\textbf{r}} \big(\theta(t)\big) \end{gather*}\]

2. and the velocity vector of the point at time \(t\) is

   \[\begin{gather*} \vecs{v} (t) = \dfrac{dr}{dt}(t)\ \hat{\textbf{r}} \big(\theta(t)\big) + r(t)\ \dfrac{d\theta}{dt}(t)\ \hat{\boldsymbol{\theta}}\big(\theta(t)\big) \end{gather*}\]

3. and the acceleration vector of the point at time \(t\) is

   \[\begin{align*} \textbf{a}(t) &= \left[\frac{\mathrm{d}^{2}r}{\mathrm{d}t^{2}}(t)\!-\!r(t)\Big(\dfrac{d\theta}{dt}(t)\Big)^2\right] \hat{\textbf{r}}{r} \big(\theta(t)\big)\\ &\hskip1in +\left[r(t)\ \frac{\mathrm{d}^{2}\theta }{\mathrm{d}t^{2}}(t) \!+\! 2 \dfrac{dr}{dt}(t)\dfrac{d\theta}{dt}(t)\right] \hat{\boldsymbol{\theta}}\big(\theta(t)\big) \end{align*}\]

It is standard to suppress the arguments \(t\) and \(\theta(t)\) and write, for example,

\[ \vecs{v} = \dfrac{dr}{dt}\ \hat{\textbf{r}} + r\ \dfrac{d\theta}{dt}\ \hat{\boldsymbol{\theta}} \nonumber \]

But it is important to remember that the arguments really are there.

Proof

The vector from the origin to the point whose polar coordinates are \((r,\theta)\) is \(\vecs{r} = r\,\hat{\textbf{r}} (\theta)\text{.}\) So if we parametrize a curve by giving the polar coordinates at time \(t\text{,}\)

\[\begin{align*} \vecs{r} (t) &= r(t)\ \hat{\textbf{r}} \big(\theta(t)\big)\\ \vecs{v} (t) &= \dfrac{dr}{dt}(t)\ \hat{\textbf{r}} \big(\theta(t)\big) + r(t)\ \dfrac{d\hat{\textbf{r}} }{d\theta}\big(\theta(t)\big)\ \dfrac{d\theta}{dt}(t) \notag\\ &= \dfrac{dr}{dt}(t)\ \hat{\textbf{r}} \big(\theta(t)\big) + r(t)\ \dfrac{d\theta}{dt}(t)\ \hat{\boldsymbol{\theta}}\big(\theta(t)\big)\\ \textbf{a}(t) & = \frac{\mathrm{d}^{2}r}{\mathrm{d}t^{2}}\ \hat{\textbf{r}} + \dfrac{dr}{dt}\ \dfrac{d\hat{\textbf{r}}}{d\theta}\ \dfrac{d\theta}{dt} + \dfrac{dr}{dt}\ \dfrac{d\theta}{dt}\ \hat{\boldsymbol{\theta}} + r\ \frac{\mathrm{d}^{2}\theta }{\mathrm{d}t^{2}}\ \hat{\boldsymbol{\theta}} + r\ \Big(\dfrac{d\theta}{dt}\Big)^2\ \dfrac{d\hat{\boldsymbol{\theta}}}{d\theta}\\ &=\Big[\frac{\mathrm{d}^{2}r}{\mathrm{d}t^{2}}-r\ \Big(\dfrac{d\theta}{dt}\Big)^2\Big] \hat{\textbf{r}} +\Big[r\ \frac{\mathrm{d}^{2}\theta }{\mathrm{d}t^{2}} + 2 \dfrac{dr}{dt}\ \dfrac{d\theta}{dt}\Big]\hat{\boldsymbol{\theta}} \end{align*}\]

Example 1.8.3

As an example, consider a bead that is sliding on a frictionless rod that has one end fixed at the origin and that is rotating about the origin at a constant \(\Omega\,\)rad/sec.

Because the rod is frictionless, it is incapable of applying to the bead any force parallel to the rod. So under Newton's law, \(m\textbf{a}=\vecs{F} \text{,}\) the radial² component of the acceleration of the particle is exactly zero. So, if the polar coordinates of the bead at time \(t\) are \(\big(r(t),\theta(t)\big)\text{,}\) then, by Lemma 1.8.2.c,

\[ \frac{\mathrm{d}^{2}r}{\mathrm{d}t^{2}}-r\ \Big(\dfrac{d\theta}{dt}\Big)^2 = 0 \nonumber \]

As the rod is rotating at \(\Omega\,\)rad/sec, \(\dfrac{d\theta}{dt}=\Omega\) and

\[ \frac{\mathrm{d}^{2}r}{\mathrm{d}t^{2}}-\Omega^2\ r = 0 \nonumber \]

The general solution to this constant coefficient second order ordinary differential equation is³

\[ r(t) = A e^{\Omega\,t} + B e^{-\Omega\,t} \nonumber \]

where \(A\) and \(B\) are arbitrary constants that are determined by initial conditions. Just as an example, if \(r(0)= 1\) and \(r'(0)= 0\text{,}\) then \(A+B=1\) and \(A\Omega-B\Omega=0\text{,}\) so that \(A=B=\frac{1}{2}\) and

\[ r(t) = \frac{1}{2}\big(e^{\Omega\,t}+e^{-\Omega\,t}\big)\qquad \nonumber \]

If, again for example, \(\theta(0) = 0\text{,}\) then \(\theta(t) = \Omega t\) and the bead follows the polar coordinate curve

\[ r(\theta) = \frac{1}{2}\big(e^{\theta}+e^{-\theta}\big) \nonumber \]

Observe that \(r(\theta)\) is \(1\) when \(\theta=0\text{,}\) increases as \(\theta\) increases, and tends to \(\infty\) as \(\theta\rightarrow+\infty\text{.}\) The curve is a spiral.

Example 1.8.4. Conic sections in polar coordinates

In this example, we derive the equation of a general conic section in polar coordinates. A conic section is the intersection of a plane with a cone. This is illustrated in the figures below.

                        

For our current purposes, it is convenient to use the equivalent⁴ (and often used) definition that a conic section is the set of points \(P\) in the \(xy\)-plane

• whose distance from a fixed point \(F\) (called the focus of the conic)
• is a constant multiple \(\varepsilon \ge 0\) (called the eccentricity of the conic)
• of the distance from \(P\) to a fixed line \(L\) (called the directrix of the conic).

Choose a coordinate system with the focus \(F\) of the conic being the origin and with the directrix \(L\) being \(x=p\) for some \(p \gt 0\text{.}\)

If \(P\) has polar coordinates \((r,\theta)\text{,}\) then \(P\) has \(x\)-coordinate \(r\cos\theta\text{.}\) The point \(Q\) on the line \(L\) in the figure above has \(x\)-coordinate \(p\text{.}\) So the distance from \(P\) to \(L\text{,}\) which is also the distance from \(P\) to \(Q\text{,}\) is \(p-r\cos\theta\text{.}\) The distance from \(P\) to \(F\) is \(r\text{.}\) We require that the distance from \(P\) to \(F\) is \(\varepsilon \) times the distance from \(P\) to \(L\text{.}\) So

\[ r=\varepsilon \big(p-r\cos\theta\big) \iff r=\frac{\varepsilon p}{1+\varepsilon \cos\theta} \nonumber \]

The numerator \(\varepsilon  p\) is usually renamed to \(\ell\) giving the equation

\[ r=\frac{\ell}{1+\varepsilon \cos\theta} \nonumber \]

Example 1.8.5. Conic sections in polar coordinates, again

We'll now take the equation \(r=\frac{\ell}{1+\varepsilon \cos\theta}\) for a conic section in polar coordinates, from the last example, and convert it to the more familiar Cartesian coordinates. Just by the definition of polar coordinates

\[\begin{align*} r\big(1+\varepsilon \cos\theta\big)=\ell &\iff r = \ell -\varepsilon  x\\ &\iff x^2+y^2 = \ell^2-2\varepsilon \ell x+\varepsilon ^2 x^2\\ &\iff (1-\varepsilon ^2) x^2 + 2\varepsilon \ell x + y^2 = \ell^2 \tag{C} \end{align*}\]

Now consider separately four different cases, depending on the value of \(\varepsilon  \ge 0\text{.}\)

• If \(\varepsilon =0\text{,}\) (C) reduces to

  \[ x^2+y^2 = \ell^2 \nonumber \]

  

  which is of course a circle of radius \(\ell\text{.}\)

• If \(0 \lt \varepsilon  \lt 1\text{,}\) completing the square in (C) gives

  \[ (1-\varepsilon ^2)\Big(x+\frac{\varepsilon \ell}{1-\varepsilon^2}\Big)^2 + y^2 = \ell^2 + \frac{\varepsilon ^2\ell^2}{1-\varepsilon ^2} = \frac{\ell^2}{1-\varepsilon ^2} \nonumber \]

  which is equivalent to

  \[ \frac{\big(x+\frac{\varepsilon \ell}{1-\varepsilon ^2}\big)^2} { \frac{\ell^2}{(1-\varepsilon ^2)^2}} +\frac{y^2}{\frac{\ell^2}{1-\varepsilon ^2}} =1 \nonumber \]

  

  and is of course an ellipse with semi-major axis \(r_M=\frac{\ell}{1-\varepsilon ^2}\) and semi-minor axis \(r_m=\frac{\ell}{\sqrt{1-\varepsilon ^2}}\text{.}\)

• If \(\varepsilon =1\text{,}\) (C) reduces to

  \[ y^2 = \ell^2-2\ell x \nonumber \]

  

  which is of course a parabola.

• If \(\varepsilon\gt 1\text{,}\) the same computation as in the \(0 \lt \varepsilon \lt 1\) case gives

  \[ \frac{\big(x-\frac{\varepsilon \ell}{\varepsilon^2-1}\big)^2} { \frac{\ell^2}{(\varepsilon ^2-1)^2}} -\frac{y^2}{\frac{\ell^2}{\varepsilon ^2-1}} =1 \nonumber \]

  

  and is of course a hyperbola.

Exercises

Stage 1

1

Consider the points

\[\begin{align*} (x_1,y_1) &= (3,0) & (x_2,y_2) &= (1,1) & (x_3,y_3) &= (0,1)\\ (x_4,y_4) &= (-1,1) & (x_5,y_5) &= (-2,0) \end{align*}\]

For each \(1\le i\le 5\text{,}\)

• sketch, in the \(xy\)-plane, the point \((x_i,y_i)\) and
• find the polar coordinates \(r_i\) and \(\theta_i\text{,}\) with \(0\le\theta_i \lt 2\pi\text{,}\) for the point \((x_i,y_i)\text{.}\)

2

1. Find all pairs \((r,\theta)\) such that

   \[ (-2,0) = \big(r\cos\theta\,,\,r\sin\theta\big) \nonumber \]

2. Find all pairs \((r,\theta)\) such that

   \[ (1,1) = \big(r\cos\theta\,,\,r\sin\theta\big) \nonumber \]

3. Find all pairs \((r,\theta)\) such that

   \[ (-1,-1) = \big(r\cos\theta\,,\,r\sin\theta\big) \nonumber \]

3

Consider the points

\[\begin{align*} (x_1,y_1) &= (3,0) & (x_2,y_2) &= (1,1) & (x_3,y_3) &= (0,1)\\ (x_4,y_4) &= (-1,1) & (x_5,y_5) &= (-2,0) \end{align*}\]

Also define, for each angle \(\theta\text{,}\) the vectors

\[\begin{gather*} \hat{\mathbf{e}}_r(\theta)=\cos\theta\ \hat{\pmb{\imath}} + \sin\theta\ \hat{\pmb{\jmath}}\qquad \hat{\mathbf{e}}_\theta(\theta) = -\sin\theta\ \hat{\pmb{\imath}} + \cos\theta\ \hat{\pmb{\jmath}} \end{gather*}\]

1. Determine, for each angle \(\theta\text{,}\) the lengths of the vectors \(\hat{\mathbf{e}}_r(\theta)\) and \(\hat{\mathbf{e}}_\theta(\theta)\) and the angle between the vectors \(\hat{\mathbf{e}}_r(\theta)\) and \(\hat{\mathbf{e}}_\theta(\theta)\text{.}\) Compute \(\hat{\mathbf{e}}_r(\theta)\times\hat{\mathbf{e}}_\theta(\theta)\) (viewing \(\hat{\mathbf{e}}_r(\theta)\) and \(\hat{\mathbf{e}}_\theta(\theta)\) as vectors in three dimensions with zero \(\hat{\mathbf{k}}\) components).
2. For each \(1\le i\le 5\text{,}\) sketch, in the \(xy\)-plane, the point \((x_i,y_i)\) and the vectors \(\hat{\mathbf{e}}_r(\theta_i)\) and \(\hat{\mathbf{e}}_\theta(\theta_i)\text{.}\) In your sketch of the vectors, place the tails of the vectors \(\hat{\mathbf{e}}_r(\theta_i)\) and \(\hat{\mathbf{e}}_\theta(\theta_i)\) at \((x_i,y_i)\text{.}\)

4 ✳

Match the following equations with the corresponding pictures. Cartesian coordinates are \((x, y)\) and polar coordinates are \((r, \theta)\text{.}\)

\[\begin{alignat*}{5} &\text{(a)}\quad& r&=2+\sin(4\theta) \qquad\qquad & &\text{(b)}\quad& r&=1+2\sin(4\theta)\\ &\text{(c)}\quad& r&=1\qquad\qquad & &\text{(d)}& r&=2\cos(\theta),\ -\tfrac{\pi}{2}\le\theta\le\tfrac{\pi}{2}\\ &\text{(e)}& r&=e^{\theta/10}+e^{-\theta/10}\qquad\qquad & &\text{(f)}& r&=\theta & \end{alignat*}\]

(A)

(B)

(C)

(D)

(E)

(F)

Stage 2

5

Recall that a point with polar coordinates \(r\) and \(\theta\) has \(x=r\cos\theta\) and \(y=r\sin\theta\text{.}\) Let \(r=f(\theta)\) be the equation of a plane curve in polar coordinates. Find the curvature of this curve at a general point \(\theta\text{.}\)

6

Find the curvature of the cardioid \(r=a(1-\cos\theta)\text{.}\)

1. As usual \(r\) is the distance from the origin to the point and \(\theta\) is angle between the \(x\)-axis and the vector from the origin to the point. The symbols \(r\text{,}\) \(\theta\) are the standard mathematics symbols for the polar coordinates. Appendix A.7 gives another set of symbols that is commonly used in the physical sciences and engineering.
2. The \(\hat{\boldsymbol{\theta}}\) component of the acceleration just tells us how much normal force the rod is applying to the bead to keep it on the rod.
3. A review of the technique used to find this solution is given in Appendix A.9. In any event, it is easy to check that \(r(t)=A e^{\Omega\,t} + B e^{-\Omega\,t}\) really does obey \(\frac{\mathrm{d}^{2}r}{\mathrm{d}t^{2}}-\Omega^2\ r = 0\text{.}\)
4. It is outside our scope to prove this equivalence.