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CLP-4 Vector Calculus

Joel Feldman, Andrew Rechnitzer, Elyse Yeager

Section 1.8 Optional — Polar Coordinates

So far we have always written vectors in two dimensions in terms of the basis vectors \(\hi\) and \(\hj\text{.}\) This is not always convenient. For example, when working in polar coordinates it is often convenient to use basis vectors \(\hat\vr(\theta)\text{,}\) \(\hat\vth(\theta)\) which depend on the value of the current polar coordinate \(\theta\) — though one usually just writes \(\hat\vr\text{,}\) \(\hat\vth\text{,}\) 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
Note that this basis has two very nice properties.
  1. \(|\hat\vr(\theta)| = |\hat\vth(\theta)| = 1\text{,}\) \(\hat\vr(\theta) \perp \hat\vth(\theta)\) (orthonormality)
  2. \(\diff{\hat\vr}{\theta}(\theta)= \hat\vth(\theta)\text{,}\) \(\diff{\hat\vth}{\theta}(\theta) = -\hat\vr(\theta)\)
That \(\diff{\hat\vr}{\theta}(\theta)\) is some scalar multiple of \(\hat\vth(\theta)\) follows just from the fact that \(|\hat\vr(\theta)| = 1\text{.}\)
\begin{align*} |\hat\vr(\theta)| = 1 &\implies \hat\vr(\theta)\cdot\hat\vr(\theta)=1\\ &\implies \hat\vr(\theta)\cdot \diff{\hat\vr}{\theta}(\theta) =\frac{1}{2}\diff{}{t}\big( \hat\vr(\theta)\cdot\hat\vr(\theta)\big) =0\\ &\implies \diff{\hat\vr}{\theta}(\theta)\perp \hat\vr(\theta) \implies \diff{\hat\vr}{\theta}(\theta)\parallel \hat\vth(\theta) \end{align*}
Similarly, that \(\diff{\hat\vth}{\theta}(\theta)\) is some scalar multiple of \(\hat\vr(\theta)\) follows just from the fact that \(|\hat\vth(\theta)| = 1\text{.}\)
It is standard to suppress the arguments \(t\) and \(\theta(t)\) and write, for example,
\begin{equation*} \vv= \diff{r}{t}\ \hat\vr + r\ \diff{\theta}{t}\ \hat\vth \end{equation*}
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 \(\vr = r\,\hat\vr(\theta)\text{.}\) So if we parametrize a curve by giving the polar coordinates at time \(t\text{,}\)
\begin{align*} \vr(t) &= r(t)\ \hat\vr\big(\theta(t)\big)\\ \vv(t) &= \diff{r}{t}(t)\ \hat\vr\big(\theta(t)\big) + r(t)\ \diff{\hat\vr}{\theta}\big(\theta(t)\big)\ \diff{\theta}{t}(t) \notag\\ &= \diff{r}{t}(t)\ \hat\vr\big(\theta(t)\big) + r(t)\ \diff{\theta}{t}(t)\ \hat\vth\big(\theta(t)\big)\\ \va(t) & = \difftwo{r}{t}\ \hat\vr + \diff{r}{t}\ \diff{\hat\vr}{\theta}\ \diff{\theta}{t} + \diff{r}{t}\ \diff{\theta}{t}\ \hat\vth + r\ \difftwo{\theta}{t}\ \hat\vth + r\ \Big(\diff{\theta}{t}\Big)^2\ \diff{\hat\vth}{\theta}\\ &=\Big[\difftwo{r}{t}-r\ \Big(\diff{\theta}{t}\Big)^2\Big] \hat\vr +\Big[r\ \difftwo{\theta}{t} + 2 \diff{r}{t}\ \diff{\theta}{t}\Big]\hat\vth \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, \(m\va=\vF\text{,}\) the radial
 2 
The \(\hat\vth\) component of the acceleration just tells us how much normal force the rod is applying to the bead to keep it on the rod.
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,
\begin{equation*} \difftwo{r}{t}-r\ \Big(\diff{\theta}{t}\Big)^2 = 0 \end{equation*}
As the rod is rotating at \(\Omega\,\)rad/sec, \(\diff{\theta}{t}=\Omega\) and
\begin{equation*} \difftwo{r}{t}-\Omega^2\ r = 0 \end{equation*}
The general solution to this constant coefficient second order ordinary differential equation is
 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 \(\difftwo{r}{t}-\Omega^2\ r = 0\text{.}\)
\begin{equation*} r(t) = A e^{\Omega\,t} + B e^{-\Omega\,t} \end{equation*}
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
\begin{equation*} r(t) = \frac{1}{2}\big(e^{\Omega\,t}+e^{-\Omega\,t}\big)\qquad \end{equation*}
If, again for example, \(\theta(0) = 0\text{,}\) then \(\theta(t) = \Omega t\) and the bead follows the polar coordinate curve
\begin{equation*} r(\theta) = \frac{1}{2}\big(e^{\theta}+e^{-\theta}\big) \end{equation*}
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 equivalent
 4 
It is outside our scope to prove this equivalence.
(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 \(\veps\ge 0\) (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 \(\veps\) times the distance from \(P\) to \(L\text{.}\) So
\begin{equation*} r=\veps\big(p-r\cos\theta\big) \iff r=\frac{\veps p}{1+\veps\cos\theta} \end{equation*}
The numerator \(\veps p\) is usually renamed to \(\ell\) giving the equation
\begin{equation*} r=\frac{\ell}{1+\veps\cos\theta} \end{equation*}

Example 1.8.5. Conic sections in polar coordinates, again.

We’ll now take the equation \(r=\frac{\ell}{1+\veps\cos\theta}\) 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+\veps\cos\theta\big)=\ell &\iff r = \ell -\veps x\\ &\iff x^2+y^2 = \ell^2-2\veps\ell x+\veps^2 x^2\\ &\iff (1-\veps^2) x^2 + 2\veps\ell x + y^2 = \ell^2 \tag{C} \end{align*}
Now consider separately four different cases, depending on the value of \(\veps \ge 0\text{.}\)
  • If \(\veps=0\text{,}\) (C) reduces to
    \begin{equation*} x^2+y^2 = \ell^2 \end{equation*}
    which is of course a circle of radius \(\ell\text{.}\)
  • If \(0 \lt \veps \lt 1\text{,}\) completing the square in (C) gives
    \begin{equation*} (1-\veps^2)\Big(x+\frac{\veps\ell}{1-\veps^2}\Big)^2 + y^2 = \ell^2 + \frac{\veps^2\ell^2}{1-\veps^2} = \frac{\ell^2}{1-\veps^2} \end{equation*}
    which is equivalent to
    \begin{equation*} \frac{\big(x+\frac{\veps\ell}{1-\veps^2}\big)^2} { \frac{\ell^2}{{(1-\veps^2)}^2}} +\frac{y^2}{\frac{\ell^2}{1-\veps^2}} =1 \end{equation*}
    and is of course an ellipse with semi-major axis \(r_M=\frac{\ell}{1-\veps^2}\) and semi-minor axis \(r_m=\frac{\ell}{\sqrt{1-\veps^2}}\text{.}\)
  • If \(\veps=1\text{,}\) (C) reduces to
    \begin{equation*} y^2 = \ell^2-2\ell x \end{equation*}
    which is of course a parabola.
  • If \(\veps \gt 1\text{,}\) the same computation as in the \(0 \lt \veps \lt 1\) case gives
    \begin{equation*} \frac{\big(x-\frac{\veps\ell}{\veps^2-1}\big)^2} { \frac{\ell^2}{{(\veps^2-1)}^2}} -\frac{y^2}{\frac{\ell^2}{\veps^2-1}} =1 \end{equation*}
    and is of course a hyperbola.

Exercises Exercises

Exercise Group.

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
    \begin{equation*} (-2,0) = \big(r\cos\theta\,,\,r\sin\theta\big) \end{equation*}
  2. Find all pairs \((r,\theta)\) such that
    \begin{equation*} (1,1) = \big(r\cos\theta\,,\,r\sin\theta\big) \end{equation*}
  3. Find all pairs \((r,\theta)\) such that
    \begin{equation*} (-1,-1) = \big(r\cos\theta\,,\,r\sin\theta\big) \end{equation*}
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*} \he_r(\theta)=\cos\theta\ \hi + \sin\theta\ \hj\qquad \he_\theta(\theta) = -\sin\theta\ \hi + \cos\theta\ \hj \end{gather*}
  1. Determine, for each angle \(\theta\text{,}\) the lengths of the vectors \(\he_r(\theta)\) and \(\he_\theta(\theta)\) and the angle between the vectors \(\he_r(\theta)\) and \(\he_\theta(\theta)\text{.}\) Compute \(\he_r(\theta)\times\he_\theta(\theta)\) (viewing \(\he_r(\theta)\) and \(\he_\theta(\theta)\) as vectors in three dimensions with zero \(\hk\) components).
  2. For each \(1\le i\le 5\text{,}\) sketch, in the \(xy\)-plane, the point \((x_i,y_i)\) and the vectors \(\he_r(\theta_i)\) and \(\he_\theta(\theta_i)\text{.}\) In your sketch of the vectors, place the tails of the vectors \(\he_r(\theta_i)\) and \(\he_\theta(\theta_i)\) at \((x_i,y_i)\text{.}\)
4. (✳).
Match the following equations with the corresponding pictures. Cartesian coordinates are \((x, y)\) and polar coordinates are \((r, \theta)\text{.}\)
(A)
(B)
(C)
(D)
(E)
(F)
\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*}

Exercise Group.

Exercises — 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 \(r=a(1-\cos\theta)\text{.}\)
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