Chapter 6Motion in Polar Coordinates and Conic Sections

Example 6.1: From the velocity components in polar coordinates to the velocity components in rectangular coordinates

At a certain instant, the radial component of the velocity is $\dot r=2\ \mathrm{m/s}$ and the angular component is $r\dot\theta=3\ \mathrm{m/s}$. Suppose that $\theta=\pi/4\ (=45^\circ)$ at this instant. Find the velocity components $v_x,v_y$ in the rectangular (Cartesian) coordinate system.

Solution. Substituting $\dot r=2,\ r\dot\theta=3,\ \theta=\pi/4$ ($\cos\dfrac\pi4=\sin\dfrac\pi4=\dfrac{\sqrt2}{2}$) into Eq. Ⓑ (Eq. \eqref{eq:6-vB}) gives

Therefore $v_x=-\dfrac{\sqrt2}{2}\approx-0.707\ \mathrm{m/s}$ and $v_y=\dfrac{5\sqrt2}{2}\approx3.54\ \mathrm{m/s}$. The reason $v_x$ is negative is that the contribution ($-r\dot\theta\sin\theta$) of the angular component $r\dot\theta=3$ to the $x$ direction is larger in magnitude than the contribution ($\dot r\cos\theta$) of the radial component $\dot r=2$ to the $x$ direction.

Check: The speed should not change under a rotation. From the rectangular components, $v_x^2+v_y^2=\dfrac12+\dfrac{25}{2}=13\ \mathrm{m^2/s^2}$; from the polar components, $\dot r^2+(r\dot\theta)^2=2^2+3^2=13\ \mathrm{m^2/s^2}$. The two agree ($\abs{\bm v}=\sqrt{13}\approx3.61\ \mathrm{m/s}$).

Example 6.2: Checking the components and signs of the basis vectors

(1) By reading off the column vectors of the rotation matrix $R(\theta)=\begin{bmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{bmatrix}$ in Eq. \eqref{eq:6-vA} (Eq. Ⓐ), write down the basis vectors $\bm e_r,\bm e_\theta$ in rectangular components, and confirm that they are mutually orthogonal unit vectors ($\bm e_r\cdot\bm e_\theta=0$, $|\bm e_r|=|\bm e_\theta|=1$).
(2) In writing down $\bm e_\theta$, a sign error was made, and it was written as $\bm e_\theta'=\sin\theta\,\bm e_x-\cos\theta\,\bm e_y$. Show that this error cannot be found by the orthogonality and unit-length checks in $(1)$, and find the error by considering the cases $\theta=0$ and $\theta=\pi/2$ geometrically.

Solution. (1) As stated in the main text, $\bm e_r$ is the first column $\begin{bmatrix}\cos\theta\\\sin\theta\end{bmatrix}$ of $R(\theta)$ and $\bm e_\theta$ is the second column $\begin{bmatrix}-\sin\theta\\\cos\theta\end{bmatrix}$, each read directly as the components along $\bm e_x,\bm e_y$:

Their magnitudes are, respectively,

so both are unit vectors of magnitude 1. The inner product is

so they are indeed orthogonal. (In general, the column vectors of an orthogonal matrix such as a rotation matrix are always mutually orthogonal unit vectors—this is also the origin of the name “orthogonal” matrix.)

(2) First, the incorrect $\bm e_\theta'=(\sin\theta,\ -\cos\theta)$ also has magnitude $\sqrt{\sin^2\theta+\cos^2\theta}=1$, and its inner product with $\bm e_r$ is $\cos\theta\sin\theta-\sin\theta\cos\theta=0$, so it satisfies both unit length and orthogonality anyway. In fact, $\bm e_\theta'=-\bm e_\theta$; it merely points in the direction exactly opposite to $\bm e_\theta$, so these two checks cannot tell them apart.

We therefore examine the direction directly. $\bm e_\theta$ is the unit vector in “the direction in which $\theta$ increases (counterclockwise).” When $\theta=0$, we have $\bm e_r=\bm e_x$, and the direction of counterclockwise motion from there is the $+\bm e_y$ direction, so we must have $\bm e_\theta=\bm e_y$. The expression in the main text gives $\bm e_\theta=(-\sin0,\ \cos0)=(0,\ 1)=\bm e_y$, which is correct, but the incorrect expression gives $\bm e_\theta'=(\sin0,\ -\cos0)=(0,\ -1)=-\bm e_y$, which points the opposite way. Likewise for $\theta=\pi/2$: the direction of counterclockwise motion from $\bm e_r=\bm e_y$ is the $-\bm e_x$ direction; the expression in the main text gives $(-\sin\frac\pi2,\ \cos\frac\pi2)=(-1,\ 0)$, which is correct, whereas the incorrect expression gives $(1,\ 0)=+\bm e_x$, which points the opposite way. Therefore the correct one is $\bm e_\theta=-\sin\theta\,\bm e_x+\cos\theta\,\bm e_y$.

Example 6.3: Comparing polar components and tangential and normal components in circular motion

Consider a particle (point mass) moving counterclockwise ($\dot\theta\gt0$) with angular velocity $\dot\theta$ on a circle of radius $r_0$ (constant). (The motion need not be uniform circular motion; $\dot\theta$ may vary with time.) Compare the polar components $a_r,a_\theta$ of this section with the tangential and normal components $\bm a=\dot v\,\bm e_t+\dfrac{v^2}{\rho}\bm e_n$ studied in Chapter 3 ($v$ is the speed, $\rho$ is the radius of curvature, $\bm e_t$ is the unit tangent vector in the direction of motion, and $\bm e_n$ is the unit principal normal vector pointing toward the center of curvature).

Solution. Since the motion is circular, $r=r_0$ (constant), so $\dot r=0,\ \ddot r=0$. Substituting into Formula 6.2, we get

These are the polar components. On the other hand, consider the tangential and normal components. For motion on a counterclockwise ($\dot\theta\gt0$) circle, the direction of motion (the tangent direction) coincides with the angular direction $\bm e_\theta$, so $\bm e_t=\bm e_\theta$, and the speed is $v=r_0\dot\theta$ (Formula 6.1 of Section 6.1 with $\dot r=0$). The tangential acceleration is therefore

This agrees exactly with $a_\theta=r_0\ddot\theta$. Moreover, the radius of curvature of circular motion is the radius of the circle itself, $\rho=r_0$ (confirmed in Chapter 3), and the principal normal $\bm e_n$ points toward the center of curvature (= the center of the circle = the origin), that is, $\bm e_n=-\bm e_r$. The magnitude of the normal acceleration is

Since its direction is $\bm e_n=-\bm e_r$, its component along $\bm e_r$ is $-r_0\dot\theta^2$. This too agrees exactly with $a_r=-r_0\dot\theta^2$.

To summarize: in the special case of circular motion, the polar components $(a_r,a_\theta)$ and the tangential and normal components $(v^2/\rho\text{ in the }-\bm e_r\text{ direction},\ \dot v\text{ in the }\bm e_\theta\text{ direction})$ are merely two ways of writing one and the same acceleration vector, under the correspondence of bases $\bm e_r=-\bm e_n,\ \bm e_\theta=\bm e_t$—both are correct, and which to use can be chosen according to which makes the problem easier to see. Note that in the clockwise case ($\dot\theta\lt0$) the direction of motion is $-\bm e_\theta$, so $\bm e_t=-\bm e_\theta$ (the signs of the tangential acceleration $\dot v$ and of $a_\theta$ are opposite), but $\bm e_n=-\bm e_r$ does not change, and the normal component gives the same result.

Example 6.4: Comparing the two decompositions for motion with a changing radius

A particle moves along a spiral given by $r=bt,\ \theta=\omega t$ ($b=1\ \mathrm{m/s}$, $\omega=2\ \mathrm{rad/s}$, with $t$ in seconds). At $t=1\ \mathrm s$, find (a) the polar components $a_r,a_\theta$ and the magnitude $\abs{\bm a}$ of the acceleration, (b) the speed $v$, the tangential acceleration $\dot v$ (the time derivative of $v$), the normal acceleration $a_n=\sqrt{\abs{\bm a}^2-\dot v^2}$, and the radius of curvature $\rho$, and (c) the polar components of the unit tangent vector $\bm e_t$ (confirm that $\bm e_t\neq\bm e_\theta$).

Solution. From $r=bt$ we have $\dot r=b=1\ \mathrm{m/s}$ and $\ddot r=0$. From $\theta=\omega t$ we have $\dot\theta=\omega=2\ \mathrm{rad/s}$ and $\ddot\theta=0$. At $t=1\ \mathrm s$, $r=1\ \mathrm m$.

(a) By Formula 6.2, at a general time $t$, $a_r=\ddot r-r\dot\theta^2=-b\omega^2t$ and $a_\theta=2\dot r\dot\theta+r\ddot\theta=2b\omega$. At $t=1\ \mathrm s$,

Thus $\abs{\bm a}=\sqrt{a_r^2+a_\theta^2}=\sqrt{16+16}=4\sqrt2\approx5.66\ \mathrm{m/s^2}$. In this motion $\ddot\theta=0,\ \ddot r=0$, but thanks to the $2\dot r\dot\theta$ term, $a_\theta\neq0$.

(b) By Formula 6.1, $v_r=\dot r=1\ \mathrm{m/s}$ and $v_\theta=r\dot\theta=bt\cdot\omega=2t\ \mathrm{m/s}$. The speed is $v=\sqrt{\dot r^2+(r\dot\theta)^2}=b\sqrt{1+\omega^2t^2}=\sqrt{1+4t^2}$, and at $t=1\ \mathrm s$, $v=\sqrt5\approx2.24\ \mathrm{m/s}$. Differentiating $v$ gives the tangential acceleration $\dot v=\dfrac{b\omega^2t}{\sqrt{1+\omega^2t^2}}=\dfrac{4t}{\sqrt{1+4t^2}}$, and at $t=1\ \mathrm s$, $\dot v=\dfrac{4}{\sqrt5}\approx1.79\ \mathrm{m/s^2}$. From the relation $\abs{\bm a}^2=\dot v^2+a_n^2$ of Chapter 3,

(Here we used the Pythagorean theorem, since $\bm e_t\perp\bm e_n$ in $\bm a=\dot v\bm e_t+a_n\bm e_n$.)

(c) Since $\bm e_t$ is the unit vector in the direction of the velocity, $\bm e_t=\bm v/v=(v_r\bm e_r+v_\theta\bm e_\theta)/v=\dfrac{1}{\sqrt5}\bm e_r+\dfrac{2}{\sqrt5}\bm e_\theta$. It is not $\bm e_t=\bm e_\theta$ as in the case of circular motion; it is tilted from $\bm e_\theta$ toward the $\bm e_r$ side by $\tan^{-1}\dfrac12\approx26.6^\circ$. As a check, the component of $\bm a$ in the $\bm e_t$ direction is $\bm a\cdot\bm e_t=\dfrac{a_r\times1+a_\theta\times2}{\sqrt5}=\dfrac{-4+8}{\sqrt5}=\dfrac{4}{\sqrt5}$, which agrees with the tangential acceleration $\dot v$. Also, $-r\dot\theta^2=-4\ \mathrm{m/s^2}$ does not agree with the normal acceleration $a_n\approx5.37\ \mathrm{m/s^2}$ ($a_r=-4\ \mathrm{m/s^2}$ is a consequence of $\bm e_r$ being oblique to both $\bm e_n$ and $\bm e_t$).

Example 6.5: Acceleration in the angular direction and conservation of angular momentum (qualitative)

As stated in the “Discussion” of this section, the acceleration in the angular direction can be written $a_\theta=\dfrac1r\dfrac{\dd}{\dd t}(r^2\dot\theta)$. Suppose that the force acting on a particle always points only along the line to the origin (the $\bm e_r$ direction) and has no component in the angular direction ($F_\theta=ma_\theta=0$). How would you expect $r$ and $\dot\theta$ to be related to each other during the motion? Explain qualitatively from the form of the equations.

Solution. We have $a_\theta=\dfrac1r\dfrac{\dd}{\dd t}(r^2\dot\theta)=0$, and as long as $r\neq0$ this is the same as $\dfrac{\dd}{\dd t}(r^2\dot\theta)=0$. That the time derivative is always $0$ means that the quantity $r^2\dot\theta$ does not change with time—it is kept constant throughout the motion. Therefore, when the particle approaches the origin ($r$ decreases), the angular velocity $\dot\theta$ must increase in order to keep $r^2\dot\theta$ constant, and conversely, when the particle moves away from the origin ($r$ increases), we can expect $\dot\theta$ to decrease—the closer to the origin, the faster the particle goes around, and the farther away, the more slowly. This is exactly the same form of relation as the phenomenon in which a figure skater spins faster when the arms are pulled in (the effective radius becomes smaller). In Chapter 8, when we treat the motion of a planet that is subject to the Sun's gravitational attraction (a force that always points toward the Sun), this expectation will be proved exactly in the form of “constant areal velocity” (Kepler's second law).

Example 6.6: Acceleration components for a given motion (numerical example)

The motion of a particle is given by $r(t)=2+\cos\omega t\ [\mathrm m]$, $\theta(t)=\omega t\ [\mathrm{rad}]$, and $\omega=1\ \mathrm{rad/s}$ (the unit of $t$ is $\mathrm s$). Find the polar components $a_r,a_\theta$ of the acceleration at (a) $t=0$ and (b) $t=\pi/2\ \mathrm s$.

Solution. Differentiating $r(t)=2+\cos\omega t$ in turn, we obtain $\dot r(t)=-\omega\sin\omega t\ [\mathrm{m/s}]$ and $\ddot r(t)=-\omega^2\cos\omega t\ [\mathrm{m/s^2}]$. From $\theta(t)=\omega t$ we have $\dot\theta=\omega=1\ \mathrm{rad/s}$ and $\ddot\theta=0$ (motion with constant angular velocity).

(a) Substituting $t=0$, we have

Substituting into Formula 6.2,

At the instant $t=0$, we have $\dot r=0$ ($r$ takes its maximum value $3\ \mathrm m$, a turning point) and $\ddot\theta=0$ (constant angular velocity), so the acceleration in the angular direction happens to be $0$, and the acceleration has only a radial component (pointing toward the origin, $a_r\lt0$). At this instant, the term $2\dot r\dot\theta$, the centerpiece of the chapter, plays no role.

(b) When $\omega t=\pi/2$, we have $\cos\dfrac\pi2=0,\ \sin\dfrac\pi2=1$, so

Since $\ddot\theta=0$, the acceleration in the angular direction, $a_\theta=-2\ \mathrm{m/s^2}$, arises entirely from the term $2\dot r\dot\theta$ alone. Physically, this can be understood as follows: if $r$ shrinks ($\dot r\lt0$) while the angular velocity $\dot\theta$ stays constant, the speed $r\dot\theta$ along the circumferential direction decreases, so an acceleration arises in the direction opposite to $\bm e_\theta$, the direction in which $\theta$ increases.

Check: At $t=\pi/2\ \mathrm s$ the particle is at $(x,y)=(0,\ 2)\ \mathrm m$, and $\bm e_r=\bm e_y,\ \bm e_\theta=-\bm e_x$. Differentiating $x=(2+\cos t)\cos t,\ y=(2+\cos t)\sin t$ twice in rectangular coordinates gives $(a_x,a_y)=(2,\ -2)\ \mathrm{m/s^2}$, so that $a_r=a_y=-2\ \mathrm{m/s^2}$ and $a_\theta=-a_x=-2\ \mathrm{m/s^2}$, which agrees with the result above.

Illustration 6.1: The case $\varepsilon=1/2,\ k=1$

Substituting the eccentricity $\varepsilon=\dfrac12$ and $k=1$ into Formula 6.3, we obtain

Multiplying both sides by the denominator and rearranging, we have

(we multiplied both sides by 2 to eliminate the fraction). To see what figure this is, we convert to rectangular coordinates. We use the basic relations between polar and rectangular coordinates, $x^2+y^2=r^2$, $x=r\cos\theta$, $y=r\sin\theta$ (that is, $r=\sqrt{x^2+y^2}$ and $r\cos\theta=x$). Substituting $r\cos\theta=x$ into $r(2+\cos\theta)=2r+r\cos\theta=2$, we obtain

which is an equation solved for $r$ (no $r$ remains on the right-hand side, so all that is left is to substitute $r=\sqrt{x^2+y^2}$ to make it an equation in $x,y$ alone). However, since $r=\sqrt{x^2+y^2}$ contains a square root, substituting it as it is would be awkward. So we use the standard technique of squaring both sides of Eq. \eqref{eq:6-ex1-step1} to eliminate the square root:

(we rewrote the right-hand side as $(2-x)^2=(x-2)^2$; squaring removes the difference in sign). Using $r^2=x^2+y^2$ here, we obtain

Expanding the right-hand side ($(x-2)^2=x^2-4x+4$), we have

Moving all the terms to the left-hand side and rearranging, we obtain

($4x^2-x^2=3x^2$; moving $-4x$ gives $+4x$; moving $-4$ gives $-4$; and $4y^2$ stays as it is). If we complete the square in $x$ in this equation, the figure it represents becomes visible in standard form. Since $3x^2+4x=3\left(x^2+\dfrac43x\right)=3\left[\left(x+\dfrac23\right)^2-\dfrac49\right]=3\left(x+\dfrac23\right)^2-\dfrac43$, Eq. \eqref{eq:6-ex1-step3} becomes

Dividing both sides by 12 (the least common multiple of $3$ and $4$), we obtain

The fraction $4/9$ on the right-hand side may look like an obstacle, but if we divide both sides further by $4/9$, we obtain

which agrees exactly with the standard form of the ellipse, $\dfrac{(x-x_0)^2}{a^2}+\dfrac{(y-y_0)^2}{b^2}=1$, learned in High School Mathematics, Section 8.2 (center $(x_0,y_0)=\left(-\dfrac23,0\right)$, semi-major axis $a=\dfrac43$, semi-minor axis $b=\dfrac{2}{\sqrt3}$). We have confirmed that the curve is indeed an ellipse.

Let us also check that the origin is a focal point. The distance from the center to the focal point is $c=\sqrt{a^2-b^2}=\sqrt{\dfrac{16}{9}-\dfrac43}=\sqrt{\dfrac49}=\dfrac23$, which agrees exactly with the distance from the center $x=-\dfrac23$ to the origin. Also, $c/a=\dfrac{2/3}{4/3}=\dfrac12=\varepsilon$. The point of the ellipse nearest to the origin is $r(0)=\dfrac{1}{1+1/2}=\dfrac23$, and the farthest point is $r(\pi)=\dfrac{1}{1-1/2}=2$, which also agrees with the length of the major axis, $\dfrac23+2=\dfrac83=2a$.

Example 6.7: A conic section of eccentricity 2 (hyperbola)

When $\varepsilon=2,\ k=3$, what kind of figure is the curve of Formula 6.3? Classify it using Theorem 6.1, and find its standard form in rectangular coordinates.

Solution. Since $\varepsilon=2\gt1$, Theorem 6.1 tells us that the curve is a hyperbola. To find the standard form, we substitute $\varepsilon=2,k=3$ into Eq. \eqref{eq:6-conic-cartesian}:

Multiplying both sides by $-1$ to make the coefficient of $x^2$ positive, we get $3x^2-12x-y^2+9=0$. We complete the square in $x$. Since $3x^2-12x=3(x^2-4x)=3\left[(x-2)^2-4\right]=3(x-2)^2-12$,

Dividing both sides by 3 gives

This agrees with the standard form $\dfrac{(x-x_0)^2}{a^2}-\dfrac{y^2}{b^2}=1$ of the hyperbola learned in High School Mathematics, Section 8.3, with center $(x_0,0)=(2,0)$, $a=1,\ b=\sqrt3$ (this also matches $a=\frac{k}{\varepsilon^2-1}=\frac33=1$, $b=\frac{k}{\sqrt{\varepsilon^2-1}}=\sqrt3$, and the center $a\varepsilon=2$ of Theorem 6.1). The distance from the center to the focal points is $c=\sqrt{a^2+b^2}=\sqrt{1+3}=2$, so the focal points are at $(2\pm2,\,0)=(0,0),(4,0)$—$(0,0)$, that is, the origin, is indeed one of the focal points of this hyperbola (which is consistent with the fact that Formula 6.3 is constructed with the origin as a focal point). Of the curve, only the branch on the side nearer the origin (the range $x\le2-1=1$) is the part that can be traced as $r(\theta)=3/(1+2\cos\theta)$ ($r\gt0$) when $\theta$ is varied from $0$.

Exercise 6.1: Solving the basis-vector relation in reverse

Eq. \eqref{eq:6-basis-vectors} (the same equation as Chapter 3, Formula 3.2) expresses $\bm e_r,\bm e_\theta$ in terms of $\bm e_x,\bm e_y$. Solve this relation in reverse to express $\bm e_x,\bm e_y$ in terms of $\bm e_r,\bm e_\theta$ and $\theta$.

Hint: Regard Eq. \eqref{eq:6-basis-vectors} as the matrix relation $[\bm e_r\ \bm e_\theta]=[\bm e_x\ \bm e_y]R(\theta)$, and use the property from Section 6.1(ii) that “the inverse of a rotation matrix equals its transpose”: multiply both sides on the right by $R(\theta)^{-1}=R(\theta)^{\mathrm T}$.

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Exercise 6.2: Perihelion and aphelion distances

For the curve $r(\theta)=k/(1+\varepsilon\cos\theta)$ of Formula 6.3, express in terms of $k,\varepsilon$ the value of $r$ at $\theta=0$ and (for $0\le\varepsilon\lt1$) the value of $r$ at $\theta=\pi$ (in Chapter 8, these correspond to the distances to the point nearest the Sun, the perihelion, and to the point farthest from it, the aphelion). Furthermore, since the sum of these two distances equals the length $2a$ of the major axis, express the semi-major axis $a$ in terms of $k,\varepsilon$, and verify that it agrees with $a=k/(1-\varepsilon^2)$ of Theorem 6.1.

Hint: Simply substitute $\theta=0$ ($\cos0=1$) and $\theta=\pi$ ($\cos\pi=-1$) into Formula 6.3.

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Exercise 6.3: An ellipse of eccentricity 0.8

Convert the conic section with $\varepsilon=0.8,\ k=2$ to the standard form in rectangular coordinates. Also find its semi-major axis $a$, its semi-minor axis $b$, and the position of its center, and verify that they agree with the values in Theorem 6.1.

Hint: Substitute the numerical values into Eq. \eqref{eq:6-conic-cartesian} and bring it to the standard form by the same procedure (completing the square) as in Example 6.7. In this case the coefficient of $x^2$ is positive, so the multiplication by $-1$ that was needed in Example 6.7 is not necessary.

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Exercise 6.4: An alternative proof of the acceleration formula using the time derivatives of the basis vectors

In this chapter we derived the acceleration formula (Formula 6.2) by comparing rotation matrices (Eqs. Ⓒ and Ⓓ), but there is another method, which uses the formulas for the time derivatives of the basis vectors $\bm e_r,\bm e_\theta$ themselves (also treated in Exercise 3.3 of Chapter 3):

Accept these formulas and, by differentiating the position vector $\bm r=r\bm e_r$ twice with respect to time, derive the velocity $\bm v=\dot r\bm e_r+r\dot\theta\bm e_\theta$ (Formula 6.1) and the acceleration $\bm a=(\ddot r-r\dot\theta^2)\bm e_r+(2\dot r\dot\theta+r\ddot\theta)\bm e_\theta$ (Formula 6.2), independently of the derivation in this chapter.

Hint: The formula $\dot{\bm e}_r=\dot\theta\bm e_\theta$ itself can be confirmed by differentiating Eq. \eqref{eq:6-basis-vectors} with respect to $t$ (note that $\theta$ is a function of $t$, so use the chain rule) and using $\dfrac{\dd}{\dd t}\cos\theta=-\dot\theta\sin\theta,\ \dfrac{\dd}{\dd t}\sin\theta=\dot\theta\cos\theta$. Then apply the product rule to $\bm v=\dfrac{\dd}{\dd t}(r\bm e_r)=\dot r\bm e_r+r\dot{\bm e}_r$ and substitute $\dot{\bm e}_r=\dot\theta\bm e_\theta$. For the acceleration, differentiate this once more.

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Exercise 6.5: Checking the case of eccentricity 1 (parabola)

Confirm, by converting to the standard form, that for $\varepsilon=1$ the curve of Formula 6.3 is a parabola ($k$ may be left as a symbol).

Hint: Substituting $\varepsilon=1$ into Eq. \eqref{eq:6-conic-cartesian}, the $x^2$ term vanishes exactly. Rearrange the remaining equation into the form $y^2=\cdots$ and compare it with the standard form $y^2=4px$ of the parabola (or the form with the vertex shifted) in High School Mathematics, Section 8.1.

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Exercise 6.6: Numerical computation of the polar components of acceleration

The motion of a particle is given by $r(t)=1+t^2\ [\mathrm m]$ and $\theta(t)=t\ [\mathrm{rad}]$ (the unit of $t$ is $\mathrm s$). Find $a_r,a_\theta$ at $t=1\ \mathrm s$ using Formula 6.2. Also, are $\ddot r$ and $a_r$ equal in this case?

Hint: First find $\dot r,\ \ddot r,\ \dot\theta,\ \ddot\theta$ and then substitute $t=1$. Compare the values of $\ddot r$ and $r\dot\theta^2$.

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Exercise 6.7: Polar components of uniform rectilinear motion

A point moves in uniform rectilinear motion in the $y$ direction with speed $u$ along the line $x=b$ ($x=b,\ y=ut$, $b,u\gt0$). (1) Express $r,\ \dot r,\ \dot\theta$ in terms of $b,u,t$. (2) Show that $\ddot r=r\dot\theta^2$ and $r\ddot\theta=-2\dot r\dot\theta$ hold, and verify that $a_r=a_\theta=0$.

Hint: From $r^2=b^2+u^2t^2$ we get $r\dot r=u^2t$. Since $\theta=\tan^{-1}(ut/b)$, we have $\dot\theta=\dfrac{bu}{b^2+u^2t^2}=\dfrac{bu}{r^2}$. Differentiating $r\dot r=u^2t$ once more gives $\dot r^2+r\ddot r=u^2$.

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Exercise 6.8: Direction of the asymptotes of a hyperbola

For $\varepsilon\gt1$, the curve of Formula 6.3 has an angle $\theta_\infty\ (0\lt\theta_\infty\lt\pi)$ at which $r\to\infty$. Express $\theta_\infty$ in terms of $\varepsilon$, and find $\cos\theta_\infty,\ \tan\theta_\infty$ for $\varepsilon=5/3$. Also verify that this agrees with the magnitude $b/a$ of the slope of the asymptotes of the hyperbola in Theorem 6.1.

Hint: $r\to\infty$ occurs when the denominator $1+\varepsilon\cos\theta$ becomes $0$. By Theorem 6.1, $b/a=\sqrt{\varepsilon^2-1}$.

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