Chapter 3Description of Curvilinear Motion — Polar Coordinates and Tangential and Normal Acceleration

Example 3.1: Converting between rectangular and polar coordinates

(a) Find the polar coordinates $(r,\theta)$ of the point $(x,y)=(3,4)$ (in units of $\mathrm m$). (b) Find the rectangular coordinates $(x,y)$ of the point whose polar coordinates are $(r,\theta)=(5\ \mathrm m,\ \pi/3)$. (c) Find the polar coordinates of the point $(x,y)=(-3,4)$ (in units of $\mathrm m$). Also note what happens if $\theta=\tan^{-1}(y/x)$ is used as it stands.

Solution. (a) By Formula 3.1, we have

$$ r=\sqrt{x^2+y^2}=\sqrt{3^2+4^2}=\sqrt{25}=5\ \mathrm m $$

The angle is $\theta=\tan^{-1}(4/3)$, and since $(x,y)=(3,4)$ lies in $x\gt0,\ y\gt0$ (the first quadrant), we may use the principal value as it is:

$$ \theta=\tan^{-1}\frac{4}{3}\approx0.927\ \mathrm{rad}\ (\approx53.1^\circ) $$

Therefore $(r,\theta)\approx(5\ \mathrm m,\ 0.927\ \mathrm{rad})$.

(b) We use the forward equations of Formula 3.1. Since $\cos(\pi/3)=1/2,\ \sin(\pi/3)=\sqrt3/2$, we have

$$ x=r\cos\theta=5\times\frac12=2.5\ \mathrm m,\qquad y=r\sin\theta=5\times\frac{\sqrt3}{2}=\frac{5\sqrt3}{2}\approx4.33\ \mathrm m $$

These are the rectangular coordinates of the point.

(c) The distance is $r=\sqrt{(-3)^2+4^2}=\sqrt{25}=5\ \mathrm m$. Since $y/x=4/(-3)=-4/3$, the principal value of $\tan^{-1}$ is

$$ \tan^{-1}\Bigl(-\frac43\Bigr)\approx-0.927\ \mathrm{rad}\ (\approx-53.1^\circ) $$

This is an angle in the fourth quadrant ($x\gt0,\ y\lt0$), whereas the actual point $(-3,4)$ lies in the second quadrant ($x\lt0,\ y\gt0$), so as it stands it is wrong. Since $\tan(\theta+\pi)=\tan\theta$, correcting by adding $\pi$ gives

$$ \theta=\tan^{-1}\Bigl(-\frac43\Bigr)+\pi\approx-0.927+3.142=2.214\ \mathrm{rad}\ (\approx126.9^\circ) $$

As a check, verify that $\cos\theta=x/r=-3/5=-0.6,\ \sin\theta=y/r=4/5=0.8$: we have $\cos2.214\approx-0.6,\ \sin2.214\approx0.8$, so this is indeed an angle in the second quadrant. Therefore $(r,\theta)\approx(5\ \mathrm m,\ 2.214\ \mathrm{rad})$. (The function $\operatorname{atan2}(4,-3)$ also returns the same $2.214$.)

Example 3.2: Components and orthogonality of the basis vectors

Find the polar-coordinate basis vectors $\bm{e}_r,\bm{e}_\theta$ at the position $\theta=\pi/3\ (=60^\circ)$ in terms of their components with respect to $\bm{e}_x,\bm{e}_y$. Also verify that $\bm{e}_r,\bm{e}_\theta$ are both unit vectors and are orthogonal to each other ($\bm{e}_r\cdot\bm{e}_\theta=0$).

Solution. Substituting $\cos(\pi/3)=1/2,\ \sin(\pi/3)=\sqrt3/2$ into Formula 3.2 gives

$$ \bm{e}_r=\frac12\bm{e}_x+\frac{\sqrt3}{2}\bm{e}_y=\Bigl[\frac12,\ \frac{\sqrt3}{2}\Bigr],\qquad \bm{e}_\theta=-\frac{\sqrt3}{2}\bm{e}_x+\frac12\bm{e}_y=\Bigl[-\frac{\sqrt3}{2},\ \frac12\Bigr] $$

The magnitudes are, respectively,

$$ \abs{\bm{e}_r}=\sqrt{\left(\frac12\right)^2+\left(\frac{\sqrt3}{2}\right)^2}=\sqrt{\frac14+\frac34}=\sqrt1=1,\qquad \abs{\bm{e}_\theta}=\sqrt{\left(-\frac{\sqrt3}{2}\right)^2+\left(\frac12\right)^2}=1 $$

so both are indeed unit vectors (this always holds for general $\theta$ as well, since $\cos^2\theta+\sin^2\theta=1$). The inner product is

$$ \bm{e}_r\cdot\bm{e}_\theta=\frac12\times\left(-\frac{\sqrt3}{2}\right)+\frac{\sqrt3}{2}\times\frac12=-\frac{\sqrt3}{4}+\frac{\sqrt3}{4}=0 $$

so orthogonality is confirmed (we will verify in Exercise 3.4 that $\bm{e}_r\cdot\bm{e}_\theta=-\cos\theta\sin\theta+\sin\theta\cos\theta=0$ also holds for general $\theta$).

Example 3.3: Velocity and tangent vector of a point moving on a parabola

The position of a point moving in a plane is given at time $t$ (in seconds) by $\bm r(t)=[t,\ t^2,\ 0]$ (in meters, $t\ge0$). This is a point moving on the parabola $y=x^2$ in the $xy$ plane (since $x=t,\ y=t^2=x^2$). (a) Find the velocity vector $\bm v(t)$ and the speed $v(t)$ as functions of time $t$. (b) Find the unit tangent vector $\bm e_t(t)$. (c) Find $\bm v,\ v,\ \bm e_t$ at $t=1$ (the instant at which the point passes through the point $(1,1)$).

Solution. (a) Differentiating each component with respect to $t$ (Definition 17.1), we obtain

$$ \bm v(t)=\dot{\bm r}(t)=[1,\ 2t,\ 0]\ \mathrm{m/s} $$

The speed is

$$ v(t)=\abs{\bm v(t)}=\sqrt{1^2+(2t)^2+0^2}=\sqrt{1+4t^2}\ \mathrm{m/s} $$

(b) By Theorem 3.1, $\bm e_t=\bm v/v$, so

$$ \bm e_t(t)=\frac{1}{\sqrt{1+4t^2}}\bigl[1,\ 2t,\ 0\bigr] $$

(Let us confirm that $\abs{\bm e_t}=1$ holds regardless of $t$: $\abs{\bm e_t}^2=\dfrac{1^2+(2t)^2}{1+4t^2}=1$.) (c) Substituting $t=1$, we obtain

$$ \bm v(1)=[1,\ 2,\ 0]\ \mathrm{m/s},\qquad v(1)=\sqrt{1+4}=\sqrt5\ \mathrm{m/s}\approx2.24\ \mathrm{m/s} $$ $$ \bm e_t(1)=\frac{1}{\sqrt5}[1,\ 2,\ 0]=\left[\frac{1}{\sqrt5},\ \frac{2}{\sqrt5},\ 0\right]\approx[0.447,\ 0.894,\ 0] $$

The vector $\bm e_n$ at this point $(1,1)$ cannot yet be found with the tools of this section alone—after we learn the decomposition of the acceleration in Section 3.4, we will return to the same point in Example 3.4 and find $\bm e_n$ as well.

Example 3.4: Unit normal vector and radius of curvature at a point on a parabola

This continues Example 3.3. For the point moving as $\bm r(t)=[t,\ t^2,\ 0]$ (in meters), find the acceleration $\bm a$, the unit normal vector $\bm e_n$, and the radius of curvature $\rho$ at $t=1$ (the point $(1,1)$).

Solution. By Example 3.3 the velocity is $\bm v(t)=[1,2t,0]$, so the acceleration is

$$ \bm a(t)=\dot{\bm v}(t)=[0,\ 2,\ 0]\ \mathrm{m/s^2}\quad(t\text{-independent, constant}) $$

On the other hand, the speed is $v(t)=\sqrt{1+4t^2}$, so the tangential component of the acceleration is

$$ \frac{\dd v}{\dd t}=\frac{4t}{\sqrt{1+4t^2}} \qquad\therefore\quad \left.\frac{\dd v}{\dd t}\right|_{t=1}=\frac{4}{\sqrt5}=\frac{4\sqrt5}{5}\ \mathrm{m/s^2}\approx1.79\ \mathrm{m/s^2} $$

(we differentiated a composite function: $\dfrac{\dd}{\dd t}\sqrt{1+4t^2}=\dfrac{8t}{2\sqrt{1+4t^2}}=\dfrac{4t}{\sqrt{1+4t^2}}$). Since $\bm e_t(1)=\bigl[1/\sqrt5,\ 2/\sqrt5,\ 0\bigr]$ by Example 3.3(c), the tangential component vector is

$$ \frac{\dd v}{\dd t}\bm e_t=\frac{4}{\sqrt5}\left[\frac{1}{\sqrt5},\frac{2}{\sqrt5},0\right]=\left[\frac45,\ \frac85,\ 0\right] $$

By Theorem 3.2, $\bm a=(\dd v/\dd t)\bm e_t+(v^2/\rho)\bm e_n$, so the normal component vector is found as what remains when the tangential component is subtracted from the whole:

$$ \frac{v^2}{\rho}\bm e_n=\bm a-\frac{\dd v}{\dd t}\bm e_t=[0,2,0]-\left[\frac45,\frac85,0\right]=\left[-\frac45,\ \frac25,\ 0\right] $$

Its magnitude is $v^2/\rho$, and its direction is $\bm e_n$. The magnitude is

$$ \frac{v^2}{\rho}=\left|\left[-\frac45,\frac25,0\right]\right|=\sqrt{\left(\frac45\right)^2+\left(\frac25\right)^2}=\sqrt{\frac{16}{25}+\frac4{25}}=\sqrt{\frac{20}{25}}=\frac{2}{\sqrt5}=\frac{2\sqrt5}{5}\ \mathrm{m/s^2} $$

Therefore, the unit normal vector is

$$ \bm e_n(1)=\frac{1}{2\sqrt5/5}\left[-\frac45,\frac25,0\right]=\left[-\frac{2}{\sqrt5},\ \frac{1}{\sqrt5},\ 0\right]\approx[-0.894,\ 0.447,\ 0] $$

(orthogonality can also be confirmed: $\bm e_t\cdot\bm e_n=\frac{1}{\sqrt5}\times\bigl(-\frac2{\sqrt5}\bigr)+\frac2{\sqrt5}\times\frac1{\sqrt5}=-\frac25+\frac25=0$). Using $v(1)^2=(\sqrt5)^2=5$, the radius of curvature is

$$ \rho=\frac{v^2}{2\sqrt5/5}=\frac{5}{2\sqrt5/5}=\frac{5\sqrt5}{2}\ \mathrm{m}\approx5.59\ \mathrm{m} $$

This result also agrees exactly with the value obtained from Formula 3.4, which we derive in Section 3.5 (the radius of curvature of $y=f(x)$, $\rho=(1+f'(x)^2)^{3/2}/\abs{f''(x)}$), by substituting $f'(x)=2x,\ f''(x)=2$ at $x=1$: $\rho=(1+4)^{3/2}/2=5\sqrt5/2\ \mathrm m$. Figure 3.7 shows $\bm e_t,\bm e_n$ at this point together with the osculating circle of radius $\rho$ (the circle that touches the curve and fits it most closely).

Illustration 3.1: Speed and radius of curvature of a point moving along a curve

The position of a particle $\mathrm P$ is given by $\bm r(t)=\left[\dfrac{t^2}{2},\ t,\ 2-t\right]$ (the components are in $\mathrm m$, and $t$ is measured in $\mathrm s$). Find its speed $v$ and radius of curvature $\rho$ at time $t$. Also find $\bm e_t,\bm e_n$ at $t=0$.

Solution. Differentiating each component gives

$$ \bm v(t)=\dot{\bm r}(t)=[t,\ 1,\ -1]\ \mathrm{m/s} \qquad\therefore\quad v=\abs{\bm v(t)}=\sqrt{t^2+1^2+(-1)^2}=\sqrt{t^2+2}\ \mathrm{m/s} $$

The tangential component of the acceleration is

$$ \frac{\dd v}{\dd t}=\frac{\dd}{\dd t}\sqrt{t^2+2}=\frac{2t}{2\sqrt{t^2+2}}=\frac{t}{\sqrt{t^2+2}}\ \mathrm{m/s^2} $$

Therefore, by Theorem 3.2, the acceleration can be decomposed into tangential and normal components and written as

\begin{equation} \bm a=\frac{\dd v}{\dd t}\bm e_t+\frac{v^2}{\rho}\bm e_n=\frac{t}{\sqrt{t^2+2}}\bm e_t+\frac{t^2+2}{\rho}\bm e_n \label{eq:3-ex31-decomp} \end{equation}

On the other hand, differentiating $\bm r(t)$ once more gives the acceleration directly:

\begin{equation} \bm a(t)=\ddot{\bm r}(t)=[1,\ 0,\ 0]\ \mathrm{m/s^2} \qquad\therefore\quad \abs{\bm a}=\sqrt{1^2+0^2+0^2}=1\ \mathrm{m/s^2} \label{eq:3-ex31-a} \end{equation}

Eq. \eqref{eq:3-ex31-decomp} is a decomposition into the mutually orthogonal unit vectors $\bm e_t,\bm e_n$, so, in the same way as with the Pythagorean theorem, $\abs{\bm a}^2$ is equal to the sum of the squares of the two components (since $\bm e_t\cdot\bm e_n=0$, the cross terms vanish on expansion). Using Eq. \eqref{eq:3-ex31-a}, namely $\abs{\bm a}^2=1$, we have

\begin{equation} \abs{\bm a}^2=\left(\frac{t}{\sqrt{t^2+2}}\right)^2+\left(\frac{t^2+2}{\rho}\right)^2=\frac{t^2}{t^2+2}+\frac{(t^2+2)^2}{\rho^2}=1 \label{eq:3-ex31-sq} \end{equation}

Solving Eq. \eqref{eq:3-ex31-sq} for $\rho^2$ gives

$$ \frac{(t^2+2)^2}{\rho^2}=1-\frac{t^2}{t^2+2}=\frac{(t^2+2)-t^2}{t^2+2}=\frac{2}{t^2+2} $$ $$ \therefore\quad \rho^2=\frac{(t^2+2)^2(t^2+2)}{2}=\frac{(t^2+2)^3}{2} $$

Taking the square root of both sides (with $\rho\gt0$) gives

$$ \rho=\frac{(t^2+2)\sqrt{t^2+2}}{\sqrt2}\ \mathrm m $$

as the radius of curvature.

Check using Formula 3.3: For $\bm v=[t,1,-1],\ \bm a=[1,0,0]$, the cross product, by the component formula (University Mathematics, Section 15.4), is

$$ \bm v\times\bm a=[\,1\times0-(-1)\times0,\ \ (-1)\times1-t\times0,\ \ t\times0-1\times1\,]=[0,\ -1,\ -1] $$ $$ \therefore\quad \abs{\bm v\times\bm a}=\sqrt{0^2+(-1)^2+(-1)^2}=\sqrt2 $$

Substituting into Formula 3.3 gives

$$ \rho=\frac{v^3}{\abs{\bm v\times\bm a}}=\frac{(t^2+2)^{3/2}}{\sqrt2}=\frac{(t^2+2)\sqrt{t^2+2}}{\sqrt2} $$

which agrees exactly with the result obtained from the system of Eqs. \eqref{eq:3-ex31-decomp} through \eqref{eq:3-ex31-sq}. Two independent calculations, one that decomposes the acceleration into tangential and normal components and one that uses the cross product, have reached the same answer, so we can be highly confident in both results.

$\bm e_t,\bm e_n$ at $t=0$: We have $\bm v(0)=[0,1,-1]$, $v(0)=\sqrt2$, and $\bm a(0)=[1,0,0]$. Since $\dd v/\dd t=t/\sqrt{t^2+2}$ is 0 at $t=0$, there is no tangential acceleration, and the acceleration is entirely a normal component: $\bm a=(v^2/\rho)\bm e_n$. Hence $\bm e_t(0)=\bm v/v=[0,1,-1]/\sqrt2$, $\bm e_n(0)=\bm a/\abs{\bm a}=[1,0,0]$, and $\rho(0)=v^2/\abs{\bm a}=2/1=2\ \mathrm m$ (this agrees with $2^{3/2}/\sqrt2=2$, obtained by substituting $t=0$ into the general expression above). Indeed, $\bm e_t\cdot\bm e_n=[0,1,-1]\cdot[1,0,0]/\sqrt2=0$, so they are orthogonal.

Interpretation: $\rho(t)=(t^2+2)^{3/2}/\sqrt2$ takes its minimum value $2\ \mathrm m$ at $t=0$ and becomes larger as $\abs t$ becomes larger. The trajectory of this motion is a parabola lying in the plane $y+z=2$, and $t=0$ corresponds to its vertex. This is consistent with the shape of a parabola: it bends most strongly at the vertex and approaches a straight line as we move away from it.

Example 3.5: Radius of curvature of uniform circular motion

The position of a point performing uniform circular motion with (constant) angular velocity $\omega$ on a circle of radius $R$ is given by $\bm r(t)=[R\cos\omega t,\ R\sin\omega t,\ 0]$. Use Formula 3.3 to verify that the radius of curvature of this motion is $\rho=R$ (equal to the radius of the circle itself).

Solution. The velocity and the acceleration are, respectively,

$$ \bm v(t)=[-R\omega\sin\omega t,\ R\omega\cos\omega t,\ 0],\qquad \bm a(t)=[-R\omega^2\cos\omega t,\ -R\omega^2\sin\omega t,\ 0] $$

The speed is $v=\abs{\bm v}=\sqrt{R^2\omega^2\sin^2\omega t+R^2\omega^2\cos^2\omega t}=R\omega$ (constant, independent of $t$—this is what “uniform” means). The $z$ component of the cross product is

$$ v_xa_y-v_ya_x=(-R\omega\sin\omega t)(-R\omega^2\sin\omega t)-(R\omega\cos\omega t)(-R\omega^2\cos\omega t) $$ $$ =R^2\omega^3\sin^2\omega t+R^2\omega^3\cos^2\omega t=R^2\omega^3 $$

so $\abs{\bm v\times\bm a}=R^2\omega^3$. Substituting into Formula 3.3, we obtain

$$ \rho=\frac{v^3}{\abs{\bm v\times\bm a}}=\frac{(R\omega)^3}{R^2\omega^3}=\frac{R^3\omega^3}{R^2\omega^3}=R $$

Thus the fact stated after Definition 3.3, that “the radius of curvature of a circle equals its radius,” has been confirmed using Formula 3.3 as well.

Incidentally, the acceleration of uniform circular motion can also be examined directly by differentiating the position vector twice (Section 5.4). The form it takes when viewed in terms of the polar-coordinate components $(a_r,a_\theta)$ is treated in Example 6.3, as a comparison with the tangential and normal components.

Example 3.6: Radius of curvature of helical motion

Find the radius of curvature $\rho$ of the motion $\bm r(t)=[\cos t,\ \sin t,\ t]$ (the components are in $\mathrm m$, and $t$ is measured in $\mathrm s$; this is helical motion in which the point goes around the unit circle in the $xy$ plane while simultaneously advancing at a constant speed in the $z$ direction).

Solution. The velocity and the acceleration are

$$ \bm v(t)=[-\sin t,\ \cos t,\ 1],\qquad \bm a(t)=[-\cos t,\ -\sin t,\ 0] $$

and the speed is $v=\sqrt{\sin^2t+\cos^2t+1^2}=\sqrt2$ (constant). Computing the cross product component by component,

$$ \bm v\times\bm a=[\,\cos t\times0-1\times(-\sin t),\ \ 1\times(-\cos t)-(-\sin t)\times0,\ \ (-\sin t)(-\sin t)-\cos t\times(-\cos t)\,] $$ $$ =[\sin t,\ -\cos t,\ \sin^2t+\cos^2t]=[\sin t,\ -\cos t,\ 1] $$

so $\abs{\bm v\times\bm a}=\sqrt{\sin^2t+\cos^2t+1^2}=\sqrt2$. By Formula 3.3,

$$ \rho=\frac{v^3}{\abs{\bm v\times\bm a}}=\frac{(\sqrt2)^3}{\sqrt2}=\frac{2\sqrt2}{\sqrt2}=2\ \mathrm m $$

Interestingly, $\rho=2\ \mathrm m$ is constant, independent of the time $t$. Projected onto the $xy$ plane, helical motion is uniform circular motion (of radius 1), but because the point also advances at a constant speed in the $z$ direction, the way it bends in three dimensions is gentler than that of the circle in the plane (radius of curvature 1), and the radius of curvature is doubled (we also check this result numerically in Exercise 3.6).

Exercise 3.1: A general formula for the radius of curvature $\rho(t)$

For the curve $\bm r(t)=[t,\ t^2,\ 0]$ treated in Examples 3.3 and 3.4 (a point moving on the parabola $y=x^2$), find the radius of curvature $\rho(t)$ at an arbitrary time $t$ as a function of $t$. Also verify that the result at $t=1$ agrees with $\rho=5\sqrt5/2$ of Example 3.4.

Hint: Use Formula 3.3, $\rho=v^3/\abs{\bm v\times\bm a}$. We have already found $\bm v(t)=[1,2t,0]$ and $\bm a(t)=[0,2,0]$ (constant, independent of $t$).

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Exercise 3.2: A car rounding a curve

A car travels at a speed of $72\ \mathrm{km/h}$ around a curve in the shape of a circular arc of radius $50\ \mathrm m$ (on a level road). (a) What is the normal acceleration $v^2/\rho$, in $\mathrm{m/s^2}$? (b) Find the minimum value of the coefficient of static friction $\mu$ required between the tires and the road surface so that the car does not skid sideways (the required centripetal force $mv^2/\rho$ must be no greater than the maximum static friction $\mu mg$; $g=9.8\ \mathrm{m/s^2}$). (c) At an instant when the car, while rounding this curve, is decelerating at a constant rate of $2.0\ \mathrm{m/s^2}$, find the magnitude of the acceleration $\abs{\bm a}$.

Hint: First convert $72\ \mathrm{km/h}$ to $\mathrm{m/s}$. In (c), the tangential and normal components are perpendicular to each other, so their magnitudes can be combined by the Pythagorean theorem.

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Exercise 3.3: Derivatives of $\bm e_r,\bm e_\theta$ when $\theta$ varies with time

When the angle $\theta=\theta(t)$ changes with time, differentiate the basis vectors of polar coordinates, $\bm e_r=\cos\theta\,\bm e_x+\sin\theta\,\bm e_y$ and $\bm e_\theta=-\sin\theta\,\bm e_x+\cos\theta\,\bm e_y$, with respect to time to show that

$$ \frac{\dd\bm e_r}{\dd t}=\dot\theta\,\bm e_\theta,\qquad \frac{\dd\bm e_\theta}{\dd t}=-\dot\theta\,\bm e_r $$

Further, use this result to derive that the velocity of the uniform circular motion ($\dot\theta=\omega$) described by $\bm r=R\,\bm e_r$ ($R$ constant) is $\bm v=R\omega\,\bm e_\theta$.

Hint: Since $\bm e_x,\bm e_y$ are constant vectors, only $\cos\theta,\sin\theta$ are differentiated with respect to time. Use the rule for differentiating a composite function: $\dfrac{\dd}{\dd t}\cos\theta=-\dot\theta\sin\theta$. We will use this result again in Chapter 6 when we derive the velocity and acceleration in polar coordinates.

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Exercise 3.4: A general proof of the orthogonality of the polar basis vectors

Show, using components, that for any angle $\theta$ the basis vectors $\bm e_r,\bm e_\theta$ of polar coordinates (Formula 3.2) are always orthogonal, that is, $\bm e_r\cdot\bm e_\theta=0$.

Hint: Compute the inner product using the component expressions $\bm e_r=[\cos\theta,\sin\theta],\ \bm e_\theta=[-\sin\theta,\cos\theta]$ of Formula 3.2, and simplify with the fundamental trigonometric identity.

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Exercise 3.5: The radius of curvature of $y=\sin x$

Find the radius of curvature $\rho(x)$ at an arbitrary point of the curve $y=\sin x$ using Formula 3.4, $\rho=(1+f'(x)^2)^{3/2}/\abs{f''(x)}$ (with $f(x)=\sin x$). Also find the value of $\rho$ at $x=\pi/2$ (the top of a hump of the curve).

Hint: We have $f'(x)=\cos x,\ f''(x)=-\sin x$. At $x=\pi/2$, substitute $\cos(\pi/2)=0,\ \sin(\pi/2)=1$.

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Exercise 3.6: Velocity and acceleration at $t=0$ in helical motion

For the helical motion $\bm r(t)=[\cos t,\ \sin t,\ t]$ of Example 3.6, find the velocity $\bm v(0)$ and the acceleration $\bm a(0)$ at $t=0$ in components. Further, compute $\rho$ at this time using Formula 3.3, and confirm that it does not contradict the result found in Example 3.6, “$\rho=2$ (constant, independent of time).”

Hint: Simply substitute $t=0$ ($\sin0=0,\cos0=1$) into the expressions for $\bm v(t),\bm a(t)$ in Example 3.6.

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