EXAMPLE 1

Calculate and plot the velocity and acceleration vectors at \(t = 1\) of \({\bf r}(t) = \langle \sin 2t, -\cos 2t,\sqrt{t+1} \rangle\).

Solution  \[ \begin{align*}{2} {\bf v}(t) = {\bf r}'(t) &= \langle 2\cos 2t, 2\sin 2t, \frac12(t+1)^{-1/2} \rangle, &\qquad{\bf v}(1)&\approx\langle -0.83,0.84,0.35\rangle\\ \bolda(t) = {\bf r}”(t) & = \langle -4\sin 2t, 4\cos 2t, -\frac14(t+1)^{-3/2} \rangle, &\qquad\bolda(1)&\approx\langle -3.64,0.54,-0.089 \rangle \end{align*} \]

The speed at \(t=1\) is \[ \|{{\bf v}(1)}\|\approx\sqrt{(-0.83)^2+(0.84)^2+(0.35)^2} \approx 1.23 \]

EXAMPLE 2

Find \({\bf r}(t)\) if \[ \bolda(t) =2{\bf i} +12t{\bf j},\qquad {\bf v}(0) = 7{\bf i}, \qquad {\bf r}(0)=2{\bf i}+9{\bf k} \]

Solution We have \[ \begin{align*} {\bf v}(t) &= \int \bolda(t)\,dt +{\bf v}_0= 2t{\bf i} +6t^2{\bf j} + {\bf v}_0 \intertext{The initial condition \({\bf v}(0) = {\bf v}_0 = 7{\bf i}\) gives us \({\bf v}(t)= 2t{\bf i} +6t^2 {\bf j} +7{\bf i}\). Then we have } {\bf r}(t) &= \int {\bf v}(t)\,dt +{\bf r}_0 = t^2{\bf i} +2t^3{\bf j} +7t{\bf i}+{\bf r}_0 \end{align*} \]

The initial condition \({\bf r}(0)={\bf r}_0=2{\bf i}+9{\bf k}\) yields \[ {\bf r}(t) = t^2{\bf i} +2t^3{\bf j} + 7t{\bf i}+(2{\bf i}+9{\bf k}) = (t^2+7t+2){\bf i}+2t^3{\bf j}+9{\bf k} \]

EXAMPLE 3

A bullet is fired from the ground at an angle of \(60^{\circ}\) above the horizontal. What initial speed \(v_0\) must the bullet have in order to hit a point 150 m high on a tower located 250 m away (ignoring air resistance)?

Solution Place the gun at the origin, and let \({\bf r}(t)\) be the position vector of the bullet (Figure 13.36).

Figure 13.36: Trajectory of the bullet.

Step 1. Use Newton’s Law.

Gravity exerts a downward force of magnitude \(mg\), where \(m\) is the mass of the bullet and \(g=9.8 \text{m/s}^2\). In vector form, \[ {\bf F} = \langle 0,-mg\rangle = m\langle 0, -g\rangle \]

Newton’s Second Law \({\bf F} = m{\bf r}”(t)\) yields \(m\langle 0, -g\rangle=m{\bf r}”(t)\) or \({\bf r}”(t)=\langle 0,- g \rangle\). We determine \({\bf r}(t)\) by integrating twice: \[ \begin{align*} {\bf r}'(t) &= \int_0^t {\bf r}”(u)\, du = \int_0^t \langle 0,-g\rangle\, du = \langle 0,- gt \rangle +{\bf v}_0\\ {\bf r}(t) &= \int_0^t {\bf r}'(u)\, du = \int_0^t \big(\langle 0,- gu \rangle +{\bf v}_0\big)\, du = \left\langle 0,- \frac12 gt^2 \right\rangle +t{\bf v}_0+{\bf r}_0 \end{align*} \]

Step 2. Use the initial conditions.

By our choice of coordinates, \({\bf r}_0 = \mathbf{0}\). The initial velocity \({\bf v}_0\) has unknown magnitude \(v_0\), but we know that it points in the direction of the unit vector \(\langle \cos 60^{\circ}, \sin 60^{\circ}\rangle\). Therefore, \[ \begin{align*} {\bf v}_0 &= v_0\langle \cos 60^{\circ}, \sin 60^{\circ}\rangle = v_0\langle \frac12, \frac{\sqrt{3}}2\rangle \\ {\bf r}(t) &= \left\langle 0,-\frac12gt^2\right\rangle +tv_0\langle \frac12, \frac{\sqrt{3}}2\rangle \]

Step 3. Solve for \(v_0\).

The bullet hits the point \(\langle 250, 150\rangle\) on the tower if there exists a time \(t\) such that \displaystyle{{\bf r}(t) = \langle 250, 150\rangle}; that is, \[ \left\langle 0,-\frac12gt^2\right\rangle +tv_0\langle \frac12, \frac{\sqrt{3}}2\rangle =\langle 250, 150\rangle \]

Equating components, we obtain \[ \frac12tv_0=250,\qquad - \frac12gt^2+ \frac{\sqrt{3}}2tv_0= 150 \]

The first equation yields \(\displaystyle{t = {500}/{v_0}}\). Now substitute in the second equation and solve, using \(g=9.8\): \[ \begin{align*} - 4.9\left(\frac{500}{v_0}\right)^2 &+\frac{\sqrt 3} 2\left(\frac{500}{v_0}\right)v_0 = 150 \\ \left(\frac{500}{v_0}\right)^2 &= \frac{250\sqrt 3 -150}{4.9}\\ \left(\frac{v_0}{500}\right)^2 &= \frac{4.9}{250\sqrt 3 -150} \approx 0.0173 \end{align*} \]

We obtain \displaystyle{v_0 \approx 500\sqrt{0.0173} \approx 66 \textrm{m/s}}.

EXAMPLE 4 Uniform Circular Motion

Find \(\displaystyle{\bolda(t)}\) and \(\displaystyle{\norm{\bolda(t)}}\) for motion around a circle of radius \(R\) with constant speed \(v\).

Solution Assume that the particle follows the circular path \(\displaystyle{{\bf r}(t) = R\langle \cos \omega t, \sin \omega t\rangle}\) for some constant \(\omega\). Then the velocity and speed of the particle are \[ {\bf v}(t) = R\omega\langle -\sin \omega t, \cos \omega t\rangle,\qquad v = \norm{{\bf v}(t)} = R|\omega| \]

Thus \(|\omega| = v/R\), and accordingly, \[ \bolda(t)= {\bf v}'(t) = -R\omega^2\langle \cos \omega t,\sin \omega t\rangle,\qquad \norm{\bolda(t)} = R\omega^2 = R \left(\frac{v}R\right)^2 = \frac{v^2}R \]

The vector \(\bolda(t)\) is called the centripetal acceleration: It has length \(\displaystyle{ {v^2}/R}\) and points in toward the origin [because \(\bolda(t)\) is a negative multiple of the position vector \({\bf r}(t)\)], as in Figure 13.37.

Note

The constant \(\omega\) (lowercase Greek “omega”) is called the {\upshape angular speed} because the particle’s angle along the circle changes at a rate of \(\omega\) radians per unit time.

Note

When you make a left turn in an automobile at constant speed, your tangential acceleration is zero (because \(v'(t)=0\)) and you will not be pushed back against your seat. But the car seat (via friction) pushes you to the left toward the car door, causing you to accelerate in the normal direction. Due to inertia, you feel as if you are being pushed to the right toward the passenger’s seat. This force is proportional to \(\kappa v^2\), so a sharp turn (large \(\kappa\)) or high speed (large \(v\)) produces a strong normal force.

Note

The normal component \(a_\NN\) is often called the centripetal acceleration, especially in the case of circular motion where it is directed toward the center of the circle.

CONCEPTUAL INSIGHT

The tangential component \(a_\TT = v'(t)\) is the rate at which speed \(\displaystyle{v(t)}\) changes, whereas the normal component \(a_\NN=\kappa(t)v(t)^2\) describes the change in \({\bf v}\) due to a change in direction. These interpretations become clear once we consider the following extreme cases:

EXAMPLE 5

The Giant Ferris Wheel in Vienna has radius \(R=30\) m (Figure 13.39). Assume that at time \(t=t_0\), the wheel rotates counterclockwise with a speed of \(40 \text{m/min}\) and is slowing at a rate of \(15 \text{m/min}^2\). Find the acceleration vector \(\bolda\) for a person seated in a car at the lowest point of the wheel.

Solution At the bottom of the wheel, \(\TT = \langle 1,0 \rangle\) and \(\NN= \langle 0,1\rangle\). We are told that \(a_{\TT} = v' = -15\) at time \(t_0\). The curvature of the wheel is \(\kappa = 1/R=1/30\), so the normal component is \(a_{\NN} = \kappa v^2 = v^2/R= (40)^2/30 \approx 53.3 \(. Therefore (Figure 13.40), \[ \bolda \approx -15\TT+ 53.3\NN = \langle -15, 53.3 \rangle \text{m/min}^2 \]

Figure 13.39: The Giant Ferris Wheel in Vienna, Austria, erected in 1897 to celebrate the 50th anniversary of the coronation of Emperor Franz Joseph I.

The following theorem provides useful formulas for the tangential and normal components.

Keep in mind that \(a_\NN\ge 0\) but \(a_\TT\) is positive or negative, depending on whether the object is speeding up or slowing down.

EXAMPLE 6

Decompose the acceleration vector \(\bolda\) of \({\bf r}(t)=\langle t^2,2t, \ln t\rangle\) into tangential and normal components at \(t = \frac12\) (Figure 13.41).

Figure 13.41: The vectors \(\TT\), \(\NN\), and \(\bolda\) at \(t = \frac12\) on the curve \({\bf r}(t)=\langle t^2,2t, \ln t\rangle\).

Solution First, we compute the tangential components \(\TT\) and \(a_\TT\). We have \[ {\bf v}(t)={\bf r}'(t) = \langle 2t, 2, t^{-1} \rangle,\qquad \bolda(t) = {\bf r}”(t) = \langle 2, 0, -t^{-2}\rangle \]

At \(t = \frac12\), \[ \begin{align*} {\bf v} &= {\bf r}'\left(\frac12\right)= \langle 2\left(\frac12\right), 2, \left(\frac12\right)^{-1} \rangle = \langle 1, 2, 2 \rangle\\ \bolda &= {\bf r}”\left(\frac12\right) = \langle 2, 0, -\left(\frac12\right)^{-2} \rangle= \langle 2, 0, -4\rangle \end{align*} \]

Thus \[ \TT = \frac{{\bf v}}{\norm{{\bf v}}} = \frac{\langle 1, 2, 2 \rangle}{\sqrt{1^2+2^2+2^2}} = \langle \frac13,\frac23,\frac23\rangle \] and by Eq. (3), \[ a_\TT = \bolda \cdot \TT = \langle2,0,-4\rangle \cdot \langle \frac{1}{3},\frac{2}{3},\frac{2}{3} \rangle = -2 \]

Next, we use Eq. (4): \[ a_\NN\NN= \bolda-a_\TT\TT = \langle 2, 0,-4\rangle - (-2) \langle \frac13,\frac23,\frac23 \rangle = \langle \frac83,\frac43,-\frac83 \rangle \]

This vector has length \[ a_\NN = \norm{a_\NN\NN} = \sqrt{\frac{64}9+ \frac{16}9 + \frac{64}9} = 4 \] and thus \[ \NN = \frac{a_\NN\NN}{a_\NN} = \frac{\langle \frac83,\frac43,-\frac83 \rangle}4 = \langle \frac23,\frac13,-\frac23 \rangle \]

Finally, we obtain the decomposition \[ \bolda = \langle 2, 0, -4\rangle = a_\TT \TT + a_\NN \NN = -2\TT + 4\NN \]

Note

Summary of steps in Example 6: \[ \begin{align*} \TT &= \frac{{\bf v}}{\norm{{\bf v}}}\\ a_\TT &= \bolda\cdot\TT\\ a_\NN\,\NN &= \bolda -a_\TT\TT\\ a_\NN &= \norm{a_\NN\,\NN}\\ \NN &= \frac{a_\NN\,\NN}{a_\NN} \end{align*} \]

EXAMPLE 7 Nonuniform Circular Motion

Figure # shows the acceleration vectors of three particles moving counterclockwise around a circle. In each case, state whether the particle’s speed \(v\) is increasing, decreasing, or momentarily constant.

Solution The rate of change of speed depends on the angle \(\theta\) between \(\bolda\) and \(\TT\): \[ v'= a_\TT = \bolda\cdot \TT = \norm{\bolda}\,\norm{\TT}\cos\theta = \norm{\bolda} \cos\theta \]

• In (A), \(\theta\) is obtuse so \(\cos\theta<0\) and \(v'<0\). The particle’s speed is decreasing.
• In (B), \(\theta=\frac{\pi}2\) so \(\cos \theta =0\) and \(v'=0\). The particle’s speed is momentarily constant.
• In (C), \(\theta\) is acute so \(\cos\theta>0\) and \(v'>0\). The particle’s speed is increasing.

Figure 13.42: Acceleration vectors of particles moving counterclockwise (in the direction of \(\TT\)) around a circle.

EXAMPLE 8

Find the curvature \(\kappa\big(\frac12\big)\) for the path \({\bf r}(t)=\langle t^2,2t, \ln t\rangle\) in Example 6.

Solution By Eq. (2), the normal component is \[ a_\NN = \kappa v^2 \]

In Example 6 we showed that \(a_\NN = 4\) and \({\bf v}=\langle1,2,2\rangle\) at \(t=\frac12\). Therefore, \displaystyle{v^2 = {\bf v}\cdot{\bf v} = 9} and the curvature is \(\kappa\big(\frac12\big) = a_\NN/v^2 = \frac49\).