A car on the ramp of a multistory parking garage travels along a curve traced out by \(\mathbf{r}( t)\;=\;10\cos t\mathbf{i}+10\sin t\mathbf{j}+3t\mathbf{k,}\) where \(t\) is the time in hours and distance is in miles. Find the tangential component \(a_{\mathbf{T}}\) and normal component \(a_{\mathbf{N}}\) of the acceleration of the car. What are the magnitude and direction of the force on the driver?

Solution We begin by finding the velocity \(\mathbf{v}\), speed \(v \), and acceleration \(\mathbf{a}\) of the car. \begin{eqnarray*} \mathbf{v}(t)&\;=\;&\mathbf{r}^{\prime} (t)=\dfrac{d}{dt}( 10\cos t) \mathbf{i}+\dfrac{d}{dt}( 10\sin t) \mathbf{j}+\dfrac{d}{dt} ( 3t) \mathbf{k}=-10\sin t\mathbf{i}+10\cos t\mathbf{j+}3 \mathbf{k} \\[6pt] v(t)&\;=\;&\left\Vert \mathbf{v}(t)\right\Vert\;=\;\left\Vert \mathbf{r}^{\prime} (t)\right\Vert\;=\;\sqrt{(-10\sin t)^{2}+(10\cos t)^{2}+9}=\sqrt{109}\approx 10.440 \enspace \textrm{mi/h}\\[6pt] \mathbf{a}(t)&\;=\;&\mathbf{r^{\prime\prime}} (t)=\dfrac{d}{dt}\mathbf{v}(t)= -10\cos t\mathbf{i}-10\sin t\mathbf{j } \end{eqnarray*}

Since the speed is constant, the tangential component of acceleration \(a_{ \mathbf{T}}\) is \begin{eqnarray*} a_{\mathbf{T}}=\dfrac{dv}{dt}=0 \end{eqnarray*}

The normal component of acceleration \(a_{\mathbf{N}}\) is \begin{equation} a_{\mathbf{N}}=v^{2}\kappa\;=\;v^{2} {{{\hbox{$$}} }{}}\dfrac{\left\Vert \mathbf{r}^{\prime} ( t) \times \mathbf{r}^{\prime \prime} (t) \right\Vert }{\left\Vert \mathbf{r}^{\prime} (t) \right\Vert ^{3}} \underset{\underset{\color{#0066A7}{v=\;\left\Vert \mathbf{r}^{\prime} (t) \right\Vert\;=\;\sqrt{109}}}{\uparrow}} {=}\dfrac{\left\Vert \mathbf{r}^{\prime} ( t) \times \mathbf{r}^{\prime \prime} (t) \right\Vert }{\sqrt{109}} \\ \notag \end{equation}

Since, \begin{equation*} \mathbf{r}^{\prime} (t)\times \mathbf{r}^{\prime \prime} (t)= \left|\begin{array}{c@{\quad}c@{\quad}c} \mathbf{i} & \mathbf{j} & \mathbf{k} \\[3pt] -10\sin t & 10\cos t & 3 \\[3pt] -10\cos t & -10\sin t & 0 \end{array}\right| =30\sin t\mathbf{i}-30\cos t\mathbf{j}+100\mathbf{k} \end{equation*}

the normal component is \begin{eqnarray*} a_{\mathbf{N}}&=&\frac{\left\Vert 30\sin t\mathbf{i}-30\cos t\mathbf{j}+100 \mathbf{k}\right\Vert }{\sqrt{109}}=\dfrac{\sqrt{900\sin ^{2}t+900\cos ^{2}t+10000} }{\sqrt{109}}\\[4pt] &=&\dfrac{\sqrt{10900}}{\sqrt{109}}=10 \end{eqnarray*}

As the car travels on the ramp at a constant speed, the force \[ \mathbf{F}=m\mathbf{a}\;=\;ma_{\mathbf{N}} \mathbf{N}\;=\;10 m \mathbf{N} \]

pulls the car and driver toward the center of the ramp, with a magnitude about \(10\) times the mass of the car.