Chapter 11.4, Problem 11.96P

The three-dimensional motion of a particle is defined by the position vector, $r=(At\cos t)i+(A\sqrt{t^2+1})j+(Bt\sin t)k$, where r and t are expressed in feet and seconds, respectively. Show that the curve described by the particle lies on the hyperboloid (y/A)²?−?(x/A)²?−?(z/B)²?=?1. For A?=?3 and B?=?1, determine (a) the magnitudes of the velocity and acceleration when t = 0, (b) the smallest nonzero value of t for which the position vector and the velocity are perpendicular to each other.

Fig. P11.96

To determine

The magnitude of the velocity (v) and acceleration (a) when time is 0 sec.

Answer to Problem 11.96P

The magnitude of the velocity (v) and acceleration (a) when time is 0 sec are $\underset{\_}{3\text{ft/s}}$ and $\underset{\_}{3.61{\text{ft/s}}^{\text{2}}}$ respectively.

Explanation of Solution

Given Information:

The three dimensional motion of a particle is defined by the position vector is $r=\left(At\cos t\right)i+\left(A\sqrt{t^2+1}\right)j+\left(Bt\sin t\right)k$.

The curve described by the particle lies on the hyperboloid is ${\left(\frac{y}{A}\right)}^2-{\left(\frac{x}{A}\right)}^2-{\left(\frac{z}{B}\right)}^2=1$.

The value of A and B are 3 and 1 respectively.

Calculation:

Write the three dimensional motion of a particle position vector equation.

$r=\left(At\cos t\right)i+\left(A\sqrt{t^2+1}\right)j+\left(Bt\sin t\right)k$ (1)

Here, x is $\left(At\cos t\right)i$, y is $\left(A\sqrt{t^2+1}\right)j$ and z is $\left(Bt\sin t\right)k$.

Consider x:

$\begin{array}{l}x=\left(At\cos t\right)\\ \cos t=\frac{x}{At}\end{array}$ (2)

Consider y:

$\begin{array}{l}y=\left(A\sqrt{t^2+1}\right)\\ \sqrt{t^2+1}=\frac{y}{A}\\ t^2={\left(\frac{y}{A}\right)}^2-1\end{array}$ (3)

Consider z:

$\begin{array}{l}z=\left(Bt\sin t\right)\\ \sin t=\frac{z}{Bt}\end{array}$ (4)

Calculate the $t^2$ using the relation:

${\cos }^{2}t+{\sin }^{2}t=1$

Substitute $\frac{x}{At}$ for $\cos t$ and $\frac{z}{Bt}$ for $\sin t$.

$\begin{array}{l}{\left(\frac{x}{At}\right)}^2+{\left(\frac{z}{Bt}\right)}^2=1\\ t^2={\left(\frac{x}{A}\right)}^2+{\left(\frac{z}{B}\right)}^2\end{array}$

Check whether the position vector equation satisfied the curve equation or not.

Substitute ${\left(\frac{x}{A}\right)}^2+{\left(\frac{z}{B}\right)}^2$ for $t^2$ in equation (3).

$\begin{array}{l}{\left(\frac{x}{A}\right)}^2+{\left(\frac{z}{B}\right)}^2={\left(\frac{y}{A}\right)}^2-1\\ {\left(\frac{x}{A}\right)}^2+{\left(\frac{z}{B}\right)}^2-{\left(\frac{y}{A}\right)}^2=-1\\ -{\left(\frac{y}{A}\right)}^2+{\left(\frac{x}{A}\right)}^2+{\left(\frac{z}{B}\right)}^2=-1\\ {\left(\frac{y}{A}\right)}^2-{\left(\frac{x}{A}\right)}^2-{\left(\frac{z}{B}\right)}^2=1\end{array}$

Hence, the equation is satisfied.

Rewrite the Equation (1).

Substitute 3 for A and 1 for B in Equation (1).

$r=\left(3t\cos t\right)i+\left(3\sqrt{t^2+1}\right)j+\left(1t\sin t\right)k$

Write the expression for velocity using the relation:

$v=\frac{dr}{dt}$

Substitute $\left(At\cos t\right)i+\left(A\sqrt{t^2+1}\right)j+\left(Bt\sin t\right)k$ for $r$.

$\begin{array}{c}v=\frac{d\left(3t\cos t\right)i+\left(3\sqrt{t^2+1}\right)j+\left(1t\sin t\right)k}{dt}\\ =3\left(\cos t-t\sin t\right)i+3\frac{t}{\sqrt{t^2+1}}j+\left(\sin t+t\cos t\right)k\end{array}$ (5)

Calculate velocity vector $\left(v\right)$ when time is 0 sec:

Substitute 0 for t in Equation (5).

$\begin{array}{c}v=3\left(\cos \left(0\right)-\left(0\right)\sin \left(0\right)\right)i+3\frac{\left(0\right)}{\sqrt{{\left(0\right)}^2+1}}j+\left(\sin \left(0\right)+\left(0\right)\cos \left(0\right)\right)k\\ =3\left(1-0\right)i+0j+\left(0\right)k\\ =3i+0j+0k\end{array}$

Here, $v_x$ is $3$, $v_y$ is $0$ and $v_z$ is $0$.

Calculate the magnitude of velocity (v) using the relation:

$v=\sqrt{v_x^2+v_y^2+v_z^2}$

Substitute $3$ for $v_x$, $0$ for $v_y$ and $0$ for $v_z$.

$\begin{array}{c}v=\sqrt{3^2+0^2+0^2}\\ =3\text{ft/s}\end{array}$

Write the expression for acceleration vector using the relation:

$a=\frac{dv}{dt}$

Substitute $3\left(\cos t-t\sin t\right)i+3\frac{t}{\sqrt{t^2+1}}j+\left(\sin t+t\cos t\right)k$ for $v$.

$\begin{array}{c}a=\frac{d\left(3\left(\cos t-t\sin t\right)i+3\frac{t}{\sqrt{t^2+1}}j+\left(\sin t+t\cos t\right)k\right)}{dt}\\ =3\left(-\sin t-\sin t-t\cos t\right)i+3\frac{\sqrt{t^2+1}-t\left(\frac{t}{\sqrt{t^2+1}}\right)}{\left(t^2+1\right)}j+\left(\cos t+\cos t-t\sin t\right)k\\ =-3\left(2\sin t+t\cos t\right)i+3\frac{1}{{\left(t^2+1\right)}^{\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}}j+\left(2\cos t-t\sin t\right)k\end{array}$

Substitute 0 sec for t.

$\begin{array}{c}a=-3\left(2\sin \left(0\right)+\left(0\right)\cos \left(0\right)\right)i+3\frac{1}{{\left({\left(0\right)}^2+1\right)}^{\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}}j+\left(2\cos \left(0\right)-\left(0\right)\sin \left(0\right)\right)k\\ =-3\left(0\right)i+\left(3\right)j+\left(2-0\right)k\\ =0i+3j+2k\end{array}$

Here, $a_x$ is 0, $a_y$ is 3 and $a_z$ is 1.

Calculate the magnitude $\left|a\right|$ of acceleration at time 0 sec.

$\left|a\right|=\sqrt{a_x^2+a_y^2+a_z^2}$

Substitute 0 is $a_x$, 3 for $a_y$ and 2 for $a_z$.

$\begin{array}{c}\left|a\right|=\sqrt{0^2+3^2+2^2}\\ =\sqrt{0+9+4}\\ =3.61{\text{ft/s}}^{\text{2}}\end{array}$

Therefore, the magnitude of the velocity (v) and acceleration (a) when time is 0 sec are $\underset{\_}{3\text{ft/s}}$ and $\underset{\_}{3.61{\text{ft/s}}^{\text{2}}}$ respectively.

To determine

The smallest nonzero value of t for which the position vector and the velocity are perpendicular to each other.

Answer to Problem 11.96P

The smallest nonzero value of t for which the position vector and the velocity are perpendicular to each other is $\underset{\_}{3.82\sec }$.

Explanation of Solution

Given Information:

The three dimensional motion of a particle is defined by the position vector is $r=\left(At\cos t\right)i+\left(A\sqrt{t^2+1}\right)j+\left(Bt\sin t\right)k$.

The curve described by the particle lies on the hyperboloid is ${\left(\frac{y}{A}\right)}^2-{\left(\frac{x}{A}\right)}^2-{\left(\frac{z}{B}\right)}^2=1$.

The value of A and B are 3 and 1 respectively.

Calculation:

Write the equation if the position vector and velocity vector are perpendicular:

$r\cdot v=0$

Substitute $3\left(\cos t-t\sin t\right)i+3\frac{t}{\sqrt{t^2+1}}j+\left(\sin t+t\cos t\right)k$ for $v$ and $\left(3t\cos t\right)i+\left(3\sqrt{t^2+1}\right)j+\left(1t\sin t\right)k$ for $r$.

$\begin{array}{l}\left\{\begin{array}{l}\left[\left(3t\cos t\right)i+\left(3\sqrt{t^2+1}\right)j+\left(t\sin t\right)k\right]\\ \times \left[3\left(\cos t-t\sin t\right)i+3\frac{t}{\sqrt{t^2+1}}j+\left(\sin t+t\cos t\right)k\right]\end{array}\right\}=0\\ \left\{\begin{array}{l}\left[\left(3t\cos t\right)\left(3\left(\cos t-t\sin t\right)\right)\right]i\\ +\left[\left(\left(3\sqrt{t^2+1}\right)\left(3\frac{t}{\sqrt{t^2+1}}\right)\right)\right]j\\ +\left[\left(t\sin t\right)\left(\sin t+t\cos t\right)\right]k\end{array}\right\}=0\\ \left(9t{\cos }^{2}t-9t^2\sin t\cos t\right)+\left(9t\right)+\left(t{\sin }^{2}t+t^2\sin t\cos t\right)=0\end{array}$

$\begin{array}{l}9t{\cos }^{2}t-9t^2\sin t\cos t+9t+t{\sin }^{2}t+t^2\sin t\cos t=0\\ 9t{\cos }^{2}t+9t+t\left(1-{\cos }^{2}t\right)-8t^2\sin t\cos t=0\\ 10t+8{\cos }^{2}t-8t^2\sin t\cos t=0\end{array}$

$\begin{array}{l}t\left(10+8{\cos }^{2}t-8t\sin t\cos t\right)=0\\ 10+8{\cos }^{2}t-8t\sin t\cos t=0\left(\because {\cos }^{2}t=\frac{1+\cos \left(2t\right)}{2},\sin \left(2t\right)=2\sin t\cos t\right)\\ 10+8\left(\frac{1+\cos \left(2t\right)}{2}\right)-4t\sin 2t=0\end{array}$$\begin{array}{l}10+4+4\cos 2t-4t\sin 2t=0\\ 14+4\cos 2t-4t\sin 2t=0\\ 2\left(7+2\cos 2t-2t\sin 2t\right)=0\\ 7+2\cos 2t-2t\sin 2t=0\end{array}$

Using trial and error method the smallest root is (t) is 4.38 sec.

Therefore, the smallest nonzero value of t for which the position vector and the velocity are perpendicular to each other is $\underset{\_}{3.82\sec }$.