Chapter 11.5, Problem 11.181P

To determine

(a)

The direction of the binomial pathat $t=0$.

Answer to Problem 11.181P

${\theta }_x=90°,{\theta }_y=123.7°,{\theta }_z=33.7°$

Explanation of Solution

Given information:

According to problem $11.96,$

$r=(Atcost)i+(A \sqrt{t^2+1})j+(Btsint)k$

$A=3,B=1$

The first derivative of motion is equal to velocity

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

The first derivative of velocity is equal to acceleration

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

Calculation:

According to the given information,

$r=(Atcost)i+(A \sqrt{t^2+1})j+(Btsint)k$

Substitute,

$r=(3tcost)i+(3\sqrt{t^2+1})j+(tsint)k$

Differentiate to find the velocity,

$v=\frac{dr}{dt}=3(cost-tsint)i+3\left(\frac{1}{2}{\left(t^2+1\right)}^{-\frac{1}{2}}\left(2t\right)\right)j+(sint+tcost)k$

$v=\frac{dr}{dt}=3(cost-tsint)i+\frac{3t}{\sqrt{t^2+1}}j+(sint+tcost)k$

Differentiate to find the acceleration,

$a=\frac{dv}{dt}=3(-sint-sint-tcost)i+\frac{3\sqrt{t^2+1}-3t\left(\frac{1}{2}{\left(t^2+1\right)}^{-\frac{1}{2}}\left(2t\right)\right)}{t^2+1}j+(cost+cost-tsint)k$

$a=\frac{dv}{dt}=3(-sint-sint-tcost)i+3\left[\frac{\sqrt{t^2+1}-t\left(\frac{t}{\sqrt{t^2+1}}\right)}{t^2+1}\right]j+(cost+cost-tsint)k$

We know that,

$e_b=\frac{v\times a}{\left|v\times a\right|}$

At $t=0$

$\begin{array}{l}v=\left(3ft/s\right)i\\ a=\left(3ft/s\right)j+\left(2ft/s\right)k\end{array}$

Then,

$\begin{array}{l}v\times a=3i\left[3j+2k\right]\\ v\times a=3\left(-2j+3k\right)\end{array}$

Find,

$\left|v\times a\right|=3\sqrt{{\left(-2\right)}^2+{\left(3\right)}^2}=3\sqrt{13}$

Therefore,

$e_b=\frac{v\times a}{\left|v\times a\right|}=\frac{3\left(-2j+3k\right)}{3\sqrt{13}}=\frac{1}{\sqrt{13}}\left(-2j+3k\right)$

According to the above result,

$\begin{array}{l}\cos {\theta }_x=0\\ \cos {\theta }_y=-\frac{2}{\sqrt{13}}\\ \cos {\theta }_z=\frac{3}{\sqrt{13}}\end{array}$

Therefore,

${\theta }_x=90°,{\theta }_y=123.7°,{\theta }_z=33.7°$

Conclusion:

The direction of binomial path at $t=0$ ,

${\theta }_x=90°,{\theta }_y=123.7°,{\theta }_z=33.7°$

To determine

(b)

The direction of the binomial path at $t=\frac{\pi }{2}$.

Answer to Problem 11.181P

${\theta }_x=103.4°,{\theta }_y=134.4°,{\theta }_z=47.4°$

Explanation of Solution

Given information:

According to problem $11.96,$

$r=(Atcost)i+(A \sqrt{t^2+1})j+(Btsint)k$

$A=3,B=1$

The first derivative of motion is equal to velocity

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

The first derivative of velocity is equal to acceleration

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

Calculation:

According to the given information,

$r=(Atcost)i+(A \sqrt{t^2+1})j+(Btsint)k$

Substitute,

$r=(3tcost)i+(3\sqrt{t^2+1})j+(tsint)k$

Differentiate to find the velocity,

$v=\frac{dr}{dt}=3(cost-tsint)i+3\left(\frac{1}{2}{\left(t^2+1\right)}^{-\frac{1}{2}}\left(2t\right)\right)j+(sint+tcost)k$

$v=\frac{dr}{dt}=3(cost-tsint)i+\frac{3t}{\sqrt{t^2+1}}j+(sint+tcost)k$

Differentiate to find acceleration,

$a=\frac{dv}{dt}=3(-sint-sint-tcost)i+\frac{3\sqrt{t^2+1}-3t\left(\frac{1}{2}{\left(t^2+1\right)}^{-\frac{1}{2}}\left(2t\right)\right)}{t^2+1}j+(cost+cost-tsint)k$

$a=\frac{dv}{dt}=3(-sint-sint-tcost)i+3\frac{\sqrt{t^2+1}-t\left(\frac{t}{\sqrt{t^2+1}}\right)}{t^2+1}j+(cost+cost-tsint)k$

We know that,

$e_b=\frac{v\times a}{\left|v\times a\right|}$

At $t=\frac{\pi }{2}$

$\begin{array}{l}v=-\left(3\frac{\pi }{2}ft/s\right)i+\left(\frac{3\pi }{\sqrt{{\pi }^2+4}}ft/s\right)j+\left(1ft/s\right)k\\ a=\frac{dv}{dt}=-\left(6ft/s^2\right)i+\left(\frac{24}{{\left({\pi }^2+4\right)}^{3/2}}ft/s^2\right)j-\left(\frac{\pi }{2}ft/s^2\right)k\end{array}$

Therefore,

$\begin{array}{l}v\times a=-\left[\frac{3{\pi }^2}{2\sqrt{{\pi }^2+4}}+\frac{24}{{\left({\pi }^2+4\right)}^{3/2}}\right]i-\left[6+\frac{3{\pi }^4}{4}\right]j+\left[-\frac{36\pi }{{\left({\pi }^2+4\right)}^{3/2}}+\frac{18\pi }{{\left({\pi }^2+4\right)}^{1/2}}\right]k\\ =-4.4398i-13.4022j+12.9946k\end{array}$

Then,

$\begin{array}{l}\left|v\times a\right|={[}^{{(}^{-}}\\ =19.1883\end{array}$

Now substitute to find,

$e_b=\frac{v\times a}{\left|v\times a\right|}$

$e_b=\frac{v\times a}{\left|v\times a\right|}=\frac{-4.4398i-13.4022j+12.9946k}{19.1883}$

Then,

$\begin{array}{l}\cos {\theta }_x=\frac{-4.4398}{19.1883}\\ \cos {\theta }_y=\frac{-13.4022}{19.1883}\\ \cos {\theta }_z=\frac{12.9946}{19.1883}\end{array}$

Finally we get,

${\theta }_x=103.4°,{\theta }_y=134.4°,{\theta }_z=47.4°$

Conclusion:

The direction of binomial path at $t=\frac{\pi }{2}$

${\theta }_x=103.4°,{\theta }_y=134.4°,{\theta }_z=47.4°$