Chapter 6, Problem 6.23P

(a)

To determine

The plane’s ground speed and plane’s position.

Answer to Problem 6.23P

The total plane’s ground speed in terms of $V$, $v_0$ and $\varphi$ is, $\sqrt{{\left(v_0\cos \varphi +V_y\right)}^2+{\left(v_0\sin \varphi \right)}^2}$

Explanation of Solution

Write the expression for the speed of the wind along east direction is,

  $v_{\min d}=Vy\widehat{x}$        (I)

Here, $x$ and $y$ are measured in east and north respectively.

Let $v_0$ be the speed of the aircraft. From the figure, the total horizontal component of the speed of the aircraft-wind system is,

  $v_x=v_0\cos \varphi +V_y$        (II)

Similarly the vertical component of the speed of the aircraft is

  $v_y=v_0\sin \varphi$        (III)

Conclusion:

Thus, the expression for the total plane’s ground speed is,

  $v=\sqrt{v_x{}^2+v_y{}^2}$        (IV)

Substitute expression (I) and (II) in equation (IV),

  $v=\sqrt{{\left(v_0\cos \varphi +V_y\right)}^2+{\left(v_0\sin \varphi \right)}^2}$

Hence the total plane’s ground speed in terms of $V$, $v_0$ and $\varphi$ is, $\sqrt{{\left(v_0\cos \varphi +V_y\right)}^2+{\left(v_0\sin \varphi \right)}^2}$

(b)

To determine

The time of flight as an integral of the form ${\displaystyle \underset{0}{\overset{D}{\int }}fdx}$

Answer to Problem 6.23P

The required expression for the integrand $f$ is $\frac{\left(1+\frac{1}{2}{y^{\prime }}^2\right)}{v_0\left(1+ky\right)}$

Explanation of Solution

Write the expr4ession for the plane’s ground speed,

  $\begin{array}{c}v=\sqrt{{\left(v_0\cos \varphi +Vy\right)}^2+{\left(v_0\sin \varphi \right)}^2}\\ =\sqrt{\left(v_0{}^2{\cos }^2\varphi +V^2{y}^2+2v_{0}Vy\cos \varphi \right)+{\left(v_0\sin \varphi \right)}^2}\\ =\sqrt{v_0{}^2\left({\cos }^2\varphi +{\sin }^2\varphi \right)+V^2{y}^2+2v_{0}Vy\cos \varphi }\\ =\sqrt{v_0{}^2+V^2{y}^2+2v_{0}Vy\cos \varphi }\end{array}$        (I)

Assume that the angel $\varphi$ is very small. From the properties of the cosine function, the cosine function is equal to one for small angles, that is $\cos \varphi =1$ Thus, the plane’s ground speed equation becomes as,

  $\begin{array}{c}v=\sqrt{v_0{}^2+2v_0{V}_y+V^2{y}^2}\\ =\sqrt{{\left(v_0+Vy\right)}^2}\\ =v_0+V_y\end{array}$        (II)

Conclusion:

The expression for the given path from the town $O$ (origin) to town $P$ is $y=y\left(x\right)$. Thus the expression for the distance traveled by the plane is,

  $ds=\sqrt{d{x}^2+d{y}^2}$        (III)

The differential form of the variable     $y$ as a function of $x$ can be written as,

  $y^{\prime }=\frac{dy}{dx}$        (IV)

Substitute expression (IV) in expression (III)

  $\begin{array}{c}ds=\sqrt{d{x}^2+{\left(y^{\prime }dx\right)}^2}\\ =\sqrt{d{x}^2+{\left(y^{\prime }dx\right)}^2}\\ =\sqrt{\left(1+{y^{\prime }}^2\right)d{x}^2}\\ =\sqrt{1+y^{\prime }}dx\end{array}$

The expression for the time taken by the plane to cover the distance from the point $O$ to point $P$ is,

  $t={\displaystyle \underset{O}{\overset{P}{\int }}\frac{ds}{v}}$

Let $D$ be the distance covered by the plane due east. Thus, the time of flight is,

    $t={\displaystyle \underset{O}{\overset{D}{\int }}\frac{ds}{v}}$

Substitute $\sqrt{1+{y^{\prime }}^2}dx$ for $ds$ and $v_o+Vy$ for $V$

  $\begin{array}{c}t={\displaystyle \underset{O}{\overset{D}{\int }}\frac{\sqrt{1+{y^{\prime }}^2}dx}{v_o+Vy}}\\ ={\displaystyle \underset{O}{\overset{D}{\int }}\frac{\left(1+\frac{1}{2}{y^{\prime }}^2\right)dx}{v_o+Vy}}\\ ={\displaystyle \underset{O}{\overset{D}{\int }}\frac{\left(1+\frac{1}{2}{y^{\prime }}^2\right)dx}{v_o\left(1+\frac{Vy}{v_0}\right)}}\end{array}$

Let $\frac{V}{v_0}=k$ the time of flight equation becomes as,

  $t={\displaystyle \underset{O}{\overset{D}{\int }}\frac{\left(1+\frac{1}{2}{y^{\prime }}^2\right)dx}{v_o\left(1+ky\right)}}$

Compare this equation with standard form of Euler-Lagrange equation

$S={\displaystyle \underset{x_1}{\overset{x_2}{\int }}f\left(y\left(x\right),y^{\prime }\left(x\right),x\right)}dx$ and the variables $x$ and $y$ are interchanged.

  $f\left(y^{\prime },y,x\right)=\frac{\left(1+\frac{1}{2}{y^{\prime }}^2\right)}{v_0\left(1+ky\right)}$

Therefore the required expression for the integrand $f$ is $\frac{\left(1+\frac{1}{2}{y^{\prime }}^2\right)}{v_0\left(1+ky\right)}$

(c)

To determine

The distance and the time by the plane along in its path

Answer to Problem 6.23P

The distance traveled by the plane due north is $\text{366miles}$ and the time saved by the plane along in its path is $27\min$.

Explanation of Solution

Write the Euler-Lagrange equation for all the values of $x$ ,

  $\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y^{\prime }}=0$        (I)

Determine the value of $\frac{\partial f}{\partial y}$,

  $\begin{array}{c}\frac{\partial f}{\partial y}=\frac{\partial }{\partial y}\left(\frac{\left(1+\frac{1}{2}{y^{\prime }}^2\right)}{v_0\left(1+ky\right)}\right)\\ =\frac{1}{v_0}\left(1+\frac{1}{2}{y^{\prime }}^2\right)\frac{\partial }{\partial y}\left(\frac{1}{1+ky}\right)\\ =-\frac{1}{v_0}\left(1+\frac{1}{2}{y^{\prime }}^2\right)k{\left(1+ky\right)}^{-2}\end{array}$

Similarly, determine the values of $\frac{\partial f}{\partial y^{\prime }}$

  $\begin{array}{c}\frac{\partial f}{\partial y^{\prime }}=\frac{\partial }{\partial y^{\prime }}\left(\frac{\left(1+\frac{1}{2}{y^{\prime }}^2\right)}{v_0\left(1+ky\right)}\right)\\ =\frac{1}{v_0\left(1+ky\right)}\frac{\partial }{\partial y^{\prime }}\left(1+\frac{1}{2}{y^{\prime }}^2\right)\\ =\frac{1}{v_0\left(1+ky\right)}\left(0+\frac{1}{2}\left(2\right)y^{\prime }\right)\\ =\frac{y^{\prime }}{v_0\left(1+ky\right)}\end{array}$        (II)

Calculate the value of $\frac{d}{dx}\frac{\partial f}{\partial y^{\prime }}$

   $\begin{array}{c}\frac{d}{dx}\frac{\partial f}{\partial y^{\prime }}=\frac{d}{dx}\left(\frac{y^{\prime }}{v_0\left(1+ky\right)}\right)\\ =\frac{1}{v_0}\frac{d}{dx}\left(\frac{y^{\prime }}{\left(1+ky\right)}\right)\\ =\frac{1}{v_0}\left(y^{″}{\left(1+ky\right)}^{-1}+y^{\prime }\left(-1\right)\left(k{y}^{\prime }\right){\left(1+ky\right)}^{-2}\right)\\ =\frac{1}{v_0}\left(y^{″}{\left(1+ky\right)}^{-1}+\left(k{y^{\prime }}^2\right){\left(1+ky\right)}^{-2}\right)\end{array}$        (III)

Conclusion:

Thus, the Euler-Lagrange equation becomes as,

  $\begin{array}{c}-\frac{1}{v_0}\left(1+\frac{1}{2}{y^{\prime }}^2\right)k{\left(1+ky\right)}^{-2}-\frac{1}{v_0}\left(y^{″}{\left(1+ky\right)}^{-1}+\left(k{y^{\prime }}^2\right){\left(1+ky\right)}^{-2}\right)=0\\ \left(1+\frac{1}{2}{y^{\prime }}^2\right)k{\left(1+ky\right)}^{-2}-\left(y^{″}{\left(1+ky\right)}^{-1}+\left(k{y^{\prime }}^2\right){\left(1+ky\right)}^{-2}\right)=0\\ \frac{1}{\left(1+ky\right)}\left(\left(1+\frac{1}{2}{y^{\prime }}^2\right)\frac{k}{\left(1+ky\right)}+y^{″}-\frac{k{y^{\prime }}^2}{\left(1+ky\right)}\right)=0\\ \left(1+\frac{1}{2}{y^{\prime }}^2\right)\frac{k}{\left(1+ky\right)}+y^{″}-\frac{k{y^{\prime }}^2}{\left(1+ky\right)}=0\\ \frac{y^{″}\left(1+ky\right)-k{y}^{\prime }+\left(k/2\right){y^{\prime }}^2+k}{\left(1+ky\right)}=0\\ y^{″}\left(1+ky\right)-\left(k/2\right){y^{\prime }}^2+k=0\end{array}$        (IV)

The path of the plane is defined as,

  $\begin{array}{c}y\left(x\right)=\lambda x\left(D-x\right)\\ =\lambda xD-\lambda x^2\end{array}$

Differentiate the equation on both sides with respect to $x$

  $y^{\prime }=\lambda D-2\lambda x$

Similarly, differentiate the equation $y^{\prime }=\lambda D-2\lambda x$ on both sides with respect to $x$.

  $y^{″}=-2\lambda$

Substitute $\lambda xD-\lambda x^2$ for $y$, $\lambda D-2\lambda x$ for $y^{\prime }$ and $-2\lambda$ for $y^{″}$ in the equation (IV),

  $\begin{array}{c}{y}^{″}\left(1+ky\right)-\left(k/2\right){y^{\prime }}^2+k=0\\ \left(-2y\right)\left(1+k\left(\lambda xD-\lambda x^2\right)-\frac{1}{2}k{\left(\lambda D-2\lambda x\right)}^2+k\right)=0\\ -2\lambda -2{\lambda }^{2}xDk+2{\lambda }^2{x}^{2}k-\frac{1}{2}k{\lambda }^2{D}^2-\frac{1}{2}k4{\lambda }^2{x}^2+\frac{1}{2}k\left(2\left(\lambda D\right)\left(d\lambda k\right)\right)+k=0\\ -2\lambda -2k{\lambda }^{2}xD+2k{\lambda }^2{x}^2-\frac{1}{2}k{\lambda }^2{x}^2-2k{\lambda }^2{x}^2+2k{\lambda }^{2}xD+k=0\\ -2\lambda -\frac{1}{2}k{\lambda }^2{D}^2+k=0\end{array}$

The equation can be written as in the form of quadratic equation,

  $\begin{array}{c}-2\lambda -\frac{1}{2}k{\lambda }^2{D}^2+k=0\\ -4\lambda -k{\lambda }^2{D}^2+2k=0\\ -\left(k{D}^2\right){\lambda }^2-4\lambda +2k=0\\ {\lambda }^2+\left(\frac{4}{k{D}^2}\right)\lambda -\left(\frac{2}{D^2}\right)=0\end{array}$

The roots of the quadratic equation ${\lambda }^2+\left(\frac{4}{k{D}^2}\right)\lambda -\left(\frac{2}{D^2}\right)=0$ are.

  $\begin{array}{c}\lambda =\frac{-\left(\frac{4}{k{D}^2}\right)\pm \sqrt{{\left(\frac{4}{k{D}^2}\right)}^2-4\left(1\right)\left(\frac{2}{D^2}\right)}}{2\left(1\right)}\\ \lambda =\frac{-\left(\frac{4}{k{D}^2}\right)\pm \sqrt{\left(\frac{4\left(4+2k^2{D}^2\right)}{{\left(k{D}^2\right)}^2}\right)}}{2\left(1\right)}\\ \lambda =\frac{-4\pm 2\sqrt{4+2k^2{D}^2}}{2k{D}^2}\\ \lambda =\frac{\pm \sqrt{4+2k^2{D}^2}-2}{k{D}^2}\end{array}$

Thus the solution $\lambda =\frac{\pm \sqrt{4+2k^2{D}^2}-2}{k{D}^2}$ satisfies the Euler-Lagrange equation.

Thus the expression for the constant $\lambda$ is

  $\lambda =\frac{\pm \sqrt{4+2k^2{D}^2}-2}{k{D}^2}$        (V)

Substitute $\frac{V}{v_0}$ for $k$ in above expression(V)

  $\lambda =\frac{\pm \sqrt{4+2{\left(\frac{V}{v_0}\right)}^2{D}^2}-2}{\left(\frac{V}{v_0}\right)D^2}$        (VI)

Substitute $\text{0}\text{.5mph/mi}$ for $V$ , $\text{500mph}$ for $v_0$ and $2000\text{mi}$ for $D$ in expression (VI)

  $\begin{array}{c}\lambda =\frac{\pm \sqrt{4+2{\left(\frac{\text{0}\text{.5mph/mi}}{\text{500mph}}\right)}^2{\left(2000\text{mi}\right)}^2}-2}{\left(\frac{\text{0}\text{.5mph/mi}}{\text{500mph}}\right){\left(2000\text{mi}\right)}^2}\\ =3.66\times {10}^{-4}\end{array}$

The expression for the maximum displacement of the plane due north is,

  $y_{\max }=\frac{\lambda D^2}{4}$

Substitute $3.66\times {10}^{-4}$ for $\lambda$ and $2000\text{mi}$ for $D$

  $\begin{array}{c}{y}_{\max }=\frac{\left(3.66\times {10}^{-4}\right){\left(2000\text{mi}\right)}^2}{4}\\ =366\text{mi}\end{array}$

Therefore, the distance traveled by the plane due north is $366\text{mi}$.

The expression for the time along the path is,

  $\begin{array}{l}t={\displaystyle \underset{O}{\overset{D}{\int }}\frac{\left(1+\frac{1}{2}{y^{\prime }}^2\right)dx}{v_o\left(1+ky\right)}}\\ t=\frac{1}{v_0}{\displaystyle \underset{O}{\overset{D}{\int }}\frac{\left(1+\frac{1}{2}{y^{\prime }}^2\right)dx}{v_o\left(1+ky\right)}}\end{array}$        (VII)

Substitute $\lambda D-2\lambda x$ for $y^{\prime }$, $\frac{V}{v_0}$ for $k$ and $\lambda x\left(D-x\right)$ for $y$ expression,

  $t=\frac{1}{v_0}{\displaystyle \underset{O}{\overset{D}{\int }}\frac{\left(1+\frac{1}{2}{\left(\lambda D-2\lambda x\right)}^2\right)dx}{\left(1+\left(\frac{V}{v_o}\right)\lambda x\left(D-x\right)\right)}}$        (VIII)

Let $x=Ds$, then $dx=Dds$ thus, the equation (VIII) becomes as,

  $t=\frac{D}{v_0}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\left(1+\frac{1}{2}\left({\lambda }^2{D}^2\right){\left(1-2s\right)}^2\right)ds}{\left(1+\left(\frac{V}{v_o}\right)\lambda D^{2}s\left(1-s\right)\right)}}$        (IX)

Substitute $\text{0}\text{.5mph/mi}$ for $V$ , $\text{500mph}$ for $v_0$ $3.66\times {10}^{-4}$ for $\lambda$ and $2000\text{mi}$ for $D$ in expression (IX)

  $\begin{array}{c}t=\frac{D}{v_0}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\left(1+\frac{1}{2}\left({\left(3.66\times {10}^{-4}\right)}^2{\left(2000\text{mi}\right)}^2\right){\left(1-2s\right)}^2\right)ds}{\left(1+\left(\frac{\text{0}\text{.5mph/mi}}{\text{500mph}}\right)\left(3.66\times {10}^{-4}\right){\left(2000\text{mi}\right)}^{2}s\left(1-s\right)\right)}}\\ t=\frac{D}{v_0}{\displaystyle \underset{0}{\overset{1}{\int }}\frac{\left(1+\left(0.268\right){\left(1-2s\right)}^2\right)ds}{\left(1+\left(1.464\right)s\left(1-s\right)\right)}}\\ t=\left(0.889\right)\frac{D}{v_0}\end{array}$

Substitute $500\text{mph}$ for $v_0$ and $200\text{mi}$ for $D$ in the equation $t=\left(0.889\right)\frac{D}{v_0}$,

  $\begin{array}{c}t=\left(0.889\right)\frac{200\text{mi}}{500\text{mph}}\\ =3.556\text{h}\end{array}$

The time taken by the plane to travel a distance along the direct path is,

  $t^{\prime }=\frac{D}{v_0}$

Substitute $500\text{mph}$ for $v_0$ and $200\text{mi}$ for $D$ in the equation $t^{\prime }=\frac{D}{v_0}$,

  $\begin{array}{c}{t}^{\prime }=\frac{200\text{mi}}{500\text{mph}}\\ =4\text{h}\end{array}$

Thus, the time saved by the plane along in its path is,

  $\begin{array}{c}\Delta t=4\text{h-3}\text{.556h}\\ \text{=0}\text{.444h}\left(\frac{60\min }{1\text{h}}\right)\\ =26.64\min \end{array}$

Rounding off to two significant figures, the time saved by the plane along in its path is $27\min$