Question

Pursuit Curve In a naval exercise a ship ${\mathbf{S}}_{\mathbf{1}}$ is pursued by a submarine${\mathbf{S}}_{\mathbf{2}}$ as shown in Figure 5.3.9. Ship ${\mathbf{S}}_{\mathbf{1}}$departs point $\left(0,0\right)$at$\mathit{t}\mathbf{=}\mathbf{0}$ and proceeds along a straight-line course (the$\mathit{y}\mathbf{-}$ axis) at a constant speed ${\mathit{v}}_{\mathbf{1}}$. The submarine${\mathbf{S}}_2$ keeps ship${\mathbf{S}}_{\mathbf{1}}$ in visual contact, indicated by the straight dashed line$\mathit{L}$ in the figure, while traveling at a constant speed${\mathit{v}}_{\mathbf{2}}$ along a curve $\mathit{C}$. Assume that ship${\mathbf{S}}_{\mathbf{2}}$ starts at the point$\mathbf{(}\mathbf{a}\mathbf{,}\mathbf{0}\mathbf{)}\mathbf{,}\mathbf{a}\mathbf{>}\mathbf{0}\mathbf{,}$ at$\mathit{t}\mathbf{=}\mathbf{0}$ and that$\mathit{L}$ is tangent to$\mathit{C}$ .

(a) Determine a mathematical model that describes the curve $\mathit{C}$.

(b) Find an explicit solution of the differential equation. For convenience define $\mathbf{r}\mathbf{=}{\mathbf{v}}_{\mathbf{1}}\mathbf{/}{\mathbf{v}}_{\mathbf{2}}$.

(c) Determine whether the paths of${\mathit{S}}_{\mathbf{1}}$ and ${\mathit{S}}_{\mathbf{2}}$will ever intersect by considering the cases$\mathit{r}\mathbf{>}\mathbf{1}\mathbf{,}\mathit{r}\mathbf{<}\mathbf{1}\mathbf{,}$ and$\mathit{r}\mathbf{=}\mathbf{1}\mathbf{.}$

Short answer

(a)${\mathrm{xy}}^{\text{'}}=\mathrm{y}-{\mathrm{v}}_1\mathrm{t}$

(b) When$\mathrm{r}\ne 1,$ then$\mathrm{y}=\frac{\mathrm{a}}{2}\left[\frac{1}{\mathrm{r}+1}{\left(\frac{\mathrm{x}}{\mathrm{a}}\right)}^{\mathrm{r}+1}+\frac{1}{\mathrm{r}-1}{\left(\frac{\mathrm{x}}{\mathrm{a}}\right)}^{-\mathrm{r}+1}\right]+\frac{\mathrm{ar}}{1-{\mathrm{r}}^2}$ where$r=1$ then$\mathrm{y}=\frac{1}{4\mathrm{a}}({\mathrm{x}}^2-{\mathrm{a}}^2)+\frac{\mathrm{a}}{2}\ln \left(\frac{\mathrm{a}}{\mathrm{x}}\right)$

(c) The paths of $S_1$and$S_2$ will only intersect of$r<1$ that is, if$v_1<v_2$ They will intersect at$(0,\mathrm{ar}/(1-{\mathrm{r}}^2)).$

Step-by-step solution

To given information

Above is the sketch of the given figure. We know that ship $S_1$departs from$\left(0,0\right)$ at $t=0$and moves at a constant speed $v_1$along the $y-$axis. Submarine $S_2$is moving at constant speed$v_2$ along the curve$C.$ The submarine $S_2$departs from$(a,0),a>0,$ at $t=0$. We will assume that the line drawn between the ships is tangent to$C.$ Denote it as $L$.

To find the coordinates of the submarine

(a)

Let $\left(x,y\right)$be the coordinates of the submarine $S_2$.

Since we assumed that the line$L$ is a tangent to the$C$ , we can find the equation that will describe$C$ using the slope of the line $L$.

The $x-$coordinate of ship$S_1$ is always equal to zero. Since$S_1$ is moving at a constant speed$V_1$ we can write its$y-$ coordinate as $v_{1}t$.Therefore, we have:

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{v}}_1\mathrm{t}-\mathrm{y}}{0-\mathrm{x}}=\frac{\mathrm{y}-{\mathrm{v}}_1\mathrm{t}}{\mathrm{x}}$

which is equivalent to

${\mathrm{xy}}^{\text{'}}=\mathrm{y}-{\mathrm{v}}_1\mathrm{t}$

To find the coordinates of the submarine

(b)

To determine an explicit solution of the differential equation (1), let us derive it with respect to $x:$

$x{y}^{\text{'}\text{'}}+y^{\text{'}}=y^{\text{'}}-v_1\frac{dt}{dx}$

${\mathrm{xy}}^{\text{'}\text{'}}+=-{\mathrm{v}}_1\frac{\mathrm{dt}}{\mathrm{dx}}$

Now use$\frac{\mathrm{dt}}{\mathrm{dx}}=\frac{\mathrm{dt}}{\mathrm{ds}}\frac{\mathrm{ds}}{\mathrm{dx}},$where$\mathrm{s}$will be the arc length measured along$C.$

$x{y}^{\text{'}\text{'}}+=-v_1\frac{dt}{ds}\frac{ds}{dx}$

Now, since$S_2$is moving along the curve$C,\frac{ds}{dt}=v_2.$That is equivalent to$\frac{\mathrm{dt}}{\mathrm{ds}}=\frac{1}{{\mathrm{v}}_2}.$

The length of the curve given by the $y=f\left(x\right)$is

$s=\int \sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx$

Therefore, in this case, we have

$\frac{\mathrm{ds}}{\mathrm{dx}}=-\sqrt{1+{\left({\mathrm{y}}^{\text{'}}\right)}^2}$

The minus comes from the fact that $S_2$is going in the negative direction considering the $x-$axis.

Substituting this into (2), we obtain

${\mathrm{xy}}^{\text{'}\text{'}}+=-{\mathrm{v}}_1\frac{1}{{\mathrm{v}}_2}\left(-\sqrt{1+{\left({\mathrm{y}}^{\text{'}}\right)}^2}\right)$

Let us define the constant $r$with $r=\frac{v_1}{v_2}$. Now we have:$x{y}^{\text{'}\text{'}}=r\sqrt{1+{\left(y^{\text{'}}\right)}^2}$

Define$u=y^{\text{'}}.$Equation (3) becomes

$\mathrm{x}\frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{r}\sqrt{1+{\mathrm{u}}^2}$

which is equivalent to

$\frac{\mathrm{du}}{\sqrt{1+{\mathrm{u}}^2}}=\mathrm{r}\frac{\mathrm{dx}}{\mathrm{x}}$

Integrating it, we obtain

$\ln (\mathrm{u}+\sqrt{1+{\mathrm{u}}^2})=\mathrm{rlnx}+\mathrm{C}$

Let$D$be constant such that $C=\ln D.$The last equation is then equivalent to

$\ln (\mathrm{u}+\sqrt{1+{\mathrm{u}}^2})=\ln \left({\mathrm{Dx}}^{\mathrm{r}}\right)\iff \mathrm{u}+\sqrt{1+{\mathrm{u}}^2}={\mathrm{Dx}}^{\mathrm{r}}$

Let us rewrite the last equality

$\sqrt{1+{\mathrm{u}}^2}={\mathrm{Dx}}^{\mathrm{r}}-\mathrm{u}$

$1+{\mathrm{u}}^2={\mathrm{D}}^2{\mathrm{x}}^{2\mathrm{r}}-2{\mathrm{Dx}}^{\mathrm{r}}\mathrm{u}+{\mathrm{u}}^2$

$1-{\mathrm{D}}^2{\mathrm{x}}^{2\mathrm{r}}=-2{\mathrm{Dx}}^{\mathrm{r}}\mathrm{u}$

$\mathrm{u}=\frac{{\mathrm{D}}^2{\mathrm{x}}^{2\mathrm{r}}-1}{2{\mathrm{Dx}}^{\mathrm{r}}}=\frac{1}{2}\left({\mathrm{Dx}}^{\mathrm{r}}-\frac{1}{{\mathrm{Dx}}^{\mathrm{r}}}\right)$

Now, we will use the context $t=0.$at both $S_1$and$S_2$have$y-$coordinate 0 , so$u=\frac{dy}{dx}=0.$Also, $x=a$at$t=0.$Substitute it into equation (4):

$0=\frac{1}{2}\left({\mathrm{Da}}^{\mathrm{r}}-\frac{1}{{\mathrm{Da}}^{\mathrm{r}}}\right)\iff \mathrm{D}=\frac{1}{{\mathrm{a}}^{\mathrm{r}}}$

Therefore, from equation (4), we have:

$\frac{\mathrm{dy}}{\mathrm{dx}}=\left[\frac{1}{2}{\left(\frac{\mathrm{x}}{\mathrm{a}}\right)}^{\mathrm{r}}-{\left(\frac{\mathrm{a}}{\mathrm{x}}\right)}^{\mathrm{r}}\right]=\frac{1}{2}\left[{\left(\frac{\mathrm{x}}{\mathrm{a}}\right)}^{\mathrm{r}}-{\left(\frac{\mathrm{x}}{\mathrm{a}}\right)}^{-\mathrm{r}}\right]$

We will get$y$by integrating (5). There will be two cases.

Case I$\mathrm{r}\ne 1$

The equation (5) becomes:

$y=\frac{a}{2}\left[\frac{1}{r+1}{\left(\frac{x}{a}\right)}^{r+1}-\frac{1}{-r+1}{\left(\frac{x}{a}\right)}^{-r+1}\right]+C$

When$\mathrm{t}=0,\mathrm{x}=\mathrm{a}$and $\mathrm{y}=0.$Substitute it into the last equation:

$0=\frac{\mathrm{a}}{2}\left[\frac{1}{\mathrm{r}+1}{\left(\frac{\mathrm{a}}{\mathrm{a}}\right)}^{\mathrm{r}+1}-\frac{1}{-\mathrm{r}+1}{\left(\frac{\mathrm{a}}{\mathrm{a}}\right)}^{-\mathrm{r}+1}\right]+\mathrm{C}$

$0=\frac{\mathrm{a}}{2}\left[\frac{1}{\mathrm{r}+1}+\frac{1}{\mathrm{r}-1}\right]+\mathrm{C}=\frac{\mathrm{ar}-1+\mathrm{r}+1}{2}+\mathrm{C}$

Therefore, $\mathrm{C}=\frac{\mathrm{ar}}{1-{\mathrm{r}}^2}$Substitute it into (6) to obtain

$\mathrm{y}=\frac{\mathrm{a}}{2}\left[\frac{1}{\mathrm{r}+1}{\left(\frac{\mathrm{x}}{\mathrm{a}}\right)}^{\mathrm{r}+1}+\frac{1}{\mathrm{r}-1}{\left(\frac{\mathrm{x}}{\mathrm{a}}\right)}^{-\mathrm{r}+1}\right]+\frac{\mathrm{ar}}{1-{\mathrm{r}}^2}$

Case II$r=1$

If $r=1$, Eq. (5) is:

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1}{2}\left[\frac{\mathrm{x}}{\mathrm{a}}-\frac{\mathrm{a}}{\mathrm{x}}\right]$

Integrating it, we obtain:

$\mathrm{y}=\frac{1}{2}\left(\frac{{\mathrm{x}}^2}{2\mathrm{a}}-\mathrm{alnx}\right)+\mathrm{C}$

Again, we will use that $\mathrm{x}=\mathrm{a}$and$y=0$ at $t=0$:

$0=\frac{1}{2}\left(\frac{{\mathrm{a}}^2}{2\mathrm{a}}-\mathrm{alna}\right)+\mathrm{C}$

From this, we have that $\mathrm{C}=\frac{1}{2}\mathrm{alna}-\frac{\mathrm{a}}{4}.$Substitute it into (8):

$\mathrm{y}=\frac{1}{2}\left(\frac{{\mathrm{x}}^2}{2\mathrm{a}}-\mathrm{alnx}\right)+\frac{1}{2}\mathrm{alna}-\frac{\mathrm{a}}{4}$

Rewrite it:

$y=\frac{1}{4a}(x^2-a^2)+\frac{a}{2}\ln \left(\frac{a}{x}\right)$

To find the three cases

In this part of the task, we will consider three cases:$\mathrm{r}=\mid 1,\mathrm{r}>1$ and$r<1.$

- Case I$\mathrm{r}=1$

From Eq.(9), we can easily see that $\mathrm{y}\to \mathrm{\infty }$when$\mathrm{x}\to 0^{+}.$ Considering the definition of the motion of ${\mathrm{S}}_2$on the curve$C$ we can conclude that$S_2$ will never catch up with$S_1$

- Case II$r>1$

In the same way as in the previous case, from Eq. (7), we see that $y\to \infty$when $x\to 0^{+}$. Therefore, we can also conclude that $S_2$will never catch up with $S_1$.

Notice that this makes sense.$r>1$ means that$v_1>v_2$ and the ships have the same starting $y-$coordinate.

- Case III$r<1$

From Eq. (7), we can see that, when $x=0,y=\frac{ar}{1-r^2}$. Therefore, when$\mathbf{r}>\mathbf{1}$ , their paths will intersect at$(0,ar/(1-r^2)).$

Final proof

(a)${\mathrm{xy}}^{\text{'}}=\mathrm{y}-{\mathrm{v}}_1\mathrm{t}$

(b) When $r\ne 1$then$\mathrm{y}=\frac{\mathrm{a}}{2}\left[\frac{1}{\mathrm{r}+1}{\left(\frac{\mathrm{x}}{\mathrm{a}}\right)}^{\mathrm{r}+1}+\frac{1}{\mathrm{r}-1}{\left(\frac{\mathrm{x}}{\mathrm{a}}\right)}^{-\mathrm{r}+1}\right]+\frac{\mathrm{ar}}{1-{\mathrm{r}}^2}$ where $r=1$then$\mathrm{y}=\frac{1}{4\mathrm{a}}({\mathrm{x}}^2-{\mathrm{a}}^2)+\frac{\mathrm{a}}{2}\ln \left(\frac{\mathrm{a}}{\mathrm{x}}\right)$

(c) The paths of $S_1$and$S_2$ will only intersect of$r<1,$ that is, if $v_1<v_2.$They will intersect at$(0,ar/(1-r^2)).$