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\noindent{\bf Math 3226
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Laboratory  2B}\\
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{\bf The Curve of Pursuit\footnote{This laboratory is based on a group project in {\em ``Fundamentals of Differential Equations''} by R. Kent Nagle and Edward B. Saff.}}
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\noindent{\em An interesting geometric problem---first considered by Leonardo da Vinci--- arises when one tries to determine the path of a pursuer chasing its prey. The simplest problem is to find the curve along which a vessel moves in pursuing another vessel that flees along a straight line, assuming that the speeds of both vessels are constant.

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Let's assume vessel A, traveling at a speed $\alpha$, is pursuing vessel B, which is traveling at speed $\beta$. In addition, assume vessel A begins at time $t=0$ at the origin and pursues vessel B, which begins at the point $(b,0)$, $b>0$, and travels up the line $x=b$.
After $t$ hours, vessel $A$ is located at the point $P=(x,y)$ and vessel B is located at the point $Q=(b,\beta t) $. The goal is to describe the locus of points $P$; that is to find $y$ as a function of $x$. }

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\item 
Vessel A is pursuing vessel B, so at time $t$, vessel A must be heading right at vessel B. Thus the tangent line to the curve of pursuit at $P$ must pass through the point $Q$. For this to be true, show that 
\[\frac{dy}{dx}=\frac{y-\beta t}{x-b}.\] 

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\item We know that the speed at which vessel A is traveling, so we know that the distance it travels in time $t$ is $\alpha t$. This distance is also the length of the pursuit curve from $(0,0)$ to $(x,y)$. Using the arclength formula, show that 
\[\alpha t=\int_0^x \sqrt{1+\left[\frac{dy}{du}\right]^2}\ du.\]

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\item Solve the formulas in {\bf 1.} and {\bf 2.} for the variable $t$ to obtain
\[\frac{y}{\beta}-\frac{(x-b)} {\beta} \frac{dy}{dx}=\frac{1}{\alpha}  \int_0^x \sqrt{1+\left[\frac{dy}{du}\right]^2}\ du.\]

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\item Set $w(x)=dy/dx$ and differentiate both sides of the equation above with respect to $x$ to obtain the first-order differential equation
\[(x-b)\frac{dw}{dx}=-\frac{\beta}{\alpha}\sqrt{1+w^2}.\]

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\item Using appropriate initial values for both $x$ and $w=dy/dx$ when $t=0$, show that the solution of the equation in {\bf 4.} is given by
\[\frac{dy}{dx}=w(x)=\frac{1}{2} \left[ \left(1-\frac{x}{b}\right)^{-\frac{\beta}{\alpha}}-
\left(1-\frac{x}{b}\right)^{\frac{\beta}{\alpha}}\right].\]

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\item For the case $\alpha > \beta$, integrate the expression in {\bf 5.} with respect to $x$ to obtain $y$ as a function of $x$. What is the initial condition for $y$ when $t=0$?

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\item  For the case $\alpha > \beta$, find the location where  vessel A will intercept vessel B.

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\item Repeat {\bf 6.} for the case $\alpha=\beta$.

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\item For the case $\alpha=\beta$, will vessel A ever reach vessel B?
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\flushright\tiny\copyright\  H. Knaust. \today .
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