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For our math class we have to do some calculations with respect to pursuit curves. The chased object starts at point $(p,0)$. Chaser starts at $(0,0)$.(x,y) Speed of the chased object is $u$. Speed chaser = $v$.

We have that for the chaser

$$ \frac{dy}{dx}=\frac{ut-y}{p-x} $$

Then the length of the path is

$$ s = \int \sqrt{1+(\frac{dy}{dx})^2}=vt=\frac{vy}{u}-\frac{v(p-x)dy}{u\times dx}$$

the first derivative of both sides gives

$$ -\frac{u}{v}\sqrt{1+(\frac{dy}{dx})^2}=\frac{(p-x)d(\frac{dy}{dx})}{dx} $$

Now, we are asked to:

exercise 1

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Does the chased object moves along the straight line $x=p$? Solving the first equation for $t$ and plugging into the second yields $$ s = \int_{0}^{x} \sqrt{1+(\frac{dy}{dX})^2}dX=vt=\frac{vy}{u}+\frac{v(p-x)}{u}\frac{dy}{dx}.$$ –  Américo Tavares Feb 15 '13 at 19:40

1 Answer 1

up vote 2 down vote accepted

You have some wrong signs in the last two equations.

Based on the equation of the derivative of the pursuit curve $y=f(x)$ described by the chaser object that you indicate $$ \frac{dy}{dx}=\frac{ut-y}{p-x}\tag{1} $$ I assume that the chased object moves along the straight line $x=p$, as I commented above. Assume further that it moves upwards. (See remark 2). Then

$$ t=\frac{y}{u}+\frac{p-x}{u}\frac{dy}{dx}.\tag{2} $$ and from $$ s=\int_{0}^{x}\sqrt{1+(f^{\prime }(\xi ))^{2}}d\xi =vt\tag{3} $$ we conclude that $$ t=\frac{1}{v}\int_{0}^{x}\sqrt{1+(f^{\prime }(\xi ))^{2}}d\xi .\tag{4} $$ Equating the two equations for $t$ $(2)$ and $(4)$ we get $$ \frac{1}{v}\int_{0}^{x}\sqrt{1+(f^{\prime }(\xi ))^{2}}d\xi =\frac{y}{u}+ \frac{p-x}{u}f(x). $$ Differentiating both sides we obtain the equation (note that the LHS is positive) $$ \frac{1}{v}\sqrt{1+(f^{\prime }(x))^{2}}=\frac{p-x}{u}f^{\prime \prime }(x).\tag{5} $$ If we let $w=\frac{dw}{dx}=f^{\prime }(x)$ this equation corresponds to the following one in $w$ and $w^{\prime }=\frac{dw}{dx}$ $$ \sqrt{1+w^{2}}=k(p-x)\frac{dw}{dx}\qquad w=f^{\prime }(x),k=\frac{v}{u},\tag{6} $$ which can be rewritten as $$ \frac{dw}{\sqrt{1+w^{2}}}=\frac{1}{k}\frac{dx}{p-x}\tag{7} $$ by applying the method of separation of variables to $x$ and $w$. The integration is easy $$ \begin{eqnarray*} \int \frac{dw}{\sqrt{1+w^{2}}} &=&\int \frac{1}{k}\frac{dx}{p-x}+\log C \\ \text{arcsinh }w &=&-\frac{1}{k}\log \left( p-x\right) +\log C. \end{eqnarray*} $$ The initial condition $x=0,w=f^{\prime }(0)=0$ yields $$ \begin{eqnarray*} -\frac{1}{k}\log p+\log C &=&\text{arcsinh }0=0 \\ &\Rightarrow &\log C =\frac{1}{k}\log p , \end{eqnarray*} $$ which means that $$ \text{arcsinh }w=-\frac{1}{k}\log \left( p-x\right) +\frac{1}{k}\log p=-\frac{1}{k}\log \frac{p-x}{p}.\tag{8} $$ Solving for $w$ we get

$$ \begin{eqnarray*} \frac{dy}{dx} &=&w=\sinh \left( -\frac{1}{k}\log \frac{p-x}{p}\right) =\frac{1}{2}\left( e^{-\frac{1}{k}\log \frac{p-x}{p}}-e^{\frac{1}{k}\log \frac{p-x}{p}}\right) \\ &=&\frac{1}{2} \left( \frac{p-x}{p}\right) ^{-1/k}-\frac{1}{2}\left( \frac{p-x}{p}\right) ^{1/k}\tag{9} \end{eqnarray*} $$ To integrate this equation consider two cases.

  • (a) $k=\frac{v}{u}>1$ $$\begin{eqnarray*} y &=&\frac{1}{2}\int \left( \frac{p-x}{p}\right) ^{-1/k}-\left( \frac{p-x}{p}\right) ^{1/k} dx \\ &=&-\frac{1}{2}\frac{pk}{k-1}\left( \frac{p-x}{p}\right) ^{1-1/k}+\frac{1}{2}\frac{pk}{k+1}\left( \frac{p-x}{p}\right) ^{1+1/k}+C. \end{eqnarray*}$$ The constant of integration $C$ is defined by the initial condition $x=0,y=0$ $$\begin{eqnarray*} 0 &=&-\frac{1}{2}\frac{pk}{k-1}+\frac{1}{2}\frac{pk}{k+1}+C \\ &\Rightarrow &C=\frac{pk}{k^{2}-1}. \end{eqnarray*}$$ Hence $$y=-\frac{1}{2}\frac{pk}{k-1}\left( \frac{p-x}{p}\right) ^{1-1/k}+\frac{1}{2}\frac{pk}{k+1}\left( \frac{p-x}{p}\right) ^{1+1/k}+\frac{pk}{k^{2}-1}$$ $$\tag{10}$$ The chaser overtakes the chased object at the point $(p,f(p))$, with $f(p)= \frac{pk}{k^{2}-1}$.
  • (b) $k=\frac{v}{u}=1$. We have $$\frac{dy}{dx}=\frac{1}{2} \left( \frac{p-x}{p}\right) ^{-1}-\frac{1}{2}\left( \frac{p-x}{p}\right) $$ and $$\begin{eqnarray*} y &=&\frac{1}{2}\int \left( \frac{p-x}{p}\right) ^{-1}-\left( \frac{ p-x}{p}\right) dx \\ &=&-\frac{1}{2}p\ln \left( p-x\right) -\frac{1}{2}x+\frac{1}{4p}x^{2}+C. \end{eqnarray*}$$ The same initial condition $x=0,y=0$ yields now $$\begin{eqnarray*} C &=&\frac{1}{2}p\ln \left( p\right) \\ && \\ y &=&-\frac{1}{2}p\ln \left( \frac{p-x}{p}\right) -\frac{1}{2}x+\frac{1}{4p}x^{2}.\tag{11} \end{eqnarray*}$$ The chaser never overtakes the chased object.

Example for (a): graph of $y=f(x)$ for $k=2,p=50$

enter image description here

Example for (b): graph of $y=f(x)$ for $k=1,p=50$

enter image description here

Remarks:

  1. This answer is similar to the answer of mine to the question Cat Dog problem using integration.

  2. It was inspired by Helmut Knaust's The Curve of Pursuit.

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