1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Calculus 4. Pursuit Curve. Dog Chases Rabbit.

  1. Sep 20, 2012 #1
    1. The problem statement, all variables and given/known data

    (a) In Example 1.18, assume that a is less than b (so that k is less than 1) and find y as a function of x. How far does the rabbit run before the dog catches him?

    (b) Assume now that a=b, and find y as a function of x. How close does the dog come to the rabbit?



    2. Relevant equations

    N/A
    There are many integration techniques that could be used.

    3. The attempt at a solution

    Example 1.18:

    $y''+k^2y=0$ Where $k$ is an unknown real constant)

    We notice that the independent variable is missing. So we let:

    $$y'=p, y''=p\frac{dp}{dy}$$

    $$p\frac{dp}{dy}+k^2y=0$$

    $$pdp=-k^2ydy$$

    $$\frac{p^2}{2}=-k^2\frac{y^2}{2}+c$$

    $$p=\pm k\sqrt{E-y^2}$$

    Now re-substitute $p=\frac{dy}{dx}$ to obtain:

    $$\frac{dy}{dx}= \pm k\sqrt{E-y^2}$$

    $$\frac{dy}{\sqrt{E-y^2}}= \pm kdx$$

    $$\sin^{-1}({\frac{y}{\sqrt{E}})}= \pm kx+F$$

    $$\frac{y}{\sqrt{E}}=\sin{( \pm kx+F)}$$

    $$y=\sqrt{E}\sin( \pm kx+F)$$

    Now we apply the sun formula for sine to rewrite the last expression as:

    $$y=\sqrt{E}\cos(F)\sin( \pm kx)+\sqrt{E}\sin(F)\cos( \pm kx)$$

    We may write a general solution as $y=A\sin(kx)+B\cos(kx)$


    There is another example in the book that I feel relates to this problem:

    Example 1.21

    A rabbit begins at the origin and runs up the $y-axis$ with speed $a$ feet per second. At the same time, a dog runs at speed $b$ from the point $(c,0)$ in pursuit of the rabbit. What is the path of the dog?

    Solution: At time $t$, measured from the instant both the rabbit and the dog start, the rabbit will be at the point $R=(0,at)$ and the dog at $D=(x,y)$. We wish to solve for $y$ as a function of $x$.

    $$\frac{dy}{dx}=\frac{y-at}{x}$$

    $$xy'-y=-at$$

    $$xy''=-a\frac{dt}{dx}$$

    Since the $s$ is a arc length along the path of the dog, it follows that $\frac{ds}{dt}=b$. Hence,

    $$\frac{dt}{dx}=\frac{dt}{ds}\frac{ds}{dx}=\frac{-1}{b}\sqrt{1+=(y')^2}$$

    $$xy''=\frac{a}{b}\sqrt{1+(y')^2}$$

    For convenience, we set $k=\frac{a}{b}$, $y'=p$, and $y''=\frac{dp}{dx}$

    $$\frac{dp}{\sqrt{1+p^2}}=k\frac{dx}{x}$$

    $$\ln({p+\sqrt{1+p^2}})=\ln(\frac{x}{c})^k$$

    Now, solve for $p$:

    $$\frac{dy}{dx}=p=\frac{1}{2}((\frac{x}{c})^k-(\frac{c}{x})^k)$$

    In order to continue the analysis, we need to know something about the relative sizes of $a$ and $b$. Suppose, for example, that $a \lt$ $b$ (so $k\lt$ $1$), meaning that the dog will certainly catch the rabbit. Then we can integrate the last equation to obtain:

    $$y(x)=\frac{1}{2}\{\frac{c}{k+1}(\frac{x}{c})^{k+1}-\frac{c}{1-k}(\frac{c}{x})^{k-1}\}+D$$

    Again, this is all I have to go on. I need to answer questions (a) and (b) stated at the top.
     
    Last edited: Sep 20, 2012
  2. jcsd
  3. Sep 20, 2012 #2
    could somebody at least give me something to go off of? This isn't making any sense.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Calculus 4. Pursuit Curve. Dog Chases Rabbit.
  1. Dog Chase (Replies: 4)

Loading...