Xrfgtjtk

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

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

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:

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
$$
s=int_{0}^{x}sqrt{1+(f^{prime }(xi ))^{2}}dxi =vttag{3}
$$
we conclude that
$$
t=frac{1}{v}int_{0}^{x}sqrt{1+(f^{prime }(xi ))^{2}}dxi .tag{4}
$$
Equating the two equations for $t$ $(2)$ and $(4)$ we get
$$
frac{1}{v}int_{0}^{x}sqrt{1+(f^{prime }(xi ))^{2}}dxi =frac{y}{u}+
frac{p-x}{u}f(x).
$$
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
$$
begin{eqnarray*}
int frac{dw}{sqrt{1+w^{2}}} &=&int frac{1}{k}frac{dx}{p-x}+log C \
text{arcsinh }w &=&-frac{1}{k}log left( p-xright) +log C.
end{eqnarray*}
$$
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
$$
text{arcsinh }w=-frac{1}{k}log left( p-xright) +frac{1}{k}log p=-frac{1}{k}log frac{p-x}{p}.tag{8}
$$
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
  $$begin{eqnarray*}
  y &=&frac{1}{2}int left( frac{p-x}{p}right) ^{-1}-left( frac{
  p-x}{p}right) dx \
  &=&-frac{1}{2}pln left( p-xright) -frac{1}{2}x+frac{1}{4p}x^{2}+C.
  end{eqnarray*}$$
  The same initial condition $x=0,y=0$ yields now
  $$begin{eqnarray*}
  C &=&frac{1}{2}pln left( pright) \
  && \
  y &=&-frac{1}{2}pln left( frac{p-x}{p}right) -frac{1}{2}x+frac{1}{4p}x^{2}.tag{11}
  end{eqnarray*}$$
  The chaser never overtakes the chased object.

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

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

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.

The intersection point can be found in the same way as here. Since the time at which $P$ overtakes $T$ is $t_0 = v p/(v^2 - u^2)$, $T$ will be at $(p, u t_0)$.