D.E with Physics Included
Thread starter dearcomp
dearcomp
I come up with a new question which I can't express as a differential equation form.
2 cars start on x axis, M at the origin and N at the point (36,0)
Suppose:
M moves along the y axis,
and N moves directly toward M at all times,
and N moves twice as fast as M.
Q) How far will M travel before being caught by N?
I think the trick is in the bold part of the question but I couldn't write it as a D.E as I said before
Klaas van Aarsen
MHB Seeker
dearcomp said:
Suppose M moves at speed v, then N moves at speed 2v.
Let (x,y) be the position of N at some time t.
Then the position of M at some time t is (0,vt).
What will be the direction of N at time t?
Suppose at time t, N drives for a small (infinitesimal) time dt in the direction of M.
What will his (infinitesimal) change dx in his x coordinate be?
And what will his change dy in his y coordinate be?
I like Serena said:
The direction of N at time t would be the vector from N to M, which is
the vector from (x,y) to (0,vt) = -x.i + (vt-y).j
I think, dx would be 2v*(x component of the direction vector) which is
2v*(-x)
respectively, dy would be 2v*(vt-y) ?
$$dx = 2v \cdot (\text{x component of the 'normalized' direction vector}) \cdot dt$$
That should give you a system of 2 first order differential equations.
So, dx = 2v*(-x) / sqrt((-x)^2+(vt-y)^2) dt
dy = 2v*(vt-y) / sqrt((-x)^2+(vt-y)^2) dt
and the question asks dy. Am I right?
Yes. That looks right.
I believe the question asks for the y such that y=vt (and x=0).
dx/dt =2v*(-x) / sqrt((-x)^2+(vt-y)^2)
dy/dt = 2v*(vt-y) / sqrt((-x)^2+(vt-y)^2)
we can get dy/(vt-y) = -dx/x with some math, and when I integrate both sides, I get
vt-y = x*e^c
and the question asks for y
so, the answer would be y = vt-x*e^c wouldn't it?
My question comes: we haven't used the point (36,0) anywhere..
we can get dy/(vt-y) = -dx/x with some math,
True enough.
and when I integrate both sides, I get
vt-y = x*e^c and the question asks for y
The left part does not integrate because y = y(t) is a function of t.
The problem is a boundary value problem.
As part of the solution, you need that x(0)=36, y(0)=0, x(T)=0, y(T)=vT, where T is the time at which N intercepts M.
At first I thought the set of DE's could be solved in closed form.
But to be honest, I failed to do so at this time.
It looks like a tractrix, which can be solved. But alas, it is not a tractrix.
Solving numerically gives us that $y(T) \approx 24$.
I have copied a solution I gave to a similar problem, which has the coordinate system defined such that the pursuer begins at the origin:
An interesting geometric model arises when one tries to determine the path of a pursuer chasing its "prey." This path is called a curve of pursuit. These problems were analyzed using methods of calculus circa 1760 (more than two centuries after Leonardo da Vinci had considered them).
The simplest problem is to find the curve along which a vessel moves when pursuing another vessel that flees along a straight line, assuming the speeds of the two vessels are constant.
Let's assume car $N$, traveling at speed $\beta$, is pursuing car $M$, which is traveling at speed $\alpha$. In addition, assume that car $N$ begins (at time $t=0$) at the origin and pursues car $M$, which begins at the point $(b,0)$ where $0<b$ and travels down the line $x=b$.
After $t$ hours, car $N$ is located at the point $P(x,y)$, and car $M$ is located at the point $Q(b,-\alpha t)$. The goal is then to describe the locus of points $P$, that is, to find $y$ as a function of $x$.
(a) Since car $N$ is pursuing car $M$, then at time $t$, car $N$ must be heading right at car $M$. That is, the tangent line to the curve of pursuit at $P$ must pass through the point $Q$. Using the point-slope formula, this implies:
\(\displaystyle y+\alpha t=\frac{dy}{dx}(x-b)\)
(1) \(\displaystyle \frac{dy}{dx}=\frac{y+\alpha t}{x-b}\)
(b) Since we know the speed at which car $N$ is traveling, we know the distance it travels is $\beta t$. This distance is also the length of the pursuit curve from $(0,0)$ to $(x,y)$. Using the arc length formula from calculus, we find that:
(2) \(\displaystyle \beta t=\int_0^x \sqrt{1+\left(y'(u) \right)^2}\,du\)
Solving for $t$ in equations (1) and (2), we find that:
(3) \(\displaystyle (x-b)\frac{dy}{dx}-y=\frac{\alpha}{\beta}\int_0^x \sqrt{1+\left(y'(u) \right)^2}\,du\)
(c) Differentiating both sides of (3) with respect to $x$, we find:
\(\displaystyle \left((x-b)\frac{d^2y}{dx^2}+\frac{dy}{dx} \right)-\frac{dy}{dx}=\frac{\alpha}{\beta}\sqrt{1+\left( \frac{dy}{dx} \right)^2}\)
Letting \(\displaystyle \omega=\frac{dy}{dx}\) we obtain the first order IVP:
\(\displaystyle (x-b)\frac{d\omega}{dx}=\frac{\alpha}{\beta}\sqrt{1+ \omega^2}\) where \(\displaystyle \omega(0)=0\)
(d) Using separation of variables, and switching the dummy variables of integration so that we may use the boundaries as the limits, we find:
\(\displaystyle \int_0^{\omega}\frac{1}{\sqrt{1+u^2}}\,du= \frac{\alpha}{\beta}\int_0^x\frac{1}{v-b}\,dv\)
Integrating, we find:
\(\displaystyle \left[\ln|u+\sqrt{1+u^2}| \right]_0^{\omega}=\frac{\alpha}{\beta}\left[\ln|v-b| \right]_0^x\)
\(\displaystyle \ln|\omega+\sqrt{1+\omega^2}|=\ln\left|\left(1-\frac{x}{b} \right)^{\frac{\alpha}{\beta}} \right|\)
Since \(\displaystyle -\infty\le \omega<0\) and \(\displaystyle 0<1-\frac{x}{b}\) we may state:
\(\displaystyle \omega+\sqrt{1+\omega^2}=-\left(1-\frac{x}{b} \right)^{\frac{\alpha}{\beta}}\)
\(\displaystyle \left(\sqrt{1+\omega^2} \right)^2=\left(\left(1-\frac{x}{b} \right)^{\frac{\alpha}{\beta}}+\omega \right)^2\)
\(\displaystyle 1+\omega^2=\left(1-\frac{x}{b} \right)^{\frac{2\alpha}{\beta}}+2\omega\left(1-\frac{x}{b} \right)^{\frac{\alpha}{\beta}}+\omega^2\)
\(\displaystyle 2\omega\left(1-\frac{x}{b} \right)^{\frac{\alpha}{\beta}}=1-\left(1-\frac{x}{b} \right)^{\frac{2\alpha}{\beta}}\)
Back-substituting for $\omega$, we have the IVP:
\(\displaystyle \frac{dy}{dx}=\frac{1}{2}\left(\left(1-\frac{x}{b} \right)^{-\frac{\alpha}{\beta}}-\left(1-\frac{x}{b} \right)^{\frac{\alpha}{\beta}} \right)\) where \(\displaystyle y(0)=0\)
Switching the dummy variables of integration so we can use the boundaries, we have:
\(\displaystyle \int_0^y\,du=\frac{1}{2}\int_0^x\left(1-\frac{v}{b} \right)^{-\frac{\alpha}{\beta}}-\left(1-\frac{v}{b} \right)^{\frac{\alpha}{\beta}}\,dv\)
\(\displaystyle _0^y=\frac{b}{2}\left[\frac{\left(1+\frac{v}{b} \right)^{1+\frac{\alpha}{\beta}}}{1+\frac{\alpha}{\beta}}-\frac{\left(1-\frac{v}{b} \right)^{1-\frac{\alpha}{\beta}}}{1-\frac{\alpha}{\beta}} \right]_0^x\)
After simplifying, we obtain:
\(\displaystyle y=\frac{b\beta}{2(\alpha^2-\beta^2)}\left((\beta-\alpha)\left(1-\frac{x}{b} \right)^{\frac{\beta+\alpha}{\beta}}-(\beta+\alpha)\left(1-\frac{x}{b} \right)^{\frac{\beta-\alpha}{\beta}}+2\alpha \right)\)
Plugging in the given data $b=36,\,\beta=2\alpha,\,x=b$ we find:
\(\displaystyle y=\frac{36(2\alpha)}{2(\alpha^2-(2\alpha)^2)}\left((2\alpha-\alpha)\left(1-\frac{36}{36} \right)^{\frac{2\alpha+\alpha}{2\alpha}}-(2\alpha+\alpha)\left(1-\frac{36}{36} \right)^{\frac{2\alpha-\alpha}{2\alpha}}+2\alpha \right)\)
\(\displaystyle y=\frac{12}{-\alpha}\left(2\alpha \right)=-24\)
Thus, the distance $d$ traveled by car $M$ is:
\(\displaystyle d=|y|=24\)