Calculating Time Difference in Pursuit Situations

[Saitama]

Homework Statement

Homework Equations

The Attempt at a Solution

Calculating $t_m$ is easy but I have no idea how would I calculate $t_s$. I understand what kind of path would Sven take but I don't know how to form equations in her case.

 

[Curious3141]

Hint: think of a circle centred at (100,0) with time-varying radius intersecting the y-axis at some point. The circle represents the locus of all the points that Margo might reach at time t. What conditions can you impose so that Margo reaches Ursula?

 

[Tanya Sharma]

What is the answer ?

 

[Saitama]

Hint: think of a circle centred at (100,0) with time-varying radius intersecting the y-axis at some point. The circle represents the locus of all the points that Margo might reach at time t. What conditions can you impose so that Margo reaches Ursula?

You are talking about Sven, right?

But I still don't have any idea about what to do.
The only thing I can write down is the equation of circle. $(x-100)^2+y^2=r^2$. Should this circle intersect the position of Ursula?

 

[Curious3141]

You are talking about Sven, right?

But I still don't have any idea about what to do.
The only thing I can write down is the equation of circle. $(x-100)^2+y^2=r^2$. Should this circle intersect the position of Ursula?

No, I'm talking about Margo (the one that knows physics). She's basically going to take a linear path that will directly intersect Ursula's. You know the speed this path will take (2m/s). Problem is - you don't know the direction (angle with x-axis). That's why you need to consider circles of time-varying radius to represent Margo's locus (think of a ripple in a pond traveling concentrically outward at fixed speed).

That equation looks good. At time t, what's r?

What's Ursula's position at time t?

Solve those simultaneously.

EDIT: I should maybe have said *set of circles* to avoid confusion.

 

[Curious3141]

You are talking about Sven, right?

But I still don't have any idea about what to do.
The only thing I can write down is the equation of circle. $(x-100)^2+y^2=r^2$. Should this circle intersect the position of Ursula?

Sorry, Pranav, maybe you should ignore what I said. I thought you had difficulty with Margo, not Sven. (I was confused because you used "her case", and Sven is a male name). Margo is definitely the easier case.

For Sven, I'll see if I can think of something (basically, the tangent to the curve always has to intersect Ursula's position).

 

[voko]

If Ursula is at (X, Y) and Sven at (x, y), what is the direction from Sven to Ursula? And if Sven always walks in that direction with a constant speed, what is Sven's velocity at that instant?

 

[Curious3141]

Homework Statement

Homework Equations

The Attempt at a Solution

Calculating $t_m$ is easy but I have no idea how would I calculate $t_s$. I understand what kind of path would Sven take but I don't know how to form equations in her case.

OK, I worked out an outline of how to handle Sven.

Set up a pair of simultaneous differential equations using what's known about Sven's trajectory:

First equation is the velocity relationship. Sven has an x component and a y component to his velocity at any point (x,y). The magnitude of the resultant (vector sum) is a constant. Hint: Pythagoras.

Second equation relates the tangent of Sven's path at (x,y) to the gradient of the line joining that point to Ursula's position (0,t).

I can reduce these equations to a slightly ungodly looking second order non-linear d.e. in terms of only y and x. Surprisingly, Wolfram assures me of a simple solution, which you can then use to calculate the t at intersection.

I don't know if this is the simplest way to do it, but it'll work.

 

[Saitama]

Sorry, Pranav, maybe you should ignore what I said. I thought you had difficulty with Margo, not Sven. (I was confused because you used "her case"...

Sorry about that.

First equation is the velocity relationship. Sven has an x component and a y component to his velocity at any point (x,y). The magnitude of the resultant (vector sum) is a constant. Hint: Pythagoras.

$v_x^2+v_y^2=4 \Rightarrow v_xdv_x=-v_ydv_y$

Second equation relates the tangent of Sven's path at (x,y) to the gradient of the line joining that point to Ursula's position (0,t).

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

I don't know what to do next.

 

[Curious3141]

Sorry about that. $v_x^2+v_y^2=4 \Rightarrow v_xdv_x=-v_ydv_y$

I didn't do quite that. I simplified that to $1 + (\frac{dy}{dx})^2 = 4(\frac{dt}{dx})^2$ by employing Chain Rule.

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

There, I brought the x over to the LHS, then differentiated wrt x to give $x\frac{d^2y}{dx^2} = -\frac{dt}{dx}$ after simplification. You can now square this, then sub it into the first equation to get a single second order d.e. However, you should be careful when you take the square root later, because you have to consider the correct root keeping in mind you squared the expression to begin with.

Wolfram solves this d.e. nicely, but there's a hitch- it gives two arbitrary constants, but there's only one point we're sure the curve passes through (100,0). So I'm not quite sure how to resolve this. But maybe you'll have better luck if you work it out by hand, because I haven't done this yet.

EDIT: Figured out how to work both constants out. That equation $x\frac{d^2y}{dx^2} = -\frac{dt}{dx}$ comes in handy here when you realize what happens right at the beginning of Sven's trajectory.

EDIT again: Darn it, I worked it out, and I'm getting a negative (and not "nice") result for Sven's time of intercept. But I can't see anything wrong with my reasoning. Anyway, it's late at night here, so I have to turn in. Hope you have better luck, and sorry I couldn't be of more help.

 

[Saitama]

But maybe you'll have better luck if you work it out by hand, because I haven't done this yet.

Sorry, but I don't know how to solve second order D.Es. :(

EDIT: Figured out how to work both constants out. That equation $x\frac{d^2y}{dx^2} = -\frac{dt}{dx}$ comes in handy here when you realize what happens right at the beginning of Sven's trajectory.

At beginning, Sven faces the origin?

 

[voko]

Sorry, but I don't know how to solve second order D.Es. :(

Before you say you can't, you should at least write it down.

At beginning, Sven faces the origin?

Why?

 

[Saitama]

Before you say you can't, you should at least write it down.

What I get is the following:
$$1+\left(\frac{dy}{dx}\right)^2=4x^2\left(\frac{d^2y}{dx^2}\right)^2$$

Why?

It is given in the question that Sven always faces Ursula until he reaches Ursula, so at t=0, Sven will face origin, right?

 

[Curious3141]

Sorry, but I don't know how to solve second order D.Es. :(At beginning, Sven faces the origin?

Yes, but from that you can deduce that $\frac{dx}{dt}$ = -2 so $\frac{dt}{dx}$ is the reciprocal of that. You can then differentiate the general solution of the d.e. twice and equate them to find out one constant (although it's not pretty).

But when I went through this, I got a negative final T, which can't be right.

This d.e. looks nasty to solve, but Wolfram did it easily. You can see the step by step solution here:

Go to link: http://www.wolframalpha.com/widgets/view.jsp?id=e602dcdecb1843943960b5197efd3f2a

and key in 2y"x = sqrt(1+(y')^2) and solve.

 

[Saitama]

Yes, but from that you can deduce that $\frac{dx}{dt}$ = -2 so $\frac{dt}{dx}$ is the reciprocal of that.

How do you even get that?

 

[voko]

What I get is the following:
$$1+\left(\frac{dy}{dx}\right)^2=4x^2\left(\frac{d^2y}{dx^2}\right)^2$$

So if you let $z = y'$, what do you get?

It is given in the question that Sven always faces Ursula until he reaches Ursula, so at t=0, Sven will face origin, right?

Sounds right.

 

[Curious3141]

How do you even get that?

If he's facing the origin, then he only has a horizontal (x) component of velocity, right? And that's equal to his speed of 2. Since it's facing along the negative x-direction, it's minus 2.

 

[Curious3141]

OK, had more time today. Solved the d.e. by hand, and it's actually very easy. Pranav - use Voko's hint to let $z = \frac{dy}{dx}$ and proceed from there.

I get a nice rational answer for Sven's time. I'm not sure why Wolfram was giving that solution, but the actual solution is much simpler by hand.

The difference in the times between Sven and Margo is about 9s (you should give the exact answer when you work it out).

 

[Saitama]

OK, had more time today. Solved the d.e. by hand, and it's actually very easy. Pranav - use Voko's hint to let $z = \frac{dy}{dx}$ and proceed from there.

Let $z=y'$, the D.E becomes
$$2x\frac{dz}{dx}=\sqrt{1+z^2}$$
Solving this gives:
$$x=c_1(\sqrt{1+z^2}+z)$$
Re-arranging and squaring both the sides
$$\left(\frac{x}{c_1}-z\right)^2=1+z^2$$
$$\Rightarrow \frac{x^2}{c_1^2}-1=\frac{2xz}{c_1}$$
$$\Rightarrow \frac{x}{c_1^2}-\frac{1}{x}=\frac{2z}{c_1}$$
Substituting z=dy/dx and solving the D.E
$$\frac{x^2}{2c_1^2}-\ln x=\frac{2y}{c_1}+c_2$$

Looks good?

How should I find the constants? Should I differentiate with respect to $t$?

 

[Curious3141]

Let $z=y'$, the D.E becomes
$$2x\frac{dz}{dx}=\sqrt{1+z^2}$$
Solving this gives:
$$x=c_1(\sqrt{1+z^2}+z)$$

There's a much simpler route to take here.

First thing - I always have a habit of imposing the boundary conditions (initial, final or whatever's known) as a bound when I do the integration. This way, you don't have to worry about constants later. You should consider doing this.

So what I had is $\int_0^z 2 d(\arcsinh z) = \int_{100}^x \frac{dx}{x}$

from which you can immediately get $2\arcsinh{z} = \ln{\frac{x}{100}}$ without worrying about constants.

Bring the 2 over to the RHS, and bring it up to the exponent: $\arcsinh{z} = \ln{(\frac{x}{100})}^{\frac{1}{2}}$

Now take sinh of both sides. Remember the exponential formula for sinh, allowing you to simplify the RHS very easily into a purely algebraic expression.

Re-arranging and squaring both the sides
$$\left(\frac{x}{c_1}-z\right)^2=1+z^2$$
$$\Rightarrow \frac{x^2}{c_1^2}-1=\frac{2xz}{c_1}$$
$$\Rightarrow \frac{x}{c_1^2}-\frac{1}{x}=\frac{2z}{c_1}$$
Substituting z=dy/dx and solving the D.E
$$\frac{x^2}{2c_1^2}-\ln x=\frac{2y}{c_1}+c_2$$

Looks good?

Didn't check this part, but you should be aware that I get a purely algebraic expression for y in terms of x (no logs). You might want to try the method I suggested, makes things far easier.

How should I find the constants? Should I differentiate with respect to $t$?

Again, try imposing the bounds when you integrate, and not plug in values later. Makes for neatness and fewer errors.

 

[Saitama]

There's a much simpler route to take here.

First thing - I always have a habit of imposing the boundary conditions (initial, final or whatever's known) as a bound when I do the integration. This way, you don't have to worry about constants later. You should consider doing this.

So what I had is $\int_0^z 2 d(\arcsinh z) = \int_100^x \frac{dx}{x}$

from which you can immediately get $2\arcsinh{z} = \ln{\frac{x}{100}}$ without worrying about constants.

Bring the 2 over to the RHS, and bring it up to the exponent: $\arcsinh{z} = \ln{(\frac{x}{100})}^{\frac{1}{2}}$

Now take sinh of both sides. Remember the exponential formula for sinh, allowing you to simplify the RHS very easily into a purely algebraic expression.

I tried your suggestion, this time I get:
$$z=5\left(\frac{\sqrt{x}}{100}-\frac{1}{\sqrt{x}}\right)$$
Substituting z=y' and integrating x from 100 to zero and y from 0 to t, I get $t_s=200/3$. Is this what you get?

Also,do you get, $t_m=100/\sqrt{3}$?

Thank you for your patience Curious!

 

[Curious3141]

I tried your suggestion, this time I get:
$$z=5\left(\frac{\sqrt{x}}{100}-\frac{1}{\sqrt{x}}\right)$$
Substituting z=y' and integrating x from 100 to zero and y from 0 to t, I get $t_s=200/3$. Is this what you get?

Exactly right.

Also,do you get, $t_m=100/\sqrt{3}$?

Also right!

The difference simplifies to $\frac{100(2 - \sqrt{3})}{3}$, which is just under 9s, as I mentioned.

Thank you for your patience Curious!

No worries, glad to help.

 

[Saitama]

Thank you so much Curious, I get a time difference of 8.93 seconds. :)

 

[Saitama]

Sorry but I should have asked this before.

I did not understand when you chose the limits in the following step:
$$\int_0^z 2d(\text{arcsinh} z)=\int_{100}^x \frac{dx}{x}$$

Can you please explain how did you choose those limits?

 

[Curious3141]

Sorry but I should have asked this before.

I did not understand when you chose the limits in the following step:
$$\int_0^z 2d(\text{arcsinh} z)=\int_{100}^x \frac{dx}{x}$$

Can you please explain how did you choose those limits?

Sure. The d(arcsinh z) is shorthand for the substitution I made (which was z = sinh θ). I didn't want to bring another variable into it, is all.

The lower bound is the initial condition. When x = 100 (start of Sven's path), the gradient of the tangent to the curve (which is z) is zero because he "points" at the origin. The upper bound is the time-variable one, this should be self-evident.

I find this technique quite concise, so I always use it rather than calculating constants after doing an indefinite integration. It's often used in Physics problems, BTW.

 

[Saitama]

Sure. The d(arcsinh z) is shorthand for the substitution I made (which was z = sinh θ). I didn't want to bring another variable into it, is all.

I got this already but I couldn't understand the way you chose the limits.

The lower bound is the initial condition. When x = 100 (start of Sven's path), the gradient of the tangent to the curve (which is z) is zero because he "points" at the origin. The upper bound is the time-variable one, this should be self-evident.

Thank you once again Curious!

 

[Curious3141]

You're welcome for both your thanks!

 

[Curious3141]

For interest's sake, here's the plot of Sven's (blue) and Margo's (red) trajectories (attached).

 

[Saitama]

For interest's sake, here's the plot of Sven's (blue) and Margo's (red) trajectories (attached).

Thanks, I too thought of the same trajectories. :)

Is it possible to find the equation of path taken by Sven? We had:
$$\frac{dy}{dx}=5\left(\frac{\sqrt{x}}{100}-\frac{1}{\sqrt{x}}\right)$$

Can I use this to find the equation of path taken by Sven?

 

[Curious3141]

Thanks, I too thought of the same trajectories. :)

Is it possible to find the equation of path taken by Sven? We had:
$$\frac{dy}{dx}=5\left(\frac{\sqrt{x}}{100}-\frac{1}{\sqrt{x}}\right)$$

Can I use this to find the equation of path taken by Sven?

Of course! I thought you had. :)

Just solve that d.e., it's a very simple separable equation, and you only need to know how to integrate algebraic expressions. Remember that when x = 100, y = 0, that's your lower bound for the integration.

I'm curious (like my name suggests), but if you didn't solve that d.e., how exactly did you arrive at the time for Sven? I solved that d.e., and that's how I found that time. That curve I plotted is exact, and uses that equation.

 

[Saitama]

I'm curious (like my name suggests), but if you didn't solve that d.e., how exactly did you arrive at the time for Sven? I solved that d.e., and that's how I found that time. That curve I plotted is exact, and uses that equation.

I did solve the D.E but when I plot the result, it doesn't look like your sketch. Solving the D.E and using x=100, y=0
$$y=\frac{1}{30}(x-300)\sqrt{x}+100$$
but plotting this on Wolfram Alpha doesn't give the sketch as shown in your attachment for Sven's path.

EDIT: Got it, I have some mistakes in the above equation. It should be 200/3 instead of 100. Thank you!

 

[haruspex]

Fwiw, I felt this question might be easier using local coordinates.
u = pursuer's speed, v = target's speed, s = distance between them, theta = angle between their current directions. Initially (s, theta) = (L, pi/2).
$s \dot \theta = - v \sin \theta$, $\dot s = v \cos \theta - u$
Whence $\frac{\dot s}s = \frac{\dot \theta}{\sin \theta} ( \frac uv - \cos \theta)$
$\ln s = \frac u{2v} \ln \left(\frac{1-\cos\theta}{1+\cos\theta}\right) + \ln L$
$s = \frac L{\sin\theta}\left(\frac{1-\cos\theta}{1+\cos\theta}\right)^{\frac u{2v}}$
Using (1) again: $\dot \theta \frac L{\sin^2\theta}\left(\frac{1-\cos\theta}{1+\cos\theta}\right)^{\frac u{2v}} = - v$
Since u = 2v,
$vT = L\int_0^{\pi/2} \frac{(1-\cos \theta).d\theta}{(1+\cos\theta)\sin^2\theta}$
$= L\int^{\pi/2}_0 \frac{d\theta}{(1+\cos\theta)^2}$
Writing 2x = theta
$= L\int^{\pi/4}_0 \frac{dx}{2 \cos^4x}$
$= L\frac16 \left[\tan^3x+3\tan x\right]^{\pi/4}_0 = 2L/3$

 

[Saitama]

$s \dot \theta = - v \sin \theta$, $\dot s = v \cos \theta - u$

Sorry but I really can't follow this, can you please explain how you got those two equations?

$\dot{s}$ represents the velocity but how do you get the RHS of the second equation? And I don't have any idea about the first equation. $s\dot{\theta}$ has dimensions of velocity but I am confused.

 

[haruspex]

Sorry but I really can't follow this, can you please explain how you got those two equations?

$\dot{s}$ represents the velocity but how do you get the RHS of the second equation? And I don't have any idea about the first equation. $s\dot{\theta}$ has dimensions of angular velocity but I am confused.

In the line joining the two, target is moving away from pursuer with speed v cos θ, while pursuer advances with speed u: $\dot s=vcosθ−u$. In the perpendicular direction, the target moves with speed $v \sin \theta$. From the pursuer's perspective that's a tangential speed, tending to reduce theta. The 'radius' is s: $v \sin \theta = -s \dot \theta$ (as in the familiar v = rω).

 

[Saitama]

In the line joining the two, target is moving away from pursuer with speed v cos θ, while pursuer advances with speed u: $\dot s=vcosθ−u$. In the perpendicular direction, the target moves with speed $v \sin \theta$. From the pursuer's perspective that's a tangential speed, tending to reduce theta. The 'radius' is s: $v \sin \theta = -s \dot \theta$ (as in the familiar v = rω).

Thanks a lot haruspex!