# Calculating time difference

1. Jun 12, 2013

### Saitama

1. The problem statement, all variables and given/known data

2. Relevant equations

3. 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.

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2. Jun 12, 2013

### 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?

3. Jun 12, 2013

### Tanya Sharma

4. Jun 12, 2013

### Saitama

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?

5. Jun 12, 2013

### Curious3141

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 travelling 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.

6. Jun 12, 2013

### Curious3141

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).

7. Jun 12, 2013

### 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?

8. Jun 12, 2013

### Curious3141

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.

9. Jun 12, 2013

### Saitama

$v_x^2+v_y^2=4 \Rightarrow v_xdv_x=-v_ydv_y$
$$\frac{dy}{dx}=\frac{y-t}{x}$$

I don't know what to do next.

10. Jun 12, 2013

### Curious3141

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

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 realise 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.

Last edited: Jun 12, 2013
11. Jun 12, 2013

### Saitama

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

At beginning, Sven faces the origin?

12. Jun 12, 2013

### voko

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

Why?

13. Jun 12, 2013

### Saitama

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

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

14. Jun 12, 2013

### Curious3141

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:

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

15. Jun 12, 2013

### Saitama

How do you even get that?

16. Jun 12, 2013

### voko

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

Sounds right.

17. Jun 12, 2013

### Curious3141

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.

18. Jun 12, 2013

### 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).

Last edited: Jun 12, 2013
19. Jun 12, 2013

### Saitama

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 $t$?

20. Jun 12, 2013

### Curious3141

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.

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.

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

Last edited: Jun 13, 2013