# Moving target

1. Nov 15, 2013

### GregoryGr

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

A boy is on a boat, at a distance H from the shore, when he sees a girl (at the point on the shore where the distance is measured) running with a constant velocity u parallel to the shore. At that time, he moves towards her, with a speed v, in such a way, that the point of the boat is always pointed at the girl (so his vector is always pointing her way). Find the time of their meeting.

2. Relevant equations

$$\vec{v}= \frac{dr}{dt}\hat{r}+r\frac{d\theta}{dt}\hat{\theta}$$

3. The attempt at a solution

I changed the system of coordinates so the girl is stationary at O' , and using polar coordinates since the vector of the boat is always in the radial direction. The velocity for the boat in polar coordinates can be written:
$$\vec{v}= -(v+ucos\theta)\hat{r}+vsin\theta\hat{\theta}$$

And since the 2 quantities in front of the singular vectors must be the same, I get 2 differential equations which give me this:

$$\frac{dr}{r}= \frac{-(v-ucos\theta)d\theta}{usin\theta}$$

I am a noob at solving differentials, so firstly, are my equations right, and second, can somebody help me figure out how to do the math?

2. Nov 16, 2013

### haruspex

Yes, I agree with your first two equations, and I think the third is right. Once you have a solution you can check it satisfies the second one.
If you expand the numerator, the second term is easy to integrate. The first becomes the integral of a cosecant, which is standard enough that you can look it up. If you're interested in doing it from first principles, multiply top and bottom by sine and convert the sin2 in the denominator to 1-cos2. You can then factorise that and expand it using partial fractions. Each fraction is readily integrated.

3. Nov 16, 2013

### ehild

I think there should be "u" instead of v in front of sinθ. Check.

ehild

Last edited: Nov 16, 2013
4. Nov 16, 2013

### haruspex

Yes, I saw that, but it must be a transcription error since it is correct in the later equation.

5. Nov 16, 2013

### ehild

A figure would be necessary. I assume that the girl runs to the left so the boy has a horizontal velocity component u to the right. See figure. That component and the boy's radial velocity component v add up and determine the boy's next position.
You see that the horizontal component u makes r increase, but θ decrease.

In case the girl run to the right, the boy would move in the second quadrant, the angle θ would increase as the boy approaches the girl.

What is your set-up? Check the signs.

ehild

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6. Nov 17, 2013

### ehild

There is a standard substitution when you have to integrate a rational function of sines and cosines. All trigonometric functions can be written in terms of tan(x/2), and then the substitution z=tan(x/2), dx = 2/(1+z2) dz can be used.

$$\sin(x)= \frac {2\tan(x/2)}{1+\tan^2(x/2)}$$
$$\cos(x)= \frac {1-\tan^2(x/2)}{1+\tan^2(x/2)}$$

$$\int {\frac{1}{\sin(x)}= \frac {1+\tan^2(x/2)}{2\tan(x/2)}dx}= \int {\frac {1+z^2}{2z}2 \frac{1}{1+z^2}dz}=\int {\frac {1}{z} dz}= \log|tan(x/2)|+C$$

ehild