# How to find the length of a curved path.

We have to find a continuous model for a curved path which you then solve. A woman is running in the positive y-direction starting at x=50 (50,0) which is orthogonal to the x axis. At this point a dog starts running toward the woman from (0,0) they are both running at constant speed, the dogs path is curved and we wish to find the length of the curve until the dog reaches the woman. We need to use the dogs position the woman's position and the gradient of the dogs location to find the model. How would I go about doing this? Thank you.

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Welcome to Math.SE. Here, we want to know what you, the poster of the question, have done to make progress on the question. Please supply some details as to what you have done toward getting a solution. – Ron Gordon Mar 31 '13 at 11:36

At this point a dog starts running toward the woman from (0,0) they are both running at constant speed, the dogs path is curved and we wish to find the length of the curve until the dog reaches the woman.

Path length is given by

$$s = \int_0^T ds = \int_0^T \frac{ds}{dt} dt = \int_0^T v(t) dt = \Big[ v(t) t \Big]_0^T - \int_0^T v'(t) t dt = v(T)T - \int_0^T v'(t)tdt,$$

and if the speed is constant we obtain

$$s = v T.$$

As the speed is constant for both the dog and the woman, we obtain

$$s_w = v_w T$$

and

$$s_d = v_d T,$$

therefore

$$s_d = \frac{v_d}{v_w} s_w,$$

i.e. the ratios of the path length is the ratios of the speeds, assuming the $v_d > v_w$.

If you want to know the path of the dog - the solution is more complex.

# The path

The situation is desribed as

$$\dot{x} = \alpha (50 - x)\\ \dot{y} = \alpha (v_wt - y)$$

And constant speed implies that

$$\dot{x}^2 + \dot{y}^2 = v_d^2,$$

so

$$\alpha = \frac{v_d}{\sqrt{\big( 50 - x \big)^2 + \big(v_wt - y\big)^2}}.$$

Whence

$$\dot{x} = v_d \frac{50 - x}{\sqrt{\big( 50 - x \big)^2 + \big(v_wt - y\big)^2}}\\ \dot{y} = v_d \frac{v_wt - y}{\sqrt{\big( 50 - x \big)^2 + \big(v_wt - y\big)^2}}$$

and I wonder if we can solve this...

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Hint : Do not try to find the equation of curve.It will be complicated and unnecessary. Try to take relative velocities with respect to the dog and the qoman and find the time taken for the dog to reach the woman. As it's moving with constant speed, you can then find the distance.

Ok, might as well post a complete solution. Let the dog's speed be $u$ and the woman's $v$ , the distance between them $D$.

Let $\theta$ be the angle made by the line joining the dog aand the woman and the direction of motion of the woman. Let $T$ be the total time.

We have, $$\int_0^Tu \cos \theta dt=vT$$ and $$\int_0^Tu-v\cos\theta dt=D$$ Solve to find $T$. What you want is $uT$. The reason we have the 1st equation is by looking at the distance moved in the direction of the woman's motion. The 2nd is arrived at by looking at the velocity of approach from the woman's point of view.

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We don't know what the speed of any of them is or anything other than the initial points of the dog and woman. – Felix Mar 31 '13 at 11:47
thank you very much but we have to include a gradient of the curve, the dogs position and the woman's position – Felix Mar 31 '13 at 12:06

Sounds like this is a variant of the prototypical pursuit curve. In this case you're asked for the length of the pursuit curve. Researching pursuit curves, you can find that the working equation (formed by stipulating that the dog is always running towards the woman's position) is:

dy/dx = (y - w t)/(x - 50), where y(x) is the position of the dog and w is the velocity of the woman.

You can also research arc lengths (s) to find:

s = int((1 + (dy/dx)^2)^.5

with these two equations (and a bunch of work) you should be able to find the length of the path (arc) s as a function of time, t.

A good reference for this problem, and related ones, is:

http://home2.fvcc.edu/~dhicketh/DiffEqns/Spring11projects/Jonah_Franchi_Katy_Steiner/Diff%20EQ%20Project.pdf

Paul Safier

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Latex is your friend :)! – Chinny84 Jul 5 '14 at 11:40