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.
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.
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: