Pursuit Curves

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Solving the physics problem pertaining to Pursuit Curves. (Krantz)

* .... Differential Equations for solving physics problems.

* .... have method that we know to solve these equations.

Formulation: Suppose a Destroyer/Warship was to pursue a submarine on a particular prescribed path. What path does the warship use to follow the submarine? Well, looking at the diagram below we see that the submarine begins at the origin and travels up the y-axis with velocity a ft/s. At the same time a warship destroyer travels at speed b ft/s from point (c,0) in pursuit of the sub. What is the destroyer's path to catch the submarine? (What is its pursuit curve?)

"The premise of a pursuit analysis is that the line through D and S is tangent to the path - that is, the destroyer will always run straight at the submarine."(Krantz)

At time t0 (initial time) the submarine will be at the pt S = (0, at) and the destroyer at D = (x, y). We seek to find y(x) the pursuit path!

The calculation depicted below may seem a little blurry, but I'll make it clear. Starting with dy/dx which is the change in position of the sub divided by the change in position of the destroyer the challenge is to manipulate the variables such that we solve for xy' and xy": so we get xy'- y = -at and taking second derivative with respect to x ... xy" = a * dt/dx

Next we use the arc length of the destroyers path as the hypotenuse of a right triangle with delta x and delta y as sides. The velocity of the destroyer is b = ds/dt and using the x component of the destroyer velocity along with the chain rule: dx/dt = (dx/ds)*(ds/dt) leading to dx/dt = -b * (dx/ds) and inverting this leads to dt/dx = -(1/b)*(ds/dx)

Plugging ds/dx into previous formula yields: dt/dx = -(1/b) (1 + (y')^^2)^^1/2

xy" = -a dt/dx (and using dt/dx above ... xy" = a/b (1 + (y')^^2)^^1/2 (

a/b = xy"/ [(1 + (y')^^2)^^1/2]

After the substitution indicated q = y', dq = y", and l = a/b we integrate the expression below ...

We now solve dy/dx for dy and integrate both sides yielding the expression for the pursuit curve y(x) below:

We have furthermore made the assumption that the destroyer will catch the submarine due to being the more powerful vessel.

Back to the Hanging Chain ... Hanging Chain

Go back to Gauss' Law Problem from here!

1st Gauss' Law Example problem ... Gauss Law Example 1.

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You can go to DE Ex2 here...D.E. Ex2.

You can go to DE Ex1 here ...D.E. Ex1.