(a) Vessel A is pursuing vessel B, so at time t, vessel A must be heading right at vessel B. That is, the tangent line to the curve of pursuit at P must pass through the point Q (see Figure 3.18). For this to be true, show that (4) (b) We know the speed at which vessel A is traveling, so we know that the distance it travels in time t is at. This distance is also the length of the pursuit curve from (0, 0) to (x, y). Using the arc length formula from calculus, show that dy dx = (5) at = (6) y - Bt x-1 = 5₁² V1 + [y' (u)]² du. Solving for t in equations (4) and (5), conclude that y-(x-1) (dy/dx) = == S V₁ + [y' (u)]³² du . В (c) Differentiating both sides of (6) with respect to x, derive the first-order equation dw В (x-1)- VI+w², dx α where w = dy/dx. P = (x₂) -BQ = (1, 3t) (1, 0) X

An interesting geometric model arises when one tries to determine the path of a pursuer chasing 
its prey. This path is called a curve of pursuit. These problems were analyzed using methods of 
calculus circa 1730 (more than two centuries after Leonardo da Vinci had considered them). The 
simplest problem is to find the curve along which a vessel moves in pursuing another vessel that 
flees along a straight line, assuming the speeds of the two vessels are constant.
Let’s assume that vessel A, traveling at a speed a, is pursuing vessel B, which is traveling at a 
speed b. In addition, assume that vessel A begins (at time t = 0) at the origin and pursues vessel 
B, which begins at the point (1, 0) and travels up the line x = 1. After t hours, vessel A is located 
at the point P = 1x, y2, and vessel B is located at the point Q = 11, bt2 (see Figure 3.18). The 
goal is to describe the locus of points P; that is, to find y as a function of x.

Please look at images for questions. The formatting isn't correct if i type it out.