Two boats are racing toward a finish marker Boat A is sailin

Solution

Let x be the distance boat B is from the finish marker, y be the distance boat A is from the finish maker, and s be the distance between boat A and B. Then: x^2 + y^2 = s^2 Take the derivative of both sides with respect to t (remember to use the chain rule!): 2x*dx/dt + 2y*dy/dt = 2s*ds/dt x*dx/dt + y*dy/dt = s*ds/dt We\'re given that dy/dt = 13, ds/dt = 17, s = 16, and x = y. Plug those in: x*dx/dt + x*13 = 16*17 x*dx/dt + 13x = 272 x(dx/dt + 13) = 272 In order to find x, remember that x = y and that at the point in time that we\'re considering, the distance between each other, s, is 16. So we can say that: x^2 + y^2 = s^2 x^2 + x^2 = 16^2 2x^2 = 256 x^2 = 128 x = ?128 = 8?2 Plugging this into the equation we have, we get: x(dx/dt + 13) = 272 8(?2)(dx/dt + 13) = 272 (?2)(dx/dt + 13) = 34 dx/dt + 13 = 34/?2 dx/dt = (34/?2) - 13 ? 11.402 dx/dt ? 11.402 and dy/dt = 13. dy/dt is bigger than dx/dt so boat A will win the race.