s57 - Scooter reaches the corner, my coordinate is (0,8)....

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Solution to Problem 57 Congratulations to this week’s winners Nathan Pauli, Ray Kremer Correct solutions were also received from Jerry Callahan, Tracy Thatcher. Fruther correct solutions were also received from William Webb, Tim Kelley, Al Zimmermann, Burkart Venzke. All solutions reasonably assumed that the leash is parallel to the ground. Without this assumption, the calculations would be similar, but would involved a vertical component as well. Let’s call the distance between me and the corner x and the distance, after his turn, between Scooter and the corner y . We can set up a coordinate system so that just as
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Unformatted text preview: Scooter reaches the corner, my coordinate is (0,8). The length of the leash is L , where L 2 = x 2 + y 2 Differentiating with respect to time gives 2 L ( dL/dt ) = 2 x ( dx/dt ) + 2 y ( dy/dt ). When Scooter reaches the corner we have x = 8 ft, y = 0 ft, L = 8 ft, dx/dt = -6 ft/sec, dy/dt = 6 ft/sec. (The negative sign on dx/dt comes from the fact that x is decreasing.) Plugging in these terms gives that, when Scooter reaches the corner, the length of his leash is decreasing at a rate of 6 ft/sec. To find the minimum length, set dL/dt = 0 and solve to get x = y = 4, L = 4 Ö2 » 5.65 ft....
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This note was uploaded on 03/01/2010 for the course MATH 301 taught by Professor Albertodelgado during the Spring '10 term at Bradley.

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s57 - Scooter reaches the corner, my coordinate is (0,8)....

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