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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.
- Spring '10