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Unformatted text preview: not be negative, so the lowest we can
make it is zero. So the absolute minimum squared p
distance (hence minimum distance) is achieved
Asst #3 Due
precisely when t = 2. The distance at t = 2 is d(2) = 2(22 ) 8(2) + 17 = 9 = 3. An alternative way is to use calculus to compute the derivative of s(t) = 2t2 8t + 17 as s0 (t) = 4t 8,
note that it is negative when t < 2, vanishes at t = 2, and is positive when t > 2. So we see that
the function decreases when 1 < t 2, and then increases when 2 t < 1. So there is a global
minimum at t = 2, and...
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