MTH 101 Study Guide - Final Guide: Pythagorean Theorem, Level Set

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18 Jun 2021
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Spring 2021: answer: let v(t) be the velocity vector of the particle. Since v(t) v(t) = kv(t)k2 and the speed kv(t)k is assumed to be constant, we can di erentiate both sides to obtain. 2(v(t) v (t)) = 0 or simply v(t) v (t) = 0. i. e. , the velocity vector and the acceleration vector are orthogonal: answer: let a(t) be the acceleration vector. Then: (r(t) v(t)) = r (t) v(t) + r(t) v (t) + r(t) a(t) = r(t) a(t) 1 + 4a2t2 (d) answer: recall that t and n are always orthogonal. Hence, by the pythagorean theorem, or (recall that an (t) 0 always. ) Hence, kr (t)k = at (t)2 + an (t)2 an (t) = pkr (t)k2 at (t)2. an (t) = s(2a)2 (cid:18) Spring 2021 (e) answer: using the fact kr (t)k = 1 + 4a2t2, we obtain.

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