# MTH 101 Chapter 5: dis 5 sol

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School
Department
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Professor Calculus II Discussion Problems (Week 5) Solutions Spring 2021
1. Answer: Let v(t) be the velocity vector of the particle. Since
v(t)·v(t) = kv(t)k2
and the speed kv(t)kis assumed to be constant, we can diﬀerentiate 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.
2. Answer: Let a(t) be the acceleration vector. Then:
(r(t)×v(t))=r(t)×v(t) + r(t)×v(t)
=v(t)×v(t)
|{z }
=0
+r(t)×a(t) = r(t)×a(t)
Hence
(r(t)×v(t))·a(t) = (r(t)×a(t)) ·a(t) = 0
as desired. (Recall that r(t)×a(t) is orthogonal to both r(t) and a(t).)
aN(t) = κ(t)kr(t)k2
where r(t) is the position vector of the particle at time t. Hence, if t=t0is the parameter
corresponding to the point Pon the curve, then
aN(t0) = κ(t0)kr(t0)k2= 2 ·32=18.
r(t) = ti+at2j
of the parabola is smooth because r(t) = i+ (2at)jis continuous and never zero. There are
lots of other smooth parametrizations of the parabola, of course.
(b) Answer: We can readily calculate r(t) = i+ (2at)j,kr(t)k=1 + 4a2t2, and r(t) = (2a)j.
aT(t) = d
dtkr(t)k=4a2t
1 + 4a2t2
(d) Answer: Recall that Tand Nare always orthogonal. Hence, by the Pythagorean theorem,
kr(t)k=aT(t)2+aN(t)2
or
aN(t) = pkr(t)k2aT(t)2.
(Recall that aN(t)0 always.) Hence,
aN(t) = s(2a)24a2t
1 + 4a2t22
=2a
1 + 4a2t2.
1
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## Document Summary

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.