# MTH 101 Chapter 5: dis 5 sol

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).)

3. Answer: Recall

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.

4. (a) Answer: The parametrization

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.

(c) Answer:

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)k2−aT(t)2.

(Recall that aN(t)≥0 always.) Hence,

aN(t) = s(2a)2−4a2t

√1 + 4a2t22

=2a

√1 + 4a2t2.

1

## 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.