# MATH 32A Study Guide - Midterm Guide: Square Root, Quotient Rule, Talking Lifestyle 1278Exam

by OC2540294

Department

MathematicsCourse Code

MATH 32AProfessor

AllStudy Guide

MidtermThis

**preview**shows pages 1-2. to view the full**6 pages of the document.**MATH 32A (Butler)

Practice for Midterm II (Solutions)

1. A particle moves through three dimensional space with velocity

v(t) = hsec2t, 2 sec ttan t, tan2ti.

At time t= 0 the particle is at h0,1,2i, ﬁnd the position function of the particle for

−π/4≤t≤π/4.

If r(t) is the position function then r0(t) = v(t). So taking antiderivatives we

have

r(t) = Zv(t)dt =Zsec2t dt, Z2 sec ttan t dt, Ztan2t dt

=tan t+C, 2 sec t+D, Z(sec2t−1) dt

=tan t+C, 2 sec t+D, tan t−t+E.

Two of the three integrals are straightforward. The last one is the trickiest

but this follows by relating tan2t(something which we cannot directly inte-

grate) to sec2t(something which is easy to integrate). Now all that is left is

to determine the constants C, D, E. We have

r(0) = hC, 2 + D, Ei=h0,1,2i

giving the constants we need. So we have that the position function of the

particle is

r(t) = tan t, 2 sec t−1,tan t−t+ 2.

(The condition for −π/4≤t≤π/4 is not needed directly, it only is used to

guarantee that we stay away from the vertical asymptotes (solutions cannot

be pushed past vertical asymptotes, something you will learn in your future

math classes).)

Only pages 1-2 are available for preview. Some parts have been intentionally blurred.

2. Find the cumulative length function s(t) (starting from a= 1) of the parametric

curve hln t, √2t, 1

2t2i.

We have that the cumulative arc length function will be

s(t) = Zt

ar0(u)du.

We are told that a= 1 and we now compute the derivative. We have

r0(t) = 1

t,√2, t.

Therefore

s(t) = Zt

11

u,√2, u

du

=Zt

1s1

u2

+√22+u2du

=Zt

1r1

u2+2+u2du

=Zt

1s1

u+u2

du

=Zt

11

u+udu

=ln u+1

2u2

t

1

= ln t+1

2t2−1

2.

(Frequently in this type of problem the functions will be chosen so that the

terms on the inside of the square root “miraculously” combine into a perfect

square. Of course this is because they have been rigged to do so and would

not happen by coincidence.)

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