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Exam 3 Version 1 with Solution - Calculus III | MATH 2203, Exams of Advanced Calculus

Material Type: Exam; Professor: Ellermeyer; Class: Calculus III; University: Kennesaw State University; Term: Spring 2011;

Typology: Exams

2010/2011

Uploaded on 06/03/2011

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MATH 2203 –Exam 2 (Version 1) Solutions
February 23, 2011
S. F. Ellermeyer Name
Instructions. Your work on this exam will be graded according to two criteria: mathe-
matical correctness and clarity of presentation. In other words, you must know what
you are doing (mathematically) and you must also express yourself clearly. In particular,
write answers to questions using correct notation and using complete sentences where
appropriate. Also, you must supply su¢ cient detail in your solutions (relevant calculations,
written explanations of why you are doing these calculations, etc.). It is not su¢ cient to just
write down an “answer” with no explanation of how you arrived at that answer. As a rule
of thumb, the harder that I have to work to interpret what you are trying to say, the less
credit you will get. You may use your calculator but you may not use any books or notes.
1. For the motion of a particle whose position at time tis given by
r(t) = 2 cos (t)i+ 3 sin (t)j+ 4tk,
show how to compute
(a) the velocity vector, v(t)
(b) the acceleration vector, a(t)
(c) the speed, v(t)
(d) the unit tangent vector (also called the direction of motion), T(t)
Solution: The velocity vector is
v(t) = dr
dt =2 sin (t)i+ 3 cos (t)j+ 4k:
The acceleration vector is
a(t) = dv
dt =2 cos (t)i3 sin (t)j:
The speed is
v(t) = jv(t)j=q4 sin2(t) + 9 cos2(t) + 16.
The unit tangent vector is
T(t) = 1
v(t)v(t) = 1
p4 sin2(t) + 9 cos2(t) + 16 (2 sin (t)i+ 3 cos (t)j+ 4k).
2. What two angles of elevation will enable a projectile to reach a target 16 km downrange
on the same level as the gun if the projectile’s initial speed is 800 m/sec? (Take the
acceleration due to gravity to be 9:8m/sec2.)
(You must include all details needed to arrive at your answer to this question.)
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MATH 2203 ñExam 2 (Version 1) Solutions February 23, 2011 S. F. Ellermeyer Name

Instructions. Your work on this exam will be graded according to two criteria: mathe- matical correctness and clarity of presentation. In other words, you must know what you are doing (mathematically) and you must also express yourself clearly. In particular, write answers to questions using correct notation and using complete sentences where appropriate. Also, you must supply su¢ cient detail in your solutions (relevant calculations, written explanations of why you are doing these calculations, etc.). It is not su¢ cient to just write down an ìanswerî with no explanation of how you arrived at that answer. As a rule of thumb, the harder that I have to work to interpret what you are trying to say, the less credit you will get. You may use your calculator but you may not use any books or notes.

  1. For the motion of a particle whose position at time t is given by

r (t) = 2 cos (t) i + 3 sin (t) j + 4tk,

show how to compute

(a) the velocity vector, v (t) (b) the acceleration vector, a (t) (c) the speed, v (t) (d) the unit tangent vector (also called the direction of motion), T (t)

Solution: The velocity vector is

v (t) = dr dt = 2 sin (t) i + 3 cos (t) j + 4k:

The acceleration vector is

a (t) = dv dt

= 2 cos (t) i 3 sin (t) j:

The speed is v (t) = jv (t)j =

q 4 sin^2 (t) + 9 cos^2 (t) + 16. The unit tangent vector is

T (t) =

v (t) v (t) =

p 4 sin^2 (t) + 9 cos^2 (t) + 16

(2 sin (t) i + 3 cos (t) j + 4k).

  1. What two angles of elevation will enable a projectile to reach a target 16 km downrange on the same level as the gun if the projectileís initial speed is 800 m/sec? (Take the acceleration due to gravity to be 9 : 8 m/sec^2 .) (You must include all details needed to arrive at your answer to this question.)

Solution: Based on formulas that we have derived in class several times, the x and y coordinates of the projectile at time t are

x = 800 cos ( ) t y = 800 sin ( ) t

gt^2

where is the Öring angle. Since we want to have y = 0 when x = 16; 000 , we must Önd t such that 800 cos ( ) t = 16; 000. Solving this gives t =

cos ( )

By putting this value of t into the above y equation and then setting y = 0 we obtain

800 sin ( )

cos ( )

g

cos ( )

This gives 16 ; 000 sin ( ) cos ( ) = 200g or 40  2 sin ( ) cos ( ) = g or sin (2 ) = g 40

From this we obtain

2 = arcsin

 (^) g 40

or 2 = 180^ arcsin

 (^) g 40

By using a calculator we see that the two angles in question are

 7 : 09 ^ and  82 : 91 .

  1. Find the length of the portion of the curve

r (t) = ti +

t^3 =^2 k

that corresponds to 0  t  8. (All details of computations must be included.) Solution: First note that v (t) = i + t^1 =^2 k and that v (t) = jv (t)j =

p 1 + t.

Solution: We have v (t) = i + 2j + 2tk

and

v (t) =

q 12 + 2^2 + (2t)^2 =

p 5 + 4t^2.

The tangential component of acceleration is thus

aT (t) = dv dt

4 t p 5 + 4t^2

Also a (t) = 2k

and thus ja (t)j = j 2 kj = 2. Therefore the normal component of acceleration is

aN (t) =

q ja (t)j^2 aT (t)^2

s 4

4 t p 5 + 4t^2

r 20 5 + 4t^2