prelim spring2006 solutions - Math 111 Prelim 1 Feb 21 2006...

Math 111 Prelim 1
Feb. 21, 2006
Name: SOLUTIONS
Instructor:
Section:
INSTRUCTIONS — READ THIS NOW
• This test has 10 problems on 14 pages worth a total of 100 points.
Look over your test package right now . If you find any missing pages
or problems please ask a proctor for another test booklet.
• Write your name, your instructor’s name, and your section number
right now .
• Show your work. To receive full credit, your answers must be neatly
written and logically organized. If you need more space, write on the
back side of the preceding sheet, but be sure to label your work clearly.
• This is a 90 minute test. You are allowed to use your calculator and
a 3 × 5 inch index card of notes. All other aids are prohibited. You
DO NOT need to SIMPLIFY your answers.
OFFICIAL
USE ONLY
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Total:

Math 111 (Spring 2006) Prelim 1 Solutions (02/21/2006) 2
1. [12pts (3pts each)] Evaluate the following limits. You must show all your work.
Note that simply plotting points is not su ffi cient to receive full credit.
(a) lim
s → 3
s
2
- s
s 2 - 2
=
3
2
- 3
3 2 - 2
=
6
7
(direct substitution property)
(b) lim
x →- 3
x 2 - 9
x
2
- x - 12
= lim
x →- 3
( x - 3)( x + 3)
( x - 4)( x + 3)
= lim
x →- 3
x - 3
x - 4
= lim
x →- 3
- 3 - 3
- 3 - 4
=
6
7
(c) lim
h → 0
e h - 1
h
= 1
Since this is the formula for the definition of derivative of f ( x ) = e
x
at the point
x = 0. Note that:
f (0) = lim
h → 0
f (0 + h ) - f (0)
h
= lim
h → 0
e
h
- 1
h
Since f ( x ) = e x , then f (0) = 1.
(d) lim
x →∞
x
3
+ 2 x + 1
7 - x
= lim
x →∞
x
3
x
+
2 x
x
+
1
x
7
x
-
x
x
= lim
x →∞
x 2 + 2 +
1
x
7
x
- 1
= -∞
(because ( x 2 + 2 + 1
x
) → ∞ and ( 7
x
- 1) → - 1 as x → ∞ )
CONTINUE TO NEXT PAGE

Math 111 (Spring 2006) Prelim 1 Solutions (02/21/2006) 3
2. [15pts (3pts each)] The women’s luge final for the 2006 Winter Olympics in Torino,
Italy was held last Wednesday. The fastest luge run was completed in 46.820 seconds
by Sylke Otto, an athlete from Germany. The table below shows the data from her
fastest run.
time (s) distance (m)
0 0
5 . 832 60
20 . 370 420
33 . 596 870
40 . 375 1140
46 . 820 1435
(a) Using the data from the table, compute her average velocity (in meters per second)
over the last 565 meters of the run.
average velocity =
1435 - 870
46 . 820 - 33 . 596
= 42 . 72 m/s
(b) The function s ( t ) = 0 . 19 t 2 + 21 . 32 t models this run reasonable well. Using this
function, give the formula for her average velocity over the interval [ t, t + h ].
average velocity =
. 19( t + h ) 2 + 21 . 32( t + h ) - ( . 19 t 2 + 21 . 32 t )
t + h - t
CONTINUE TO NEXT PAGE

Math 111 (Spring 2006) Prelim 1 Solutions (02/21/2006) 4
(c) Using the function, s ( t ) = 0 . 19 t 2 + 21 . 32 t given in (b), set up the limit as h → 0
to find her instantaneous velocity at time t .