Math 1151: Calculus
I
Autumn 2016
Dr. John E. Harper
The Ohio State University, Newark
Exam 2
Name:
There are a total of 10 problems. The problems are worth 10 points each.
points.
Problem 1
(10 points). Describe in words the mea^ning of
TL *L;*
-Ql. ^
x
L
Math 1151: Calculus I Autumn 2016
Dr. John E. Harper
The Ohio State University, Newark
Exam 6
Name:
There are a total of 10 problems. The problems are worth 10 points each. A perfect score is 100
points.
Problem 1 (10 points). Determine slug] 1_3(:(
Math L151: Calculus
I
Autumn 2016
Dr. John E. Harper
The Ohio State University, Newa,rk
Exam 5
Name:
There are a total of 10 problems. The problems are worth 10 points each. A perfect score is
100
points.
Each side of a square is increasing at a rate of 6
Math 1151: Calculus I
Dr. John E. Harper
The Ohio State University, Newark
Autumn 2016
Exam 3
Name:
There are a total of 10 problems. The problems are worth 10 points each. A perfect score is 100
points.
Problem 1 (10 points). Consider the function
2
x ,
Math 1151: Calculus I
Dr. John E. Harper
The Ohio State University, Newark
Autumn 2016
Exam 1
Name:
There are a total of 10 problems. The problems are worth 10 points each. A perfect score is 100
points.
Problem 1 (10 points). Find the exact value of log1
Math 1151: Calculus I
Dr. John E. Harper
The Ohio State University, Newark
Autumn 2016
Exam 5
Name:
There are a total of 10 problems. The problems are worth 10 points each. A perfect score is 100
points.
Problem 1 (10 points). Each side of a square is inc
Math 1151: Calculus I
Dr. John E. Harper
The Ohio State University, Newark
Autumn 2016
Exam 4
Name:
There are a total of 10 problems. The problems are worth 10 points each. A perfect score is 100
points.
Problem 1 (10 points). If y =
p
(x + 3)3 + 2x, find
Math 1151: Calculus I
Dr. John E. Harper
The Ohio State University, Newark
Autumn 2016
Exam 2
Name:
There are a total of 10 problems. The problems are worth 10 points each. A perfect score is 100
points.
Problem 1 (10 points). Describe in words the meanin
Math 1151: Calculus
Autumn 2016
I
Dr. John E. HarPer
The Ohio State UniversitY, Newa^rk
Exam 3
Name:
There are a total of 10 problems. The problems
a,re
worth 10 points each' A perfect score is
100
points.
Problem 1 (10 points). Consider the function
*t'
Math 1151: Calculus I
Dr. John E. Harper
The Ohio State University, Newark
Autumn 2016
Exam 6
Name:
There are a total of 10 problems. The problems are worth 10 points each. A perfect score is 100
points.
Problem 1 (10 points). Determine lim
x0
1 cos(x)
3x
Math 1151: Calculus I
Autumn 2016
Dr. John E. Harper
The Ohio State University Newark
Exam 4
Narne:
There are a total of 10 problems. The problems
a^re
worth 10 points each. A perfect score is
find
ffi.
100
points.
Problem 1
(10 points).
If y :
JGTTT2T,
&
Math 1"151: Calculus
Dr. John E. Harper
I
Autumn
2016
The Ohio State University, Newark
Exam I
Name:
There are a total of 10 problems. The problems are worth 10 points each. A perfect score is
points.
Problem 1 (10 points). Find the exact value of log1s(/
Math 1151
Autumn 2016
Final Exam Review Problems
Final Exam on 12/13/16
Name (Print):
Time Limit on Final: 105 Minutes
Go on carmen.osu.edu to see where your final exam will be. NOTE: This is not a sample final.
This set of problems is much longer than wh
228
27.64
27.65
27.66
27.67
27.68
27.69
[7 CHAPTER 27
Use integration to show that the circumference of a circle of radius r is 21rr.
I Find the arc length of the part of the circle x2 + y2 = r2 in the ﬁrst quadrant and multiply it by 4. Since
y= Vr2 —x2,
230
27.76
27.77
27.78
27.79
27.80
[7 CHAPTER 27
25(1— 1)
47
L
Fig. 27-8 Fig. 27-9
See Fig. 27-9. Two ships sail from A at the same time. One sails south at 15 mi/ h; the other sails east at 25 mi/ h
for 1 hour until it reaches point B, and then sails no
226 L7 CHAPTER 27
dx 1116 1 dx
' f = i '— = ’ I M. From the formula in Problem 27 46 t
xV9x2—16 x\/9(x2—199) 3 x x2_(g)2 - , we ge
1; any“; C
3 4§)sec % —4sec 4
dx
27.48
xV3x2 - 2
x
dx dx 1 dx
I I = J’ — = — I —_. From the formula in Proble 27.46 w bt ‘
222 L7 CHAPTER 27
27.12
27.13
27. 14
27.15
27.16
27.17
27.18
27.19
27.20
27.21
27.22
27.23
csc"l (-V2).
I By deﬁnition, the value 0 must be an angle in either the ﬁrst or third quadrant. Since csc 0 = —\/2, 0 is in
the third quadrant, sin 0 = -1/\/2, and
216 0 CHAPTER 26
26. 10
26.11
26.12
26.13
26.14
26.15
26. 16
26.17
26.18
26.19
I By Problem 26.8, y=yo(1+ r/100n)" if the money is compounded n times per year. If we
let n approach inﬁnity, we get lim yo(1 + r/100n)" = y0 lim (1 + r/100n)" = y0 [hlim (1 +
218 [7 CHAPTER 26
26.30
26.31
26.32
26.33
26.34
26.35
26.36
26.37
I y =200eK’, where t=0 at 6am. Then, 500=200e3", e3K= %. At noon, t=6 and y=
200e6K = 200(e3")2 = 200( g )2 = 1250 grams.
A radioactive substance decreases from 8 grams to 7 grams in 1 hour
214 0 CHAPTER 25
25.56 Graph y = x"e "‘ for positive even integers n.
I See Fig. 253. y’ = -x"e_’ + ruc"e"r = x"e"‘(n - x). Hence, x = n and x = 0 are critical num-
bers. The ﬁrst-derivative test tells us that there is a relative maximum at x = n and a re
CHAPTER 27
Inverse Trigonometric Functions
1
27.1 Draw the graph of y = sin' x.
I By deﬁnition, as x varies from —1 to 1, y varies from —1-r/ 2 to 1r/ 2. The graph of y = sin"1 x is obtained
from the graph of y = sin x [Fig. 27-1(a)] by reﬂection in the l
210 L7 CHAPTER 25
_ 1 — sin x
25'“ JET/2 x - 1r/2 '
. 1—sinx_ . ~cosx_0_
' xl‘vrfl/2x_7T/2—XEI:IIZ 1 —1—0.
25.19 lim x5”.
x—tO+
sinx . l . A . . . l
I Let yl= x . Then ln y = sm x - In x = csc: , to which L’Hopital’s rule applies. lim CEC: =
.
CHAPTER 25
L’Hﬁpital’s Rule
25.1
25.2
25.3
25.4
25.5
25.6
25.7
208
State L’Hopital’s rule.
I First, let us state the zero-over—zero case. Under certain simple conditions, if Iim f(x) = lim g(x) = 0 and
I x—sb x—vb
lim fig = L, then lim jg% = L. Here, x-
248 0 CHAPTER 30
dx
30-14 I (x2 + l)(x2 + 4).
1 _ Ax + B + Cx + D
(x2+1)(x2+4) x2+1 x2+4 '
B)(x2 + 4) + (Cx + D)(x2 + 1). Equate coefﬁcients of x3: (*) 0 = A + C. Equate coefﬁcients of x:
0 = 4A + C. Subtracting (*) from this equation, we get 3A = 0, A =
250 H CHAPTER 30
30.21
30.22
30.23
30.24
I dx
1+e"
1 A B
I Let u=1+e", du=exdx. Then I dxx= du . Let ——=—+ . Then 1=
1+e u(u—1) u(u—1) u u—l1 1
A(u—1)+Bu. Let u=0. Then 1=—A, A=—1. Let u=1. Then 1=B. So m=—;
1 I du _ _ x x _ x
234 [7 CHAPTER 28
28.19 ln—z" dx.
x
'1 Let urilnx, dv=(1/x2)dx, du=(1/x)dx, v=—1/x. Then t—fdx=~:x+fizdx=‘:lx—
._ x
;+C=T(lnx+1)+C.
28.20 I x2e” dx.
I Let 3x = ~y and use Problem 28.1: Ix2e3‘ d = *2177- f yZe—y dy = 717e_“'(y2 + 2y + 2) + C =
#919):2
244 0 CHAPTER 29
29.42 Assume that an object A, starting at (a, 0), is connected by a string of constant length a to an objecth, starting at
(0, 0). As B moves up the y—axis, A traces out a curve called a tractrL'x. Find its equation.
I Let A be at (x, y)