Thomas Calculus Early Transcendentals 14th Edition Hass SOLUTIONS MANUAL

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Thomas Calculus Early Transcendentals 14th Edition Hass SOLUTIONS
MANUAL




CHAPTER 2 LIMITS AND CONTINUITY

2.1 RATES OF CHANGE AND TANGENTS TO CURVES
f f (3)f (2) f f (1)f (1)
1. (a) x  32
 289  19
1
(b) x  1(1)  20 2
1


g g (3) g (1) g g (4) g ( 2)
2. (a) x
 3 1
 3 (21)  2 (b) x
 4  ( 2)  8 68  0


3. (a) h 
h 34 h4   11   4  h 2h 6  0 3   3 3
(b) h
t  3  4   t      
4 2 2 6 3


g g ()  g (0) (21)(21) g g ()  g () (21)(21)
4. (a) t   0

 0
  2

(b) t   ( )  2
0


5. R  20  81
R(2) R(0)

1 31
1
2 2


6. P  21 
P (2) P (1) (81610)(145)
1
 22  0

2 2
y
 ((2h ) 5)(2 5) 44h h 2 51 2
7. (a) x h
 h
 4h h  4  h. As h  0, 4  h  4  at P(2, 1) the slope is 4.
h
(b) y  (1)  4( x  2)  y  1  4 x  8  y  4 x  9
2 2
y
 (7(2h )h )(72 )  744hh h 3  4hh
2 2
8. (a) x h
 4  h. As h  0, 4  h  4  at P(2, 3) the slope
is 4.
(b) y  3  (4)( x  2)  y  3  4 x  8  y  4 x  11

y ((2h)2 2(2h)3)(22 2(2)3) 44h h 2 42h3(3) 2
9. (a) x  h

h
 2hh  2  h. As h  0, 2  h  2  at
h
P(2,  3) the slope is 2.
(b) y  (3)  2( x  2)  y  3  2 x  4  y  2x  7.

y ((1h) 4(1h))(1 4(1)) 12h h 44h(3)
2 2 2
10. (a) x   h 2h  h  2. As h  0, h  2  2  at P(1,  3) the
2
 h
h h
slope is 2.
(b) y  (3)  (2)( x  1)  y  3  2 x  2  y  2x 1.
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y (2h)3 23  h 8  12h 4h h  12  4h  h 2 . As h  0, 12  4h  h 2  12,  at P(2, 8)
2 3 2 3
11. (a) x  h
 812h 4h
h h
the slope is 12.
(b) y  8  12( x  2)  y  8  12 x  24  y  12x 16.

y 2(1 h)3 (213 )
 213h3h h 1  3h3h h  3  3h  h 2 . As h  0, 3 3h  h 2  3,  at
2 3 2 3
12. (a) x  h h h
P(1, 1) the slope is 3.
(b) y  1  (3)( x  1)  y  1  3x  3  y  3x  4.




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62 Chapter 2 Limits and Continuity
62
y (1h) 12(1h)(1 12(1)) 13h3h h 1212h(11)
3 3 2 3
13. (a) x   9h3h h  9  3h  h 2 .
2 3

h h h
As h  0, 9  3h  h 2  9  at P(1,  11) the slope is 9.
(b) y  (11)  (9)( x  1)  y  11  9 x  9  y  9x  2.

y (2h) 3(2h) 4(2 3(2) 4)
3 2 3 2
 812h 6h h 1212h 3h 40  3h  h  3h  h2 .
2 3 2 2 3
14. (a) x  h h h
As h  0, 3h  h 2  0  at P(2, 0) the slope is 0.
(b) y  0  0( x  2)  y  0.

y
1 2
1
2(2h)
15. (a) x
 2hh  2(2h)  h1  2(2h)
1 .

As h  0, 2(2
1  41 ,  at P 2, 1  
the slope is 1 .
 h) 2 4

(b) y   1   1 ( x  (2))  y  1  1 x  1  y  1 x  1
2 4 2 4 2 4

(4h )
24
 
4
y 4 h  2  1  4h2(2h)  1   1  1 .
  2h
2(4h )
16. (a) x 1 h  h 2h 2h
h 2  h
As h  0, 21 h  12 ,  at P(4,  2) the slope is 12 .
(b) y  (2)  12 ( x  4)  y  2  12 x  2  y  12 x  4

y
 4hh  4  4hh 2  4h  2 
(4 h)4
17. (a)  1 .
x 4h  2 h( 4h 2) 4h 2

As h  0, 1  1  1,  at P(4, 2) the slope is 14 .
4h 2 4 2 4
(b) y  2  14 ( x  4)  y  2  14 x  1  y  14 x  1

y 7(2h)  7(2) 3 3  9h 3  (9h)9  1
18. (a) x
  9h  9h .
h h h 9h 3 h( 9h 3) 9h 3

As h  0,  1   1  1 ,  at P(2, 3) the slope is 1 .
9h 3 9 3 6 6

(b) y  3  1 ( x  (2))  y  3  1 x  1  y  1 x  8
6 6 3 6 3

p
19. (a) Q Slope of PQ 
 t
Q1 (10, 225) 650225
2010
 42.5 m/sec
Q2 (14, 375) 650375
2014  45.83 m/sec
Q3 (16.5, 475) 650475
 50.00 m/sec
2016.5
Q4 (18, 550) 650550
2018
 50.00 m/sec
(b) At t  20, the sportscar was traveling approximately 50 m/sec or 180 km/h.

p
20. (a) Q Slope of PQ t
Q1 (5, 20) 8020  12 m/sec
105

107  13.7 m/sec
Q2 (7, 39) 8039

Q3 (8.5, 58) 8058
 14.7 m/sec
108.5
Q4 (9.5, 72) 8072
109.5
 16 m/sec
(b) Approximately 16 m/sec




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Section 2.1 Rates of Change and Tangents to Curves 63

21. (a)
p

200


Profit (1000s)
160

120

80

40
0 t
2010 2011 2012 2013 2014
Ye ar
p 17462
(b) t
 20142012  112
2
 56 thousand dollars per year

(c) The average rate of change from 2011 to 2012 is p
t
 20122011
6227  35 thousand dollars per year.

p
The average rate of change from 2012 to 2013 is t  20132012
11162  49 thousand dollars per year.

So, the rate at which profits were changing in 2012 is approximately 12 (35  49)  42 thousand dollars
per year.
22. (a) F ( x)  ( x  2)/( x  2)
x 1.2 1.1 1.01 1.001 1.0001 1
F ( x) 4.0 3.4 3.04 3.004 3.0004 3
F  4.0(3) F  3.4 (3)
x
 5.0; 1.21
 4.4; x 1.11
F  3.04(3)  4.04; F  3.004(3)  4.004;
x 1.011 x 1.0011
F
x
 3.0004(3)  4.0004;
1.00011
(b) The rate of change of F ( x ) at x  1 is 4.
g g (2) g (1) g g (1.5) g (1) 1
23. (a) x  21
 21
21
 0.414213 x
 1.51
 1.5
0.5
 0.449489
g g (1h) g (1) 1

x  (1h)1
 1hh

(b) g ( x)  x
1 h 1.1 1.01 1.001 1.0001 1.00001 1.000001
1 h 1.04880 1.004987 1.0004998 1.0000499 1.000005 1.0000005
 1 h  1 /h  0.4880 0.4987 0.4998 0.499 0.5 0.5
(c) The rate of change of g ( x ) at x  1 is 0.5.
1 1
(d) The calculator gives lim 1h
h
 2.
h0

11 1
f (3)f (2)
24. (a) i)  3 1 2  16   61
32
1 1 2  T
f(T)f (2)
ii)  TT 22  2TT 2
T 2
2T
 2T2T  2T   2T
(T 2) 2T (2T )
1 ,T  2

(b) T 2.1 2.01 2.001 2.0001 2.00001 2.000001
f (T ) 0.476190 0.497512 0.499750 0.4999750 0.499997 0.499999
( f (T )  f (2))/(T  2) 0.2381 0.2488 0.2500 0.2500 0.2500 0.2500
(c) The table indicates the rate of change is 0.25 at t  2.
(d) lim 1   1
T 2 2T 4  
NOTE: Answers will vary in Exercises 25 and 26.
25. (a) [0, 1]: s  150  15 mph; [1, 2.5]: s  2015  10 mph; [2.5, 3.5]: s  3020  10 mph
t 10 t 2.51 3 t 3.52.5




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64 Chapter 2 Limits and Continuity Section 2.2 Limit of a Function and Limit Laws 64
64 64


2 
(b) At P 1 , 7.5 : Since the portion of the graph from t  0 to t  1 is nearly linear, the instantaneous rate of
change will be almost the same as the average rate of change, thus the instantaneous speed at t  12 is
157.5  15 mi/hr. At P(2, 20): Since the portion of the graph from t  2 to t  2.5 is nearly linear, the
10.5
instantaneous rate of change will be nearly the same as the average rate of change, thus v  2020
2.52  0 mi/hr.
For values of t less than 2, we have
s
Slope of PQ  t
 Q
Q1 (1, 15) 1520  5 mi/hr
12
Q2 (1.5, 19) 1920
1.52
 2 mi/hr
19.920
Q3 (1.9, 19.9)
1.92
 1 mi/hr
Thus, it appears that the instantaneous speed at t  2 is 0 mi/hr.
At P(3, 22):
s
Slope of PQ  t Q Slope of PQ  s
 Q t
Q1 (2, 20) 2022 
Q1 (4, 35) 3522
43
 13 mi/hr 23
2 mi/hr
Q2 (3.5, 30) 3022
 16 mi/hr Q2 (2.5, 20) 2022
2.53
 4 mi/hr
3.53
21.622
Q3 (3.1, 23) 2322  10 mi/hr Q3 (2.9, 21.6)  4 mi/hr
3.13 2.93
Thus, it appears that the instantaneous speed at t  3 is about 7 mi/hr.
(c) It appears that the curve is increasing the fastest at t  3.5. Thus for P (3.5, 30)
s
Slope of PQ  t
 Q Q Slope of PQ  s
t
Q1 (4, 35) 3530  10 mi/hr 2230  16 mi/hr
43.5 Q1 (3, 22)
33.5
Q2 (3.75, 34) 3430
3.753.5
 16 mi/hr Q2 (3.25, 25) 2530
 20 mi/hr
3.253.5
Q3 (3.6, 32) 3230
3.63.5
 20 mi/hr Q3 (3.4, 28) 2830
 20 mi/hr
3.43.5
Thus, it appears that the instantaneous speed at t  3.5 is about 20 mi/hr.
26. (a) [0, 3]: tA  1015  1.67 day ; [0, 5]: tA  3.915  2.2 day; [7, 10]: t
gal gal
A  01.4  0.5 gal
30 50 107 day
(b) At P(1, 14) :
A
 Q Slope of PQ  t Q Slope of PQ  tA
Q1 (2, 12.2) 12.214  1.8 gal/day Q1 (0, 15) 1514  1 gal/day
21 01
Q2 (1.5, 13.2) 13.214 14.614
1.51
 1.6 gal/day Q2 (0.5, 14.6)  1.2 gal/day
0.51
Q3 (1.1, 13.85) 13.8514  1.5 gal/day
1.11
Q3 (0.9, 14.86) 14.8614  1.4 gal/day
0.91
Thus, it appears that the instantaneous rate of consumption at t  1 is about 1.45 gal/day.
At P(4, 6):
Q Slope of PQ  tA
Q Slope of PQ  tA

Q (5, 3.9)
1
3.96  2.1 gal/day Q1 (3, 10) 106
 4 gal/day34
54
Q2 (4.5, 4.8) 4.86
 2.4 gal/day Q2 (3.5, 7.8) 7.86
3.54
 3.6 gal/day
4.54
Q3 (4.1, 5.7) 5.76
 3 gal/day Q3 (3.9, 6.3) 6.36
3.94
 3 gal/day
4.14
Thus, it appears that the instantaneous rate of consumption at t  1 is 3 gal/day.

(solution continues on next page)



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