Pre-Calculo 5Ta Edición

Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.1 The Tangent and Velocity Problems

1. (a) Using P ( 15,250 ) , we construct the following table:
t

Q
5

( 5,694 )

10

( 10,444 )

20

( 20,111 )

25

( 25,28 )

30

( 30,0 )

slope=m

PQ

694 250
444
=
= 44.4
5 15
10
444 250
194
=
= 38.8
10 15
5
111 250
139
=
= 27.8
20 15
5
28 250222
=
= 22.2
25 15
10
0 250
250
=
= 16.6
30 15
15

(b) Using the values of t that correspond to the points closest to P ( t =10 and t =20 ), we have
38.8+ ( 27.8 )
= 33.3
2
(c)

From the graph, we can estimate the slope of the tangent line at P to be

300
= 33.3 .
9

2.
2948
42
2948
(c) Slope =
42
(a) Slope =

2530 418
2948
=
69.67 (b) Slope =
36
6
42
2806 1423080
(d) Slope =
=
=71
40
2
44

2661 287
=
=71.75
38
4
2948 132
=
=66
42
2

From the data, we see that the patient’s heart rate is decreasing from 71 to 66 heartbeats / minute after
42 minutes. After being stable for a while, the patient’s heart rate is dropping.
1

Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.1 The Tangent and Velocity Problems

3.(a) For the curve y=x/ ( 1+x ) and the point P 1,
x
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)

m

Q

0.5
0.9
0.99
0.999
1.1
1.5
1.01
1.001

1
2

PQ

( 0.5,0.333333)
( 0.9,0.473684 )
( 0.99,0.497487)
( 0.999,0.499750 )
( 1.5,06 )
( 1.1,0.523810 )
( 1.01,0.502488 )
( 1.001,0.500250 )

0.333333
0.263158
0.251256
0.250125
0.2
0.238095
0.248756
0.249875

1.
4
11
1
1
(c) y
= (x 1) or y= x+ .
24
4
4
(b) The slope appears to be

4. For the curve y=ln x and the point P(2, ln 2) :
(a)
x
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
(viii)

1.5
1.9
1.99
1.999
2.5
2.1
2.01
2.001

PQ

( 1.5,0.405465)
( 1.9,0.641854 )
( 1.99,0.688135)
( 1.999,0.692647)
( 2.5,0.916291 )
( 2.1,0.741937)
( 2.01,0.698135)
( 2.001,0.693647)

(b)The slope appears to be
(c) y ln 2=

m

Q

0.575364
0.512933
0.501254
0.500125
0.446287
0.487902
0.498754
0.499875

1
.
2

1
1
(x 2) or y= x 1+ln 2
2
2

(d)

2

Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.1 The Tangent and Velocity Problems

2

2

5. (a) y= y(t )=40t 16t . At t =2 , y=40(2) 16(2) =16 . The average velocity between times 2 and2+h
2

2

y(2+h) y(2)
40(2+h) 16(2+h) 16
24h 16h
is v =
=
=
= 24 16h , if h 0 .
ave
(2+h) 2
h
h
(i) [2,2.5] : h=0.5 , v = 32 ft / s
(ii) [2,2.1] : h=0.1 , v = 25.6 ft / s
ave

ave

(iii) [2,2.05] : h=0.05 , vave= 24.8 ft / s (iv) [2,2.01] : h=0.01 , vave= 24.16 ft / s
(b) The instantaneous velocity when t =2 ( h approaches 0 ) is 24 ft / s.
6. The average velocity betweent and t +h seconds is
2

(

2

)

2

58(t +h) 0.83(t +h) 58t 0.83t
58h 1.66th 0.83h
=
=58 1.66t 0.83h if h 0 .
h
h
(a) Here t =1 , so the average velocity is 58 1.66 0.83h=56.34 0.83h .
(i)
(ii) 1,1.5 : h=0.5 , 55.925 m / s
1,2 : h=1,55.51 m / s
(iii) 1,1.1 : h=0.1 , 56.257 m / s
(iv) 1,1.01 : h=0.01 , 56.3317 m / s
(v)
1,1.001 : h=0.001 , 56.33917 m / s
(b) Theinstantaneous velocity after 1 second is 56.34 m / s.
7. (a)
(i)
(iii)

13
(ii)
ft / s
ave 6
19
1,1.5 : h=0.5 , v =
ft / s (iv)
ave 24
1,3 : h=2 , v

=

7
ft / s
ave 6
331
1,1.1 : h=0.1 , v =
ft / s
ave 600
1,2 : h=1 , v

=

(b) As h approaches 0 , the velocity approaches
3

Stewart Calculus ET 5e 0534393217;2. Limits and Derivatives; 2.1 The Tangent and Velocity Problems31
= ft / s.
62

(c)

(d)
8. Average velocity between times t =2 and t =2+h is given by
(a)
(i)
(ii)
(iii)

h=3
h=2
h=1

s(5)
av
5
s(4)
v=
av
4
s(3)
v=
av
3

v=

s(2+h) s(2)
.
h

s(2) 178 32 146
=
=
48.7 ft / s
2
3
3
s(2) 119 32 87
=
=
=43.5 ft / s
2
2
2
s(2) 70 32
=
=38 ft / s
2
1

(b) Using the points ( 0.8,0 ) and ( 5,118 ) from the...