Solutions Manual for Calculus Early
Transcendentals 8th Edition by Stewar
Ibsn
2 LIMITS AND DERIVATIVES
2.1 The Tangent and Velocity Problems
Using the values of t that correspond to the points
1. (a) Using P (15 250), we construct the following table:
closest to P (t = 10 and t = 20), we have
t slope = mPQ
−388 + (−278)
694−250 = − 444 = −444 = −333
5 (5 694) 5−15 10 2
10 (10 444) 444−250 = − 194 = −388
10−15 5
20 (20 111) 111−250 = − 139 = −278
20−15 5
25 (25 28) 28−250 = − 222 = −222
25−15 10
30 (30 0) 0−250
30−15
= − 250
15
= −166
(c) From the graph, we can estimate the slope of the
tangent line at P to be −300
9
= −333.
(b)
2. (a) Slope = 2948 − 2530 = 418
42 − 36 6 ≈ 6967 (b) Slope = 2948 − 2661
42 − 38
= 287
4 = 7175
(d) Slope = 3080 − 2948 = = 66
(c) Slope = 2948 − 2806 = 142 = 71 132
42 − 40 2 44 − 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
3. (a) y = , P (2 1)
− (b) The slope appears to be 1.
1−x
(c) Using m = 1, an equation of the tangent line to the
x (x 1 (1 − x)) mPQ
curve at P (2 −1) is y − (−1) = 1(x − 2), or
(i) 15 (15 −2) 2
(ii) 19 (19 −1111 111) 1111 111 y = x − 3.
(iii) 199 (199 −1010 101) 1010 101
(iv) 1999 (1999 −1001 001) 1001 001
(v) 25 (25 −0666 667) 0666 667
(vi) 21 (21 −0909 091) 0909 091
(201 −0990 099)
,
,68 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
4. (a) y = cos x, P (05 0) (b) The slope appears to be − .
x mPQ (c) y − 0 = − (x − 05) or y = − x + 1 .2
(i) 0 (0 1) −2
(d)
(ii) 04 (04 0309017) −3090170
(iii) 049 (049 0031411) −3141076
(iv) 0499 (0499 0003142) −3141587
(v) 1 (1 −1) −2
(vi) 06 (06 −0309017) −3090170
(vii) 051 (051 −0031411) −3141076
(viii) 0501 (0501 −0003142) −3141587
5. (a) y = y(t) = 40t − 16t2. At t = 2, y = 40(2) − 16(2)2 = 16. The average velocity between times 2 and 2 + h is
y(2 + h) − y(2) 40(2 + h) 16(2 + h)2 − 16 2
vave = = = −24h − 16h = −24 − 16h, if h6= 0.
(2 + h) − 2 h − h
(i) [2 25]: h = 05, vave = −32 ft s (ii) [2 21]: h = 01, vave = −256 ft s (iii)
[2 205]: h = 005, vave = −248 ft s (iv) [2 201]: h = 001, vave = −2416 ft s
(b) The instantaneous velocity when t = 2 (h approaches 0) is −24 ft s.
6. (a) y = y(t) = 10t − 186t2 . At t = 1, y = 10(1) − 186(1)2 = 814. The average velocity between times 1 and 1 + h is
y(1 + h) − y(1) 10(1 + h) 186(1 + h)2 − 814 628h 186h2
vave = = = − = 628 − 186h, if h 6= 0.
(1 + h) − 1 h − h
(i) [1 2]: h = 1, vave = 442 m s (ii) [1 15]: h = 05, vave = 535 m s
(iii) [1 11]: h = 01, vave = 6094 m s (iv) [1 101]: h = 001, vave = 62614 m s
(v) [1 1001]: h = 0001, vave = 627814 m s
(b) The instantaneous velocity when t = 1 (h approaches 0) is 628 m s.
7. (a) (i) On the interval [2 4] , v s(4) − s(2) 792 − 206
= = = 293 ft s.
ave
4−2 2
(ii) On the interval [3 4] , v s(4) − s(3) 792 − 465
= = = 327 ft s.
ave
4−3 1
(iii) On the interval [4 5] , v s(5) − s(4) 1248 − 792
= = = 456 ft s.
ave
5−4 1
(iv) On the interval [4 6] , v s(6) − s(4) 1767 − 792
= = = 4875 ft s.
ave
6−4 2
, °
c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
Transcendentals 8th Edition by Stewar
Ibsn
2 LIMITS AND DERIVATIVES
2.1 The Tangent and Velocity Problems
Using the values of t that correspond to the points
1. (a) Using P (15 250), we construct the following table:
closest to P (t = 10 and t = 20), we have
t slope = mPQ
−388 + (−278)
694−250 = − 444 = −444 = −333
5 (5 694) 5−15 10 2
10 (10 444) 444−250 = − 194 = −388
10−15 5
20 (20 111) 111−250 = − 139 = −278
20−15 5
25 (25 28) 28−250 = − 222 = −222
25−15 10
30 (30 0) 0−250
30−15
= − 250
15
= −166
(c) From the graph, we can estimate the slope of the
tangent line at P to be −300
9
= −333.
(b)
2. (a) Slope = 2948 − 2530 = 418
42 − 36 6 ≈ 6967 (b) Slope = 2948 − 2661
42 − 38
= 287
4 = 7175
(d) Slope = 3080 − 2948 = = 66
(c) Slope = 2948 − 2806 = 142 = 71 132
42 − 40 2 44 − 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
3. (a) y = , P (2 1)
− (b) The slope appears to be 1.
1−x
(c) Using m = 1, an equation of the tangent line to the
x (x 1 (1 − x)) mPQ
curve at P (2 −1) is y − (−1) = 1(x − 2), or
(i) 15 (15 −2) 2
(ii) 19 (19 −1111 111) 1111 111 y = x − 3.
(iii) 199 (199 −1010 101) 1010 101
(iv) 1999 (1999 −1001 001) 1001 001
(v) 25 (25 −0666 667) 0666 667
(vi) 21 (21 −0909 091) 0909 091
(201 −0990 099)
,
,68 ¤ CHAPTER 2 LIMITS AND DERIVATIVES
4. (a) y = cos x, P (05 0) (b) The slope appears to be − .
x mPQ (c) y − 0 = − (x − 05) or y = − x + 1 .2
(i) 0 (0 1) −2
(d)
(ii) 04 (04 0309017) −3090170
(iii) 049 (049 0031411) −3141076
(iv) 0499 (0499 0003142) −3141587
(v) 1 (1 −1) −2
(vi) 06 (06 −0309017) −3090170
(vii) 051 (051 −0031411) −3141076
(viii) 0501 (0501 −0003142) −3141587
5. (a) y = y(t) = 40t − 16t2. At t = 2, y = 40(2) − 16(2)2 = 16. The average velocity between times 2 and 2 + h is
y(2 + h) − y(2) 40(2 + h) 16(2 + h)2 − 16 2
vave = = = −24h − 16h = −24 − 16h, if h6= 0.
(2 + h) − 2 h − h
(i) [2 25]: h = 05, vave = −32 ft s (ii) [2 21]: h = 01, vave = −256 ft s (iii)
[2 205]: h = 005, vave = −248 ft s (iv) [2 201]: h = 001, vave = −2416 ft s
(b) The instantaneous velocity when t = 2 (h approaches 0) is −24 ft s.
6. (a) y = y(t) = 10t − 186t2 . At t = 1, y = 10(1) − 186(1)2 = 814. The average velocity between times 1 and 1 + h is
y(1 + h) − y(1) 10(1 + h) 186(1 + h)2 − 814 628h 186h2
vave = = = − = 628 − 186h, if h 6= 0.
(1 + h) − 1 h − h
(i) [1 2]: h = 1, vave = 442 m s (ii) [1 15]: h = 05, vave = 535 m s
(iii) [1 11]: h = 01, vave = 6094 m s (iv) [1 101]: h = 001, vave = 62614 m s
(v) [1 1001]: h = 0001, vave = 627814 m s
(b) The instantaneous velocity when t = 1 (h approaches 0) is 628 m s.
7. (a) (i) On the interval [2 4] , v s(4) − s(2) 792 − 206
= = = 293 ft s.
ave
4−2 2
(ii) On the interval [3 4] , v s(4) − s(3) 792 − 465
= = = 327 ft s.
ave
4−3 1
(iii) On the interval [4 5] , v s(5) − s(4) 1248 − 792
= = = 456 ft s.
ave
5−4 1
(iv) On the interval [4 6] , v s(6) − s(4) 1767 − 792
= = = 4875 ft s.
ave
6−4 2
, °
c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.
