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2 2.1
LIMITS AND DERIVATIVES The Tangent and Velocity Problems
1. (a) Using
(15 250), we construct the following table:
(b) Using the values of that correspond to the points closest
( = 10 and = 20), we have
slope = 5
(5 694)
694−250 5−15
= − 444 10 = −44 4
10
(10 444)
444−250 10−15
= − 194 5 = −38 8
20
(20 111)
111−250 20−15
= − 139 5 = −27 8
25
(25 28)
28−250 25−15
30
(30 0)
0−250 30−15
−38 8 + (−27 8) = −33 3 2
= − 222 10 = −22 2
= − 250 15 = −16 6
(c) From the graph, we can estimate the slope of the tangent line a
to be
2. (a) Slope =
2948 − 2530 42 − 36
(c) Slope =
2948 − 2806 42 − 40
−300 9
=
418 6
=
142 2
= −33 3.
≈ 69 67 = 71
(b) Slope =
2948 − 2661 42 − 38
(d) Slope =
3080 − 2948 44 − 42
=
287 4
= 71 75
=
132 2
= 66
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) = (b) The slope appears to be 1. , (2 −1) 1− (c) Using = 1, an equation of the tangent line to the ( 1 (1 − )) curve at(2 (i)
15
(1 5 −2)
2
(ii)
19
(1 9 −1 111 111)
1 111 111