12-3 Tangent Lines and Velocity
Find the slope of the lines tangent to the graph of each function at the given points.
5. y = x3 + 8; (−2, 0) and (1, 9)
SOLUTION: Find the slope of the line tangent to the graph at (–2, 0).
Find the slope of the line tangent to the graph at (1, 9).
ANSWER: 12; 3
6. y =
; (2, 0.25) and (−1, 1)
SOLUTION: Find the slope of the line tangent to the graph at (2, 0.25).
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6. y =
; (2, 0.25) and (−1, 1)
12-3SOLUTION: Tangent Lines and Velocity
Find the slope of the line tangent to the graph at (2, 0.25).
Find the slope of the line tangent to the graph at (–1, 1).
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12-3 Tangent Lines and Velocity
ANSWER: Find an equation for the slope of the graph of each function at any point.
7. y = 4 − 2x
SOLUTION: An equation for the slope of the graph at any point is m = −2.
ANSWER: m = −2
8. y = −x2 + 4x
SOLUTION: An equation for the slope of the graph at any point is m = −2x + 4.
ANSWER: m = −2x + 4
12. y = 2x2
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An equation for the slope of the graph at any point is m = −2x + 4.
ANSWER: 12-3 Tangent Lines and Velocity
m = −2x + 4
12. y = 2x2
SOLUTION: An equation for the slope of the graph at any point is m = 4x.
ANSWER: m = 4x
The position of an object in miles after t minutes is given by s(t). Find the average velocity of the object
in miles per hour for the given interval of time. Remember to convert from minutes to hours.
18. for 3 ≤ t ≤ 5
SOLUTION: First, find the position of the object for a =3 and b = 5.
Now use the formula for average velocity.
The average velocity of the object is 0.75 miles per minute. Thus, the average velocity of the object is 45 miles per
hour.
ANSWER: 45 mi/h
24. TYPING The number of words w a person has typed after t minutes is given by
a. What was the average number of words per minute the person typed between the 2nd and 4th minutes?
What was the average number of words per minute the person typed between the 3rd and 7th minutes?
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SOLUTION: Page 4
The average velocity of the object is 0.75 miles per minute. Thus, the average velocity of the object is 45 miles per
hour.
ANSWER: 12-345Tangent
Lines and Velocity
mi/h
24. TYPING The number of words w a person has typed after t minutes is given by
a. What was the average number of words per minute the person typed between the 2nd and 4th minutes?
b. What was the average number of words per minute the person typed between the 3rd and 7th minutes?
SOLUTION: a. First, find the number of words typed for a =2 and b = 4.
Now use the formula for average number of words.
The average number of words typed between the 2nd and 4th minutes is 46 words/min.
b. Find the number of words typed for a =3 and b = 7.
Now use the formula for average number of words.
The average number of words typed between the 3rd and 7th minutes is 60.5 words/min.
ANSWER: a. 46 words/min
b. 60.5 words/min
The distance d an object is above the ground t seconds after it is dropped is given by d(t). Find the
instantaneous velocity of the object at the given value for t.
26. d(t) = 38t − 16t2; t = 0.8
SOLUTION: To find the instantaneous velocity, let t = 0.8 and apply the formula for instantaneous velocity.
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The average number of words typed between the 3rd and 7th minutes is 60.5 words/min.
ANSWER: 46 words/min
12-3a.Tangent
Lines and Velocity
b. 60.5 words/min
The distance d an object is above the ground t seconds after it is dropped is given by d(t). Find the
instantaneous velocity of the object at the given value for t.
26. d(t) = 38t − 16t2; t = 0.8
SOLUTION: To find the instantaneous velocity, let t = 0.8 and apply the formula for instantaneous velocity.
The instantaneous velocity of the object is 12.4 feet per second.
ANSWER: 12.4 ft/s
32. d(t) = 853 − 48t − 16t2; t = 1.3
SOLUTION: To find the instantaneous velocity, let t = 1.3 and apply the formula for instantaneous velocity.
The instantaneous velocity of the object is −89.6 feet per second.
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ANSWER: −89.6 ft/s
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The instantaneous velocity of the object is 12.4 feet per second.
ANSWER: 12-312.4
Tangent
Lines and Velocity
ft/s
32. d(t) = 853 − 48t − 16t2; t = 1.3
SOLUTION: To find the instantaneous velocity, let t = 1.3 and apply the formula for instantaneous velocity.
The instantaneous velocity of the object is −89.6 feet per second.
ANSWER: −89.6 ft/s
Find an equation for the instantaneous velocity v(t) if the path of an object is defined as s(t) for any point
in time t.
34. s(t) = t − 3t2
SOLUTION: Apply the formula for instantaneous velocity.
The instantaneous velocity of the object at time t is v(t) = 1 − 6t.
ANSWER: v(t) = 1 − 6t
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40. s(t) = 12t2 − 2t3
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The instantaneous velocity of the object is −89.6 feet per second.
12-3ANSWER: Tangent Lines and Velocity
−89.6 ft/s
Find an equation for the instantaneous velocity v(t) if the path of an object is defined as s(t) for any point
in time t.
34. s(t) = t − 3t2
SOLUTION: Apply the formula for instantaneous velocity.
The instantaneous velocity of the object at time t is v(t) = 1 − 6t.
ANSWER: v(t) = 1 − 6t
40. s(t) = 12t2 − 2t3
SOLUTION: Apply the formula for instantaneous velocity.
2
The instantaneous velocity of the object at time t is v(t) = 24t − 6t .
ANSWER: 2
v(t) = 24t − 6t
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The instantaneous velocity of the object at time t is v(t) = 1 − 6t.
ANSWER: 12-3v(t)
Tangent
= 1 − 6t Lines and Velocity
40. s(t) = 12t2 − 2t3
SOLUTION: Apply the formula for instantaneous velocity.
2
The instantaneous velocity of the object at time t is v(t) = 24t − 6t .
ANSWER: 2
v(t) = 24t − 6t
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