1. No category

Chapter 3
Average velocity, Average Rates of
Change, and Secant Lines
3.1
A certain function takes values given in the table below.
t
0 0.5 1.0 1.5 2.0
f (t) 0 1
0
-1
0
Find the average rate of change of the function over the intervals
(a) 0 < t < 0.5,
(b) 0 < t < 1.0,
(c) 0.5 < t < 1.5,
(d) 1.0 < t < 2.0.
Detailed Solution:
(a) For 0 < t < 0.5, the average rate of change is ∆f /∆t = 1/0.5 = 2.
(b) 0 < t < 1.0, the average rate of change is ∆f /∆t = 0.
(c) 0.5 < t < 1.5, the average rate of change is ∆f /∆t = (−1 − 1)/(1.5 − 0.5) = −2.
(d) 1.0 < t < 2.0, the average rate of change is ∆f /∆t = 0.
3.2
Consider the functions f1 (x) = x, f2 (x) = x2 , f3 (x) = x3 .
Find the average rate of change of these functions over each of the following intervals.
(a) Over 0 ≤ x ≤ 1.
(b) Over −1 ≤ x ≤ 1.
(c) Over 0 ≤ x ≤ 2.
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Math 102 Problems
Chapter 3
Detailed Solution:
(a) Over 0 ≤ x ≤ 1: We find
∆f1
∆f2
∆f3
1−0
=
=
=
=1
∆x
∆x
∆x
1−0
(b) Over −1 ≤ x ≤ 1: We have
∆f3
(1 − (−1))
∆f1
=
=
=1
∆x
∆x
(1 − (−1))
and
∆f2
1−1
=
=0
∆x
1 − (−1)
(c) Over 0 ≤ x ≤ 2: We have
(2 − 0)
∆f1
=
=1
∆x
(2 − 0)
(22 − 0)
4
∆f2
=
= =2
∆x
(2 − 0)
2
∆f3
(23 − 0)
8
=
= =4
∆x
(2 − 0)
2
Figure 3.1 shows the points of intersection of the above three power functions. This helps to
understand why they have identical rates of change over some intervals, but not others.
1
0.5
–1
–0.8 –0.6 –0.4 –0.2
0.2
0.4
x
0.6
0.8
1
–0.5
–1
Figure 3.1: Figure for Problem 3.2
3.3
Find the average rate of change for each of the following functions over the given interval.
(a) y = f (x) = 3x − 2 from x = 3.3 to x = 3.5.
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2
Math 102 Problems
Chapter 3
(b) y = f (x) = x2 + 4x over [0.7, 0.85].
(c) y = − x4 and x changes from 0.75 to 0.5.
Detailed Solution:
(a)
f (3.5) − f (3.3)
8.5 − 7.9
=
=3
3.5 − 3.3
0.2
(b)
(0.852 + 4 · 0.85) − (0.72 + 4 · 0.7)
0.8325
f (0.85) − f (0.7)
=
=
= 5.55
0.85 − 0.7
0.15
0.15
(c)
4
)
− 4 − (− 0.75
−8 +
f (0.5) − f (0.75)
= 0.5
=
0.5 − 0.75
−0.25
− 14
3.4
16
3
=
32
3
Trig Minireview
Consider the following table of values of the trigonometric functions sin(x) and cos(x):
x sin(x) cos(x)
0
0
1
√
π
6
π
4
π
3
π
2
1
√2
2
√2
3
2
3
√2
2
2
1
2
1
0
Find the average rates of change of the given function over the given interval. Express your
answer in terms of square roots and π. Do not compute decimal expressions.
(a) Find the average rate of change of sin(x) over 0 ≤ x ≤ π/4.
(b) Find the average rate of change of cos(x) over π/4 ≤ x ≤ π/3.
(c) Is there an interval over which the functions sin(x) and cos(x) have the same average rate
of change? (Hint: consider the graphs of these functions over one whole cycle, e.g. for
0 ≤ x ≤ 2π. Where do they intersect?)
Detailed Solution:
(a) The average rate of change of sin(x) over 0 ≤ x ≤ π/4 is given by
√
√
2/2 − 0
∆ sin(x)
sin(π/4) − sin(0)
2 2
=
=
=
∆x
π/4 − 0
π/4
π
(b) The average rate of change of cos(x) over π/4 ≤ x ≤ π/3 is
√
√
√
∆ cos(x)
cos(π/3) − cos(π/4)
1/2 − 2/2
(1 − 2)/2
6(1 − 2)
=
=
=
=
∆x
π/3 − π/4
π/3 − π/4
4π/12 − 3π/12
π
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3
Math 102 Problems
Chapter 3
1
0.5
0
1
2
3
x
4
5
6
–0.5
–1
Figure 3.2: Figure for problem 3.4
(c) The two functions are shown plotted in Figure 3.2. They intersect at x = π/4 and at x = 5π/4.
This means that the average rate of change of the functions over this interval will be the same.
There are, of course, many other such intervals, since these functions are periodic.
3.5
(a) Consider the function y = f (x) = 1 + x2 . Consider the point (1, 2) on its graph and some
point nearby, for example (1 + h, 1 + (1 + h)2 ). Find the slope of a secant line connecting these
two points.
(b) Use this slope to figure out what the slope of the tangent line to the curve at (1, 2) would be.
(c) Find the equation of the tangent line through the point (1, 2).
Detailed Solution:
(a) The slope of the secant line is
f (1 + h) − f (1)
1 + (1 + h)2 − 2
2h + h2
=
=
= 2 + h.
1+h−1
h
h
(b) The slope of the tangent line at the point (1, 2) is
1 + (1 + h)2 − 2
= lim (2 + h) = 2.
h→0
h→0
h
lim
(c) The tangent line has the equation
y−2
= 2, that is y = 2x.
x−1
3.6
Given the function y = f (x) = 2x3 + x2 − 4, find the slope of the secant line joining the points
(4, f (4)) and (4 + h, f (4 + h)) on its graph, where h is a small positive number. Then find the slope
of the tangent line to the curve at (4, f (4)).
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4
Math 102 Problems
Chapter 3
Detailed Solution:
f (4) = 2 · 43 + 42 − 4 = 140.
The slope of secant line from x = 4 to x = 4 + h is
f (4 + h) − f (4)
2(4 + h)3 + (4 + h)2 − 4 − 140
=
(4 + h) − 4
h
2(64 + 48h + 12h2 + h3 ) + (16 + 8h + h2 ) − 4 − 140
=
h
3
2
2h + 25h + 104h
=
h
= 2h2 + 25h + 104 since h 6= 0
The slope of the graph at (4, 140) can be found by letting h go to zero in the last expression.
lim (2h2 + 25h + 104) = 104
h→0
3.7
Average rate of change
Consider the function f (x) = x2 − 4x and the point x0 = 1.
(a) Sketch the graph of the function.
(b) Find the average rate of change over the intervals [1, 3], [−1, 1], [1, 1.1], [0.9, 1] and [1 − h, 1],
where h is some small positive number.
(c) Find f 0 (1).
Detailed Solution:
(a) See Figure 3.3
b2 − 4b − a2 + 4a
f (b) − f (a)
=
. The average
b−a
b−a
rates of change over the intervals are summarized below:
(b) To find the average rates of change we take
Interval
[1, 3]
[−1, 1]
[1, 1.1]
[0.9, 1]
[1 − h, 1]
Average rate
0
−4
−1.9
−2.1
−2 − h
(c) We find f 0 (1) by taking the limh→0 of the average rate of change between x = 1 and a point
very close to x = 1, in this case x = 1 − h. Taking the limit of the average rate of change
calculated in (b), we get f 0 (1) = lim (−2 − h) = −2.
h→0
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5
Math 102 Problems
Chapter 3
y
0
4
x
Figure 3.3: Figure for Problem 3.7
3.8
Given y = f (x) = x2 − 2x + 3.
(a) Find the average rate of change over the interval [2, 2 + h].
(b) Find f 0 (2).
(c) Using only the information from (a), (b) and f (2) = 3, approximate the value of y when
x = 1.99, without substituting x = 1.99 into f (x).
Detailed Solution:
(a)
((2 + h)2 − 2(2 + h) + 3) − (22 − 2 · 2 + 3)
f (2 + h) − f (2)
=
(2 + h) − 2
h
2
(4 + 4h + h − 4 − 2h + 3) − (4 − 4 + 3)
=
h
h2 + 2h
=
h
= h+2
(b) f 0 (2) can be found by letting h go to zero in the last expression.
f 0 (2) = lim (h + 2) = 2
h→0
(c) f (2) = 3. Since going from x = 2 to x = 1.99 is a small change, the tangent line at x = 2
can be used to approximate the secant line connecting the points (2, f (2)) and (1.99, f (1.99).
If we use the slope of the tangent line, f 0 (2), as the slope of the secant line, we get slope
(2)
∆y
= ∆x
= f (1.99)−f
. ∆x = 1.99 − 2 = −0.01. So ∆y = slope ×∆x ≈ 2 · (−0.01) = −0.02.
1.99−2
y = f (1.99) = f (2) + ∆y ≈ 2.98.
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6
Math 102 Problems
Chapter 3
3.9
Find the average rates of the given function over the given interval. Express your answer in terms
of square roots and π. Do not compute the decimal expressions.
(a) Find the average rate of change of tan(x) over 0 ≤ x ≤
(b) Find the average rate of change of cot(x) over
π
4
≤x≤
π
4
(Hint: tan(x) =
sin(x)
).
cos(x)
π
3
(Hint: cot(x) =
cos(x)
).
sin(x)
Detailed Solution:
sin( π4 )
π
(a) tan( ) =
=
4
cos( π4 )
√
2
2
√
2
2
= 1. tan(0) =
sin(0)
cos(0)
=
0
1
= 0. The average rate of change is
tan( π4 ) − tan(0)
4
1−0
= π =
π
−0
π
4
4
cos( π4 )
π
=
(b) cot( ) =
4
sin( π4 )
√
2
2
√
2
2
cos( π3 )
π
= 1. cot( ) =
=
3
sin( π3 )
cot( π3 ) − cot( π4 )
=
π
− π4
3
3.10
√
3
3
1
2
√
3
2
=
−1
π
12
√
3
. The average rate of change is
3
√
4 3 − 12
=
π
Average velocity
The vertical height of a ball, d (in meters) at time t (seconds) after it was thrown upwards was
found to satisfy d(t) = 14.7t − (1/2)gt2 where g = 9.8 m/s2 for the first 3 seconds of its motion.
(a) What happens after 3 seconds?
(b) When is the ball at its highest position ?
(c) What is the average velocity of the ball between the times t = 0 and t = 1 second?
Detailed Solution:
We know that d(t) = 14.7t − 4.9t2 for the first 3 seconds. The graph of this function is shown in
Figure 3.4.
(a) The vertical height d(t) becomes negative after 3 seconds. It means that the ball is below its
original height at t = 0. (Another interpretation is that the model is no longer valid, e.g. if
the ball has landed on the ground it will not keep falling.)
(b) To find the highest position, i.e. maximal value of d(t), we set the derivative of d with respect
to t equal to zero i.e. d0 (t) = 14.7 − 9.8t = 0. Solving for t, we get t = 14.7/9.8 = 1.5 seconds.
(c) The average velocity of the ball between t = 0 and t = 1 can be found as follows: at
t = 0, d = 0 and at t = 1, d = 14.7 − 4.9 = 9.8. Thus the average velocity between these times
is: vav = (9.8 − 0)/1 = 9.8 m/s.
v.2005.1 - September 4, 2009
7
Math 102 Problems
Chapter 3
d(t) = 14.7t-4.9t^2
10
5
0
0
1
2t
3
4
-5
-10
-15
-20
Figure 3.4: Vertical height of the ball for Problem 3.10
.
3.11
A ball is dropped from height h0 = 490 meters above the ground. Its height, y, at time t is known
to follow the relationship y(t) = h0 − 21 gt2 where g = 9.8 m /s2 .
(a) Find the average velocity of the falling ball between t = 1 and t = 2 seconds.
(b) Find the average velocity between t sec and t+ where 0 < < 1 is some small time increment.
(Assume that the ball is in the air during this time interval.)
(c) Determine the time at which the ball hits the ground.
Detailed Solution:
(a) At time t = 1 the ball is at y(1) = 490 − 4.9(1) = 485.1 and at t = 2 it is at y(2) = 470.4 so
the average velocity between time t = 1 and t = 2 seconds is
v¯ =
470.4 − 485.1
= −14.7 m/s.
2−1
(The minus sign signifies that the ball is losing height.)
(b)
v¯ =
h0 − g2 (t + )2 − (h0 − 2g t2 )
g
g
g
= − (t2 + 2t + 2 − t2 ) = − (2t + ) = −gt −
2
2
2
(c) The ball hits the ground when y = 0, i.e. when 0 = 490 − 4.9t2 . Thus t2 = 100, so t = 10
seconds.
3.12
A ball is thrown from the top of a building of height h0 . The height of the ball at time t is given by
1
h(t) = h0 + v0 t − gt2
2
v.2005.1 - September 4, 2009
8
Math 102 Problems
Chapter 3
where h0 , v0 , g are positive constants. Find the average velocity of the ball for the time interval
0 ≤ t ≤ 1 assuming that it is in the air during this whole time interval. Express your answer in
terms of the constants given in the problem.
Detailed Solution:
The average velocity over the time interval 0 ≤ t ≤ 1 is
vav =
h(1) − h(0)
h0 + v0 · 1 − (g · 12 /2) − h0
=
= v0 − g/2.
1−0
1
3.13
(a) Find the slope of the secant line to the graph of y = 2/x between the points x = 1 and x = 2.
(b) Find the average rate of change of y between x = 1 and x = 1 + where > 0 is some positive
constant.
(c) What happens to this slope as → 0 ?
(d) Find the equation of the tangent line to the curve y = 2/x at the point x = 1.
Detailed Solution:
(a) The slope of the secant line is
− 22
= −1.
s=
(1 − 2)
2
1
(b) The average rate of change between x = 1 and x = 1 + is
s=
2
1
2
− 1+
(1 + − 1)
−2
=2
=
.
−
−(1 + )
1+
(c) As → 0 this slope approaches -2, which is the slope of the tangent line to the curve at the
point x = 1.
(d) The tangent line goes through (1, 2) and has slope −2, so the equation of this line is
or y = −2x + 4.
y−2
x−1
= −2
3.14
For each of the following motions where s is measured in meters and t is measured in seconds, find
the velocity at time t = 2 and the average velocity over the given interval.
(a) s = 3t2 + 5 and t changes from 2 to 3s.
(b) s = t3 − 3t2 from t = 3s to t = 5s.
(c) s = 2t2 + 5t − 3 on [1, 2].
v.2005.1 - September 4, 2009
9
Math 102 Problems
Chapter 3
Detailed Solution:
(a) v(2) = s0 (2) where s0 (t) = 6t, so v(2) = 6 · 2 = 12 m/s. Average velocity on [2, 3] is
s(3) − s(2)
32 − 17
=
= 15 m/s
3−2
1
(b) v(2) = s0 (2) where s0 (t) = 3t2 − 6t, so v(2) = 3 · 22 − 6 · 2 = 0 m/s. Average velocity on [3, 5]
is
50 − 0
s(5) − s(3)
=
= 25 m/s
5−3
2
(c) v(2) = s0 (2) where s0 (t) = 4t + 5, so v(2) = 4 · 2 + 5 = 13 m/s. Average velocity on [1, 2] is
15 − 4
s(2) − s(1)
=
= 11 m/s
2−1
1
3.15
Shown in Figure 3.5 is the graph of some function f (x). Sketch the graph of its derivative, f 0 (x).
1.8
1.6
1.4
1.2
1
0.8
0.6
0.4
0.2
–3
–2
–1
0
1
2
3
x
Figure 3.5: Figure for Problem 3.15
Detailed Solution:
The derivative is shown in Figure 3.6
2
1
–3
–2
–1
0
1
x
2
3
–1
–2
Figure 3.6:
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10
Math 102 Problems
Chapter 3
3.16
Shown in Figure 3.7 below are three functions, f (x) (dotted lines). Sketch the derivatives of these
functions, f 0 (x).
y
y
y
x
x
x
Figure 3.7: Figure for problem 3.16
Detailed Solution:
See solid lines in Figure 3.8. Note that the peaks in f (x) (dashed curve) coincide with the places
where f 0 (x) = 0. In (c) note that there are sharp corners in the function at which the value of the
derivative is not defined. In fact, the graph of the derivative is a set of straight lines with jump
discontinuities.
y
y
y
x
x
x
Figure 3.8: Figure for solution to 3.16
3.17
The velocity v of an object attached to a spring is given by v = −Aω sin(ωt + δ), where A, ω and
δ are constants. Find the average change in velocity (“acceleration”) of the object for the time
.
interval 0 ≤ t ≤ 2π
ω
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11
Math 102 Problems
Chapter 3
Detailed Solution:
) − v(0)
+ δ) − (−Aω sin(ω · 0 + δ))
−Aω sin(ω · 2π
v( 2π
ω
ω
=
2π
2π
−0
ω
ω
−Aω sin(2π + δ) + Aω sin δ
=
2π
ω
Since sin(2π + δ) = sin δ, the above expression can be simplified to
−Aω sin δ + Aω sin δ
2π
ω
=0
The average rate of change is zero, so the object has the same velocity at t = 0 and t =
which are separated by exactly one period of oscillation.
2π
,
ω
3.18
Shown in Figure 3.9 is the graph of the velocity of a particle moving in one dimension. Indicate
directly on the graph any time(s) at which the particle’s acceleration is zero.
v
t
Figure 3.9: Figure for Problem 3.18
Detailed Solution:
The acceleration is the derivative of the velocity, and satisfies a = dv/dt. This means that acceleration is the slope of the velocity curve. Thus, acceleration is zero whenever the tangent is
horizontal to this curve, i.e. at the critical points. There are three such places, at the minima and
the maximum of the graph.
3.19
Use the definition of derivative to calculate the derivative of the function
1
f (x) =
x+1
(intermediate steps required).
v.2005.1 - September 4, 2009
12
Math 102 Problems
Chapter 3
Detailed Solution:
f (x + h) − f (x)
1
1
1
f (x) = lim
= lim
−
h→0
h→0 h x + h + 1
h
x+1
1 [x + 1 − (x + h + 1)]
−1
−1
= lim
= lim
=
h→0 h
h→0 (x + h + 1)(x + 1)
(x + h + 1)(x + 1)
(x + 1)2
0
3.20
Concentration gradient
Certain types of tissues, called epithelia are made up of thin sheets of cells. Substances are taken up
on one side of the sheet by some active transport mechanism, and then diffuse down a concentration
gradient by a mechanism called facilitated diffusion on the opposite side. Shown in Figure 3.10 is the
concentration profile c(x) of some substance across the width of the sheet (x represents distance).
Sketch the corresponding concentration gradient, i.e. sketch c0 (x), the derivative of the concentration
with respect to x.
c(x)
facilitated
diffusion
active
transport
distance across the sheet
x
Figure 3.10: Figure for Problem 3.20
Detailed Solution:
See Figure 3.11 for the sketch.
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13
Math 102 Problems
Chapter 3
c’(x)
x
distance across the sheet
Figure 3.11: Sketch of concentration gradient for Solution to problem 3.20
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14