Page 29 - u4

3
30. f(x) =
2
2
x − 9
2x − 1
2
29. f(x) = x
e + e
−x
e − e
x
−x
x
and y = cosh x =
graphs of y = sinh x =
.
1
2
2
32. f(x) = sin x −
−2x
sin x
31. f(x) = e
sin 2x
2
56. On the same axes as the graphs of exercise 55, sketch in the
4
2
33. f(x) = x − 16x + 42x − 39 6x + 14
3
1 x
graphs of y = e and y = e . Explain why these graphs
1 −x
2
2
2
34. f(x) = x + 32x − 0 02x − 0 8x
4
3
serve as an envelope for the graphs in exercise 55. (Hint: As
x
±∞, what happens to e and e ?)
−x
x
25 − 50 x + 0 25
2
1
36.
−1
P1: OSO/OVY 35. f(x) =
QC: OSO/OVY f(x) = tan
T1: OSO
P2: OSO/OVY
x − 1
x
2
............................................................
17:24
UAE-Math-Grade-12-Vol-1-SE-718383-ch4
GO01962-Smith-v1.cls
October 25, 2018
APPLICATIONS
In exercises 37–42, the “family of functions” contains a param-
eter c. The value of c affects the properties of the functions.
57. In a variety of applications, researchers model a
Determine what differences, if any, there are for c being zero, 55. Find all extrema and inflection points, and sketch the
phenomenon whose graph starts at the origin, rises to a
positive or negative. Then determine what the graph would look single maximum and then drops off to a horizontal asymp-
CHAPTER 4 • •
222
Applications of Differentiation
4-12
like for very large positive c’s and for very large negative c’s. tote of y = 0 For example, the probability density function
4
4
2
3−w∕2
37. f(x) = x + cx 2 38. f(x) = x + cx + x of events such as the time from conception to birth of an
≈
. Based on this, describe how changes if
APPLICATIONS animal and Dthe amount of time surviving after contracting
2
x
there is an increase in (i) w or (ii) D.
2
39. f(x) = 40. f(x) = e −x ∕c a fatal disease might have these properties. Show that the
x + c
2
2
55. A water wave of length L meters in water of depth d meters family of functions xe −bx has these properties for all positive
D
41. f(x) = sin(cx) 42. f(x) = x 2 c − x 2 constants b What effect does b have on the location of the
2
has velocity v satisfying the equation
............................................................ maximum? In the case of the time since conception, what
ϕ
−2 d∕L
2 d∕L
w
v = 4.9L e − e . would b represent? In the case of survival time, what would
2
e
−2 d∕L
+ e
2 d∕L
A function f has a slant asymptote y mx b (m 0) b represent?
0
(mx
0 and/or
2
b)]
(mx
b)]
if Treating L as a constant and thinking of v as a function
[f(x)
[f(x)
x x 58. The “FM” in FM radio stands for frequency d modulation,
f(d), use a linear approximation to show that f(d) ≈ 9.8d for
In exercises 43–48, find the slant asymptote. (Use long divi- a method of transmitting information encoded in a radio
small values of d. That is, for small depths, the velocity of
sion to rewrite the function.) Then, graph the function and its wave by modulating (or varying) the frequency. A basic ex-
√
asymptote on the same axes. 9.8d and is independent of the ample of such a modulated wave is f(x) = cos (10x + 2 cos x).
the wave is approximately
Exercise 61
wavelength L. 2 Use computer-generated graphs of f(x), f (x) and f (x)totry
′
′′
2
(b) The shooter in part (a) is assumed to be in the center
43. f(x) = 3x − 1 44. f(x) = 3x − 1 to locate all local extrema of f(x)
56. Planck’s law states that the energy density of blackbody ra-
x
x − 1
of the ice. Suppose that the line from the shooter to
diation of wavelength x is given by x − 1 the center of the goal makes an angle of with cen-
3
2
3
x − 2x + 1
45. f(x) = 46. f(x) = 59. The angle for a field goal kicked from a hash mark at a
ter line. For the goalie to completely block the goal, he
2
x 2 f(x) = 8 hcx −5 x − 1 distance of x meters is A = tan −1 29 25 − tan −1 10 75 .
.
must stand d feet away from the net where d = D(1 −
x
x
e
hc∕(kTx)
− 1x − 1
47. f(x) = x 4 48. f(x) = 4 Find x to maximize the angle A. A 4.5-meter penalty
w∕6 cos ). Show that for small angles, d ≈ D(1 − w∕6).
x + x
x + 1
3
Use the linear approximation in exercise 44 to show that increases x from 18.3 to 22.9. How does this change A?
3
............................................................ 62. In Einstein’s theory of relativity, the length of an object
f(x) ≈ 8 kT∕x , which is known as the Rayleigh-Jeans law.
4
depends on its velocity. If L is the length of the object at
60. A knuckleball thrown with 0 rotation rate (in rad/s)
57. Newton’s theory of gravitation states that the weight of
rest, v is the object’s velocity and c is the speed of light, the
In exercises 49–52, find a function whose graph has the given has lateral position x(t) = 2 5 t − 2 5 sin 4 t at time t, for
a person at elevation x feet above sea level is W(x) =
4
2
asymptotes. 2 Lorentz contraction formula for the length of the object is
PR ∕(R + x) , where P is the person’s weight at sea level and 0 t √ 0 68. Explore the effect on the graph of changing
2
2
2
0.
R is the radius of the earth (approximately 20,900,000 feet).
49. x = 1,x = 2 and y = 3 50. x =−1,x = 1 and y = 0 L = L 0 1 − v ∕c . Treating L as a function of v, find the
linear approximation of L at v = 0.
Find the linear approximation of W(x) at x = 0. Use the lin-
51. x =−1,x = 1,y =−2 and y = 2
ear approximation to estimate the elevation required to re-
duce the weight of a 120-pound person by 1%.
52. x = 1,y = 2 and x = 3
............................................................ EXPLORATORY EXERCISES
EXPLORATORY EXERCISES
58. One important aspect of Einstein’s theory of relativity is
that mass is not constant. For a person with mass m at
0
1. An important question involving Newton’s method is how
53. It can be useful to identify asymptotes other than verti- 1. One of the natural enemies of the balsam fir tree is the
√
rest, the mass will equal m = m ∕ 1 − v ∕c at velocity v
2
2
fast it converges to a given zero. Intuitively, we can dis-
2
0
cal and horizontal. For example, the parabola y = x is an spruce budworm, which attacks the leaves of the fir tree
(where c is the speed of light). Thinking of m as a function
2
asymptote of f(x) if lim[f(x) − x ] = 0 and/or lim [f(x) − in devastating outbreaks. Define N(t) to be the number
2
tinguish between the rate of convergence for f(x) = x − 1
of v, find the linear approximation of m(v) at v = 0. Use of worms 2
x ∞
x −∞
(with x = 1.1) and that for g(x) = x − 2x + 1 (with x =
0 on a particular tree at time t. A mathematical
2
4
0
the linear approximation to show that mass is essentially
2
x ] = 0 Show that x is an asymptote of f(x) = x − x + 1 . model of the population dynamics of the worm must in-
2
1.1). But how can we measure this? One method is to
2
x − 1
constant for small velocities.
take successive approximations x
n rate due to
and x and compute
Graph y = f(x) and zoom out until the graph looks like a clude a term to indicate the worm’s death
n−1
59. The spruce budworm is an enemy of the balsam fir tree. In its predators (e.g., n n−1 . To discover the importance
the difference Δ = x − xThe form of this term is often
n birds).
parabola. (Note: The effect of zooming out is to emphasize
one model of the interaction between these organisms, pos-
of this quantity, run Newton’s method with x = 1.5 and
large values of x ) taken to be B[N(t)] 2 for positive constants A and B
0
sible long-term populations of the budworm are solutions then compute the ratios Δ ∕Δ , Δ ∕Δ , Δ ∕Δ and so on,
A + [N(t)]
2
2
4
2
3
5
3
4
54. For each function, find a polynomial p(x) such that Graph the functions 4 + x 3 2 , 1 + x 2 , 3 9 + x 2 2 and 1 + x 2 2 for
2
of the equation r(1 − x∕k) = x∕(1 + x ), for positive con-
for each of the following functions:
2x
2
2
x
2
x
3x
2
Copyright © McGraw-Hill Education (a) x + 1 (b) x − 1 (c) x − 2 x is a plausible model for the 3 4 death rate by predation.
lim[f(x) − p(x)] = 0
stants r and k. (a) Find all positive solutions of the equation
x ∞
4
F (x) = (x − 1)(x + 2) = x + 5x + 6x − 4x − 8,
B[N(t)]
with r = 0.5 and k = 7. (b) Repeat with r = 0.5 and k = 7.5.
1
6
5
0 Based on
2 these graphs, discuss why
x
4
2
4
3
2
For a small change in the environmental constant k (from
A +
F (x) = (x − 1) (x + 2) = x + 2x − 3x − 4x + 4, [N(t)]
2
2
2
x + 1
x + 1
7 to 7.5), how did the solution change? The largest solution
3
3
2
F (x) = (x − 1) (x + 2) = x − x − 3x + 5x − 2 and
3
4constants A and B play? The possi-
Show by zooming out that f(x) and p(x) look similar
corresponds to an “infestation” of the spruce budworm. for
What role do the
2
4
F (x) = (x − 1) = x − 4x + 6x − 4x + 1.
4
large x
ble stable population levels for the spruce budworms are
60. Suppose that a species reproduces as follows: with prob-
ability p , an organism has no offspring; with probability
0
n→∞ Δ
p , an organism has one offspring; with probability p , In each case, conjecture a value for the limit r = lim Δ n+1 .
n
2
1
an organism has two offspring and so on. The probability If the limit exists and is nonzero, we say that Newton’s
429
that the species goes extinct is given by the smallest non- method converges linearly. How does r relate to your in-
2
negative solution of the equation p + p x + p x + ⋅⋅⋅ = x. tuitive sense of how fast the method converges? For f(x) =
0
1
2
4
Find the positive solutions of the equations 0.1 + 0.2x + (x − 1) , we say that the zero x = 1 has multiplicity 4. For
3
3
2
2
3
0.3x + 0.4x = x and 0.4 + 0.3x + 0.2x + 0.1x = x. Ex- f(x) = (x − 1) (x + 2),x = 1 has multiplicity 3 and so on.
plain in terms of species going extinct why the first equation
420_430_ADVM_G12_S_C06_L06_v2_718384 How does r relate to the multiplicity of the zero? Based on
October 8, 2016 10:46 AM
has a smaller solution than the second. this analysis, why did Newton’s method converge faster for
2
2
Program: UAE Component: MATH f(x) = x − 1 than for g(x) = x − 2x + 1? Finally, use New-
61. (a) In the diagram, a hockey player is D feet from the net ton’s method to compute the rate r and hypothesize the
1st Pass
on the central axis of the rink. The goalie blocks off multiplicity of the zero x = 0 for f(x) = x sin x and g(x) =
GRADE: 12
Vendor: MPS
a segment of width w and stands d feet from the net. x sin x .
2
The shooting angle to one side of the goalie is given
[ 3(1 − d∕D) − w∕2 ] 2. This exercise looks at a special case of the three-body prob- Copyright © McGraw-Hill Education
by = tan −1 . Use a linear approx-
D − d lem, in which there is a large object A of mass m ,a
A
imation of tan −1 x at x = 0 to show that if d = 0, then much smaller object B of mass m ≪ m and an object C of
A
B
238 | Lesson 4-1 | Linear Approximations and Newton’s Method