Recall that a differential equation is an equation involving a derivative and a function is a solution to the differential equation if it satisfies the differential equation.. For each o
Trang 118 A wheel of radius 5 meters is oriented vertically and spinning counterclockwise at a rate of 7 revolutions per minute If the origin is placed at the center of the wheel a point on the rim has a horizontal position of (7, 0) at time t = 0 What is the horizontal component of the point’s velocity at t = 2?
19 In this problem you will show that y = C1sin kx + C2cos kx is a solution to the differential equation
y= −k2y Recall that a differential equation is an equation involving a derivative and a function
is a solution to the differential equation if it satisfies the differential equation (a) Show that y1= sin kx is a solution to y= −k2y To do this, first find y1 Then write
y1= −k? 2y1 and verify that the two sides are indeed equal
(b) Show that y2= C1sin kx is a solution to y= −k2y
(c) Show that if y1and y2 are solutions to y= −k2y, then y3= C1y1+ C2y2 is
a solution to y= −k2y as well Conclude that y = C1sin kx + C2cos kx is a solution to the differential equation y= −k2y
20 For each of the functions below, determine whether the function is a solution to differential equation (i), differential equation (ii), or neither Differential equations (i) and (ii) are given below
i.y= 16y ii y= −16y
(a) y1(t ) = sin 16t
(b) y2(t ) = e4t
(c) y3(t ) = 3 cos 4t
(d) y4(t ) = sin 4t + 1
(e) y5(t ) = e−16t
(f ) y6(t ) = −3e−4t
(g) y7(t ) = e4t+ 3
(h) y8(t ) = − sin 4t
(Hint: Four of the eight functions given are solutions to neither differential equation.)
21 Each of the functions below is a solution to one of the differential equations below.
i y= 9 ii y= 9y iii y= −9y
For each function, determine which of the three differential equations it satisfies (a) y1(t ) = 5 sin 3t
(b) y2(t ) = e3t
(c) y3(t ) = 2 cos 3t
(d) y4(t ) = 4.5t2+ 3t + 8
(e) y5(t ) = 4e−3t
(f ) y6(t ) = 4.5t2− t + 2
Trang 222 After having done the previous problem, make up a solution to each of the three differential equations below
i y= 9 ii y= 9y iii y= −9y Your answers must be different from the solutions given in the preceding problem, but you can use those answers for inspiration
Check that your answers are right by“plugging them back” into the differential equation For instance, if you guess that y = e3t+ 1 is a solution to the differential equation y= 9y, test it out as follows First calculate y Since y= e3t· 3, y= 9e3t Now see if it satisfies the differential equation
y ?
= 9y 9e3t ?
= 9(e3t+ 1) 9e3t= 9e3t+ 9
So y = e3t+ 1 is not a solution to y= 9y
23 As you’re riding up an elevator inside the Hyatt Hotel right next to the Charles River, you watch a duck swimming across the Charles, swimming straight toward the base
of the elevator The elevator is rising at a speed of 10 feet per second, and the duck is swimming at 5 feet per second toward the base of the elevator As you pass the eighth floor, 100 feet up from the level of the river, the duck is 200 feet away from the base
of the elevator
(a) At this instant, is the distance between you and the duck increasing or decreasing?
At what rate?
(b) As you’re watching the duck, you have to look down at more and more of an angle
to see it At what rate is this angle of depression increasing at the instant when you are at a height of 100 feet? Include units in your answer
(Problem written by Andrew Engelward)
24 Approximating the function f (x) = sin x near x = 0 by using polynomials.
The point of this problem is to show you how the values of sin x can be approxi-mated numerically with a very high degree of accuracy It is an introduction to Taylor polynomials
(a) Find the equation of the line tangent to f (x) = sin x at x = 0
(b) Find the equation of a quadratic Q(x) = a + bx + cx2 such that the function Q(x)and its nonzero derivatives match those of sin x at x = 0 In other words, Q(0) = f (0), Q(0) = f(0), and Q(0) = f(0) The quadratic that you found
is the quadratic that best “fits” the sine curve near x = 0 In fact, the “quadratic” turns out not to really be a quadratic at all Sine is an odd function, so there is no parabola that “fits” the sine curve well at x = 0
(c) Find the equation of a cubic C(x) = a + bx + cx2+ dx3such that the function C(x)and its nonzero derivatives match those of sin x at x = 0 In other words, C(0) = f (0), C(0) = f(0), C(0) = f(0), and C(0) = f(0) The cubic that you found is the cubic that best “fits” the sine curve near x = 0
Using a calculator, on the same set of axes graph sin x, the tangent line to sin x at x = 0, and C(x), the cubic you found Now “zoom in” around x = 0 Can
Trang 3you see that near x = 0 the line is a good fit to the sine curve but the cubic is an even better fit for small x and the cubic hugs the sine curve for longer? The next set of questions asks you to investigate how good the fit is
(d) Use C(x) from part (c) to estimate the following, and then compare with the actual value using a calculator
sin(0.01) sin(0.1) sin(0.5) sin(1) sin(3)
(e) (Challenge) Find the “best” fifth degree polynomial approximation of sin(x) for x near 0 by making sure that the first five derivatives of the polynomial match those
of sin(x) when evaluated at x = 0 Graph sin x along with its first, third, and fifth degree polynomial approximations on your graphing calculator The higher the degree of the polynomial, the better the fit to sin x near x = 0, right?
(f ) Using a calculator, on the same set of axes graph sin x and the polynomial given below
x − x 3
6 + x
5
120− x
7
5040+ x
9 362880 This polynomial is an even better fit than the last one, right?
Now graph the difference between sin x and this polynomial; in other words, graph
y = sin x −
x −x 3
6 + x
5
120 − x
7
5040+ x
9 362880
On approximately what interval is the difference between sin x and this polynomial less than 0.005?
FUNCTIONS
We can find the derivatives of inverse trigonometric functions using implicit differentiation
The Derivative of arcsin x
Let y = arcsin x Our goal is to find dydx
y = arcsin x yis the angle between − π
2 and
π
2 whose sine is x. sin y = x Therefore, the sine of y is x
(sine and arcsine are inverse functions.) d
dx sin y = d
dxx Differentiate each side with respect to x
(cos y)dy
dx = 1 Remember the Chain Rule on the left-hand side,
since y is a function of x
dy
cos y Solve for
dy
dx
Trang 4We’d like to expressdydx in terms of x, not y We know that x = sin y and that sin y and cos y are related by sin2y + cos2y = 1 Solving for cos y gives us
cos y = ±
1 − sin2y
= ±1 − x2
Do we want the positive square root, or the negative square root? y is an angle between
−π2 andπ2, so cos y is nonnegative; we choose√
1 − x2 Therefore, dydx=√1
1−x 2 d
dx arcsin x =√ 1
1 − x2 Alternatively, once we knowdydx =cos y1 , we can replace y by arcsin x obtaining
dy
cos(arcsin x). cos(arcsin x) is the cosine of the angle between −π/2 and π/2 whose sine is x We can draw a triangle with an acute angle θ whose sine is x, use the Pythagorean Theorem to solve for the length of the third side, and find cos(arcsin x) =√1 − x2, as illustrated in Figure 21.12.12
1
θ
x
√1–x2
Figure 21.12
The Derivative of arctan x
We can use the same method to computedxd arctan x
y = arctan x yis the angle between −π
2 and
π
2 whose tangent is x. tan y = x
d
dx tan y =dxd x Use the Chain Rule on the left-hand side
(sec2y)dy
dx = 1 dy
dx = cos2y
12 Knowing that θ ∈ [−π/2, π/2] assures us that cos θ is nonnegative.
Trang 5To express dydx in terms of x we can replace y with arctan x cos(arctan x) is the cosine
of the angle between −π/2 and π/2 whose tangent is x Notice that cosine is positive on this interval We can draw a triangle with an acute angle θ whose tangent is x, use the Pythagorean Theorem to solve for the length of the third side, and find cos(arctan x) = 1
√ 1+x 2, as illustrated in Figure 21.13
1 θ
x
√1+x 2
Figure 21.13
cos y =√1
1+x 2, so dydx = cos2y =1+x12
d
dx arctan x = 1
1 + x2
EXERCISE 21.4 Using the same method demonstrated in the previous two examples, show the following
d
dx arccos x =√−1
1 − x2
COMMENT Surprisingly, the derivatives of the inverse trigonometric functions involve neither trigonometric functions nor inverse trigonometric functions We will find it handy to keep inverse trigonometric functions and their derivatives in mind later on, when searching for a function whose derivative is 1
1+x 2 or √1
1−x 2 We mention this here because the inverse
trigonometric functions are probably not the first things you might think of when looking
for such functions!
EXAMPLE 21.4 Differentiate
(a) f (x) = x arcsin(x2) (b) j (x) = 5 tan−1x2 (c) g(x) = tan√x · tan−1√x
SOLUTIONS (a) Use the Product Rule The Chain Rule tells us that
d
dxarcsin(mess) = √ 1
1−(mess) 2 · (mess) Here mess = x2
f(x) = x · d
dx[arcsin(x2)] + arcsin(x2)
= x · 1
1 − (x2)2· 2x + arcsin(x2)
2
√
1 − x4+ arcsin(x2)
Trang 6(b) The Chain Rule tells us thatdxd tan−1(mess) =1+(mess)1 2 · (mess), where mess = x2.
j(x) = 5
1 + (x2)2· 2x
1 + x4 CAUTION (mess)is the derivative of x2, not of x4
(c) g(x) = tan√x · tan−1√x
Again we need the Product Rule together with the Chain Rule
g(x) =
sec2√
x ·1 2
1
√x
· tan−1√x + tan√x · 1
1 + (√x)2·1
2
1
√x
2√ x
sec2√
x · tan−1√x + tan
√x
1 + x
P R O B L E M S F O R S E C T I O N 2 1 4
For Problems 1 through 6, differentiate the function given.
1 y = 3 tan x − 4 tan−1x
2 f (x) = 3 arctan2√x
3 y = sin x · arcsin x
4 y =√tan−1x
5 y = x tan−1x
6 y = arctan(ee x)
7 (a) Show that the derivative of arccos x is √−1
1−x 2 (b) What is the domain of arccos x?
(c) What is the range of arccos x?
(d) Where is the graph of arccos x decreasing?
(e) Where is the graph of arccos x concave up? Concave down? If there is a point of inflection, where is it?
(f ) Graph f (x) = arccos x
8 Differentiate f (x) = 3 cos x21+1 + x arctan
1
x
9 Compute dxd sin−1−1x Is it the same as dxd tan−1x?
Trang 721.5 BRIEF TRIGONOMETRY SUMMARY
Unit Circle Definitions of Trigonometric Functions
sin x = v
cos x = u
tan x = sin x
cos x =uv
2 1
1 0
– 0.42
v
u (cos 2, sin 2) = P(2)
u2 + v2 = 1
Right-Triangle Definitions of Trigonometric Functions
sin x =opp
hyp csc x = 1
sin x cos x = adj
cos(x) tan x =opp
tan(x)
x
hyp
opp
adj
Special triangles
1
2
1
2
π
4
√ 2
2
1 2
2 2
π
3
π
6
√ 3 2
Note:The shortest side is opposite the smallest angle and the largest side is opposite the largest angle
45◦-45◦-90◦ 30◦-60◦-90◦
Trang 8Graphs of the Trigonometric Functions
π –π
Graph of f(x) = tan(x) Graph of f(x) = cos(x)
x
Domain: x ≠ + πn Range: (– ∞, ∞)
2
x
1
–1
π
–2π
Graph of f(x) = sin(x)
Domain: (– ∞, ∞) Range: [–1, 1]
x
1
–1
π
–2π
The Relationship Between Graphs and Equations
y = A sin(Bx) + K
Acos(Bx) + K amplitude = |A|, period =
2π
|B|, balance value = K
y = A sin(Bx + C) should be rewritten as y = A sin B x +CB ; to read off amplitude = |A|, period =2π|B|, shift the graph of y = A sin(Bx) to the left CB units
Inverse Trigonometric Functions
We must restrict the domain of sine, cosine, and tangent in order to construct inverse functions, because a function must be one-to-one to have an inverse function
arcsin(x) is the angle between −π/2 and π/2 whose sine is x domain of arcsin x: [−1, 1] arctan(x) is the angle between −π/2 and π/2 whose tangent is x domain of arctan x: (−∞, ∞) arccos(x) is the angle between 0 and π whose cosine is x domain of arccos x: [−1, 1]
NOTEIf A is positive then arcsin(A), arccos(A), and arctan(A) are all angles between 0 and π/2 (provided A is in the function’s domain)
If A is positive and in the function’s domain, then arcsin(−A) = − arcsin(A),
arccos(−A) = π − arccos(A), and arctan(−A) = − arctan(A)
Arc Length The arc length of a circle of radius R subtended by an angle of x radians is
Rx
Derivatives
f (x) = sin(x) f(x) = cos(x) f (x) = arcsin(x) f(x) =√1
1−x 2
f (x) = cos(x) f(x) = − sin(x) f (x) = arccos(x) f(x) = −√1
1−x 2
f (x) = tan(x) f(x) = sec2(x) f (x) = arctan(x) f(x) =1+x12
Trang 9Remember the Chain Rule
d
dx sin[g(x)] = cos[g(x)] · dgdx dxd arcsin[g(x)] = 1
1 − [g(x)]2 · dgdx d
dx cos[g(x)] = − sin[g(x)] · dgdx dxd arccos[g(x)] = −1
1 − [g(x)]2 · dgdx d
dx tan[g(x)] = sec2[g(x)] · dgdx dxd arctan[g(x)] = 1
1 + [g(x)]2 · dgdx
Trigonometric Identities
Pythagorean Identity
sin2A + cos2A = 1
It follows that
sin2A cos2A+cos
2A cos2A= 1
cos2A so tan2A + 1 = sec2A sin2A
sin2A+cos
2A sin2A = 1
sin2A so 1 + cot2A = csc2A
Addition Formulas and the Double-Angle Formulas
sin(A + B) = sin A cos B + sin B cos A cos(A + B) = cos A cos B − sin A sin B
It follows that
sin(2A) = 2 sin A cos A and cos(2A) = cos2A − sin2A
General Triangles
A
B
C
b
Law of Cosines
c2= a2+ b2− 2ab cos C You can think of −2ab cos C as a “correction term” for the Pythagorean Theorem; it is positive when C > π/2 and negative when C < π/2
Law of Sines
a sin A= b
sin B = c
sin C
... domain of sine, cosine, and tangent in order to construct inverse functions, because a function must be one -to- one to have an inverse functionarcsin(x) is the angle between −π/2 and π/2... x domain of arcsin x: [−1, 1] arctan(x) is the angle between −π/2 and π/2 whose tangent is x domain of arctan x: (−∞, ∞) arccos(x) is the angle between and π whose cosine is x domain of arccos...
arccos(−A) = π − arccos(A), and arctan(−A) = − arctan(A)
Arc Length The arc length of a circle of radius R subtended by an angle of x radians is
Rx
Derivatives