Using the Chain Rule and either the Product or Quotient Rule allows us to find the derivatives of tan x, csc x, sec x, and cot x.. The third degree polynomial approximation of sin x arou
Trang 1Proof that limh →0 cos(h)h−1 = 0
lim
h →0
cos(h)− 1
h →0
cos(h)− 1
h · (cos(h)(cos(h)+ 1)
+ 1)
= lim
h →0
cos2(h)− 1
h · (cos(h)1
+ 1)
= lim
h →0
1− cos2(h)
h · (cos(h)−1
+ 1)
= lim
h →0
sin2(h)
(cos(h)+ 1)
= lim
h →0
sin(h)
h · − sin(h) (cos(h)+ 1) But
lim
h→0
sin(h)
h→0
− sin(h) (cos(h)+ 1)=
0
1+ 1 = 0, so
lim
h →0
cos(h)− 1
h = (1)(0) = 0
We have now shown thatdxd sin x= cos x, as conjectured!
The Chain Rule tells us thatdxd sin[g(x)]= cos[g(x)]g(x) Now that we have proven
d
dx sin x= cos x, the derivative of cos x is easy to tackle by using the fact that cos x and sin x are related to one another by a horizontal shift Looking back at the graphs of cos x and its derivative it is easy to speculate that the derivative of cos x is− sin x
Proof that the Derivative of cos x is − sin x
Observe that sinx +π
2 = cos x; replacing x by x +π
2 shifts the sine graph leftπ
2 units Similarly, cosx +π
2 = − sin x; replacing x by x +π
2 shifts the cosine graph left π
2
units
x
1
–1 π
2
–π
2
3 π
2 –3 π
2
π
1
–1 π
2
–π
2
3 π
2 –3 π
2
π
–2π
Figure 21.8
Trang 2692 CHAPTER 21 Differentiation of Trigonometric Functions
d
dx cos x=dxd sin
x+π2
= cos
x+π 2
· 1 (by the Chain Rule)
= − sin x Combining this result with the Chain Rule gives us d
dx sin[g(x)]= cos[g(x)] · g(x) or informally d
dx sin[mess]= cos[mess] · [mess] d
dx cos[g(x)]= − sin[g(x)] · g(x) d
dx cos[mess]= − sin[mess] · [mess],
where mess is a function of x.
Using the Chain Rule and either the Product or Quotient Rule allows us to find the derivatives of tan x, csc x, sec x, and cot x
dx tan x= sec2x
dx sec x= sec x tan x
EXAMPLE 21.1 Differentiate the following
(a) y= 3x sin(x2) (b) y= 7 cos2(3x+ 5) (c) y= tan(x2)
y= 3 sin(x2)+ 3x cos(x2)(2x)
= 3 sin(x2)+ 6x2cos(x2) (b) 7 cos2(3x+ 5) = 7[cos(3x + 5)]2, so basically this is 7[mess]2and its derivative is 14[mess] · [mess], where the mess is cos(3x+ 5) Then the Chain Rule must be applied to cos(3x+ 5)
y= 14 cos(3x + 5)(− sin(3x + 5)) · (3)
= −42 cos(3x + 5) sin(3x + 5) (c) y= tan(x2)=tan(x2)1/2 This is basically [mess]1/2, so its derivative is
1
2[mess]−1/2· [mess] We know mess= tan(stuff ), so [mess]= sec2(stuff ) · (stuff )
y= (1/2)tan(x2)−1/2sec2(x2)2x
y=xsec
2(x2) tan(x2)
Trang 3
An Excursion: Polynomial Approximations of Trigonometric Functions
We spent a fair amount of energy proving that limx →0 sin xx = 1 Now that this fact is ours,
we can get some mileage out of it From limx →0 sin xx = 1 it follows that sin x ≈ x for very small values of x.9This approximation is excellent for very small values of x, but
as x gets increasingly far from zero the approximation becomes poor and eventually is useless For example, sin 0.1= 0.0998334 ≈ 0.1 and sin 0.02 = 0.0199986 ≈ 0.02, but sin 3= 0.1411200 , which is not close to 3
y
y = x
x sin x
Figure 21.9
Historically the approximation sin x≈ x for very small values of x was used in ancient times, well before the development of calculus Using it, sin 1◦can be approximated with a great degree of accuracy From there, using addition formulas and knowledge about specific triangles, trigonometric tables can be built up The approximation sin x≈ x, when used to estimate sin 1◦, gives
sin 1◦= sin
π 180
≈180π = 0.01745329 , where the actual value of sin 1◦≈ 0.0174524
The approximation sin x≈ x, the tangent line approximation of sin x at x = 0, gives an estimate for sin x that is too large if x is positive and too small if x is negative (See Figure 21.9) This approximation can be improved upon The approximation sin x≈ x − x3/6, the best cubic approximation to sin x around x= 0, was used well before the seventeenth century This gives a higher degree of accuracy for x near zero Using it to approximate sin 1◦gives sin 1◦= sin(π/180) ≈ π/180 −1
6(π/180)3
≈ 0.0174524064
This is identical to the numerical approximation supplied by a calculator
The third degree polynomial approximation of sin x around x= 0 can be improved upon by using a fifth degree approximation, which can in turn be improved upon by using
a seventh degree polynomial (We use only polynomials of odd degree because sine is
an odd function.) As the degree of the polynomial used to approximate sine increases, the accuracy of the approximation near x= 0 increases and the interval around x = 0 for which the approximation is reasonable enlarges as well Amazingly enough, if we continue along in this way we can come up with an infinite “polynomial” that is exactly
equal to sin x everywhere These polynomial expansions are known as Taylor series,
after the British mathematician Brook Taylor (We take up Taylor series in Chapter 30) 9
Trang 4694 CHAPTER 21 Differentiation of Trigonometric Functions
By the seventeenth century mathematicians were using infinite polynomial expansions of functions, trigonometric and others The polynomial expansion of sin x is given by
sin x= x − x
3
3· 2+
x5
5· 4 · 3 · 2−
x7
7· 6 · 5 · 4 · 3 · 2+ · · ·
If we use factorial notation, letting n!= n(n − 1)(n − 2) 3 · 2 · 1, this can be written as
sin x= x −x
3
3! +x
5
5! −x
7
7! + · · · The first several terms can be used to approximate sin x for x small For instance,
sin(0.1)= 0.1 −(0.1)
3
3! +(0.1)
5
5! − · · · , so
sin(0.1)≈ 0.1 −(0.1)
3
3! = 0.09983 Compare this with sin(0.1)≈ 0.0998334166 Using one more term of the polyno-mial gives
sin(0.1)≈ 0.1 −(0.1)
3
3! +(0.1)
5
5! ≈ 0.0998334167
A good match!
P R O B L E M S F O R S E C T I O N 2 1 2
1 Using the derivatives of sine and cosine and either the Product Rule or the Quotient Rule, show that dxd tan x= sec2x
2 Show that dxd sec x= sec x tan x
3 Find the first and second derivatives of the following
(a) f (x)= 5 cos x (b) g(x)= −3 sin(2x) (c) h(x)= 0.5 tan x (d) j (x)= 2 sin x cos x
4 Differentiate the following
(a) y= cos2x (b) y= cos(x2) (c) y= x tan2x (d) y= sin3(x4) (e) y= 7[cos(5x) + 3]x
5 Consider the function f (x)= e−0.3xsin x
(a) For what values of x does f (x) have its local maxima and local minima? (b) Is f (x) a periodic function?
(c) Sketch the graph of f (x)= e−0.3xsin x
Trang 5(d) What is the maximum value of for e−0.3xsin x for x≥ 0? At what x-value is this maximum attained? Your answers must be exact, not numerical approximations from a calculator Give justification that this value is indeed the maximum
Evaluate the following derivatives u(x) is a differentiable function.
6 (a) dxd sin (u(x)) (b) dxd cos(u(x)) (c) dxdu(x)(sin x)
7 (a) dxdu(x)(cos x) (b) dxd tan(u(x)) (c) dxdu(x)(tan x)
Evaluate.
8 dxd sin(x3+ ln 3x)
9 dxd cos2(sin x)
10 dxd 1
sin 3 (cos 2x)
11 d
dx sin(2x3)
12 dxd √ 4
2−cos(x/7)
13 dxd e3xcos2(7x)
14 Find dy/dx in terms of x and y
sin(xy)+ y = y cos x
15 Find y (a) y= x
3 sec(3x)
16 Why have we been telling you that radians are more appropriate than degrees when using calculus? Suppose x is measured in degrees Then cos x◦= cosx ◦πradians
180 ◦
= cosπ x
180 where the argument is now in radians Find the derivative Is the derivative
− sin x◦?
Optimization
EXAMPLE 21.2 What angle of launch will propel an object (such as a cannonball or a baseball) farthest
horizontally?
This question is vitally important to engineers and sportsmen alike If we consider only the force of gravity (ignoring air resistance, the Coriolis effect, etc.), then it can be shown that the path the object will take is a parabola In Section 20.7 we showed that if an object is
Trang 6696 CHAPTER 21 Differentiation of Trigonometric Functions
launched at ground level at an angle θ and with an initial velocity of v0, then the horizontal distance it will travel is given by
R(θ )=2v
2
0cos θ sin θ
where g is the acceleration due to gravity We want to find θ such that R(θ ) is maximum
SOLUTION Using the trigonometric identity sin 2θ= 2 sin θ cos θ, R(θ) can be rewritten as
R(θ )=v
2
0sin(2θ )
2 0
g sin(2θ ), where g and v0are constants
One approach to this problem is to find the critical points of R(θ ) For our purposes
θmust be between 0 and π/2 Therefore, the critical points of R are the endpoints θ= 0 and θ= π/2, both resulting in R(θ) = 0, and the values of θ between 0 and π/2 such that
dR
dR
dθ =v
2 0
g(2 cos(2θ )) Setting dR
dθ equal to 0 gives
0= 2v
2 0
g cos(2θ ) 2v
2 0
g is a constant
0= cos(2θ) Let u= 2θ When 0 ≤ θ ≤π
2, 2· 0 ≤ 2θ ≤ 2 · π
2, so
0≤ u ≤ π
cos u= 0 when u = π/2 This is the only value of u ∈ [0, π] that satisfies the equation
2θ= π/2 (Substitute 2θ= u.)
θ= π/4
θ= π/4 is a candidate for the maximum
We can show it is actually the maximum by looking at R(θ ), ord2R
dθ 2
R(θ )=2v
2 0
g
d
dθ(cos 2θ )
=2v
2 0
g (−2 sin 2θ)
=−4v
2 0
g sin 2θ
R(π/4)=−4v
2 0
g sin(π/2)
=−4v
2 0
g
Trang 7R(π/4)=−4vg2 <0, so the graph of R(θ ) is concave down at θ= π/4 and R(θ) has a maximum at π4 Therefore, an angle of π4 radians, or 45◦, is the angle that will give the greatest horizontal distance
NOTE An alternative approach would be to do this optimization entirely without calculus R(θ )=v
2 sin(2θ )
g ; we need to find the angle θ∈0, π
2 to make sin(2θ) greatest Either this can be calculated directly, or we can again let u= 2θ
If θ∈0, π
2, then u = 2θ ∈ [0, π] On [0, π] sin u is maximum at u =π
2, so θ=u
4
We have shown that for any fixed initial speed the projectile will travel the farthest horizontal distance if it is launched at a 45◦angle
EXAMPLE 21.3 A lighthouse is located 3 kilometers away from a long, straight beach wall The beacon of
light is rotating steadily at a rate of 112revolutions per minute
(a) A lone soul is sitting on the beach wall 5 kilometers from the lighthouse, staring into the sea and contemplating the universe At what rate is the ray of light moving along the beach wall when it passes the thinker?
(b) At what point along the beach wall is the beam moving most slowly?
SOLUTION (a) Begin with a picture Do this on your own, thinking carefully about what is known and
what you are trying to find.10Then compare your work with what is given below
beach wall
lighthouse 3
x
z
θ
lone soul
lighthouse snapshot 3 4
5 θ
Figure 21.10
What We Want: dxdt, the rate at which the ray of light is moving along the beach wall when
z= 5
What We Know: The light makes 112revolutions per minute, so it goes through
3
2revolutionsminute · 2π radians
1 revolution= 3πradians
minute dθ
dt = 3π radians/minute
Strategy We knowdθdt and we want to find dxdt Our strategy is to write an equation relating
θand x.11We then differentiate with respect to time to get a relationship between dθdt and
10 A common error is to assume the beam of light has a fixed length.
11 xvaries It is not always 4 It is only after we have differentiated that we can substitute in values for quantities that vary
Trang 8698 CHAPTER 21 Differentiation of Trigonometric Functions
dx
dt We try to find a trigonometric function involving only sides we know or are concerned about
x
3 = tan θ d
dt
x 3
=dtd (tan θ ) Differentiate each side with respect to time
1 3
dx
dt = (sec2θ )dθ
dt xand θ are functions of t, so use the Chain Rule dx
dt = 3(sec2θ )dθ
dt Solve for dx
dt, the rate we want to find
Now we can substitute the values we know
We knowdθdt = 3π To find sec θ, we use the fact that z = 5 at the moment in question
sec θ=hyp
adj =5 3 Therefore, at the moment when the beam passes the thinker, z= 5 and
dx
dt = 3(sec2θ )dθ
dt
= 3 25 9
· 3π
= 25π miles per minute
lone soul
lighthouse
Figure 21.11
(b) At what point along the cliffs is the spot of light moving most slowly?
For our purposes, the angle θ must be between−π
2andπ
2or the light will not be shining
on the cliffs We know the rate at which the spot of light moves:
dx
dt = 3(sec2θ )dθ
dt = 3 · 3π(sec2θ )radians/minute
To find the place where the light moves most slowly, we want to find the angle θ that minimizesdxdt Since sec2θ= 1
cos 2 θ, we want to make cos2θas large as possible This will occur when cos θ is at its maximum, which occurs at θ= 0, or when the light is directly across from the lighthouse, when the beam is shortest
Notice that the further away an object on the beach wall is from the lighthouse, the faster the beam will sweep past it
Trang 9P R O B L E M S F O R S E C T I O N 2 1 3
1 Let f (x)= x + 2 sin x
(a) Find all of the critical points
(b) Where is f (x) increasing? Decreasing?
(c) Where does f (x) have local maxima? Local minima?
(d) Does f (x) have global maxima? Global minima? If so, what are the absolute maximum and minimum values?
(e) Where is f (x) concave up? Concave down?
(f ) Sketch a graph of f (x)
2 Consider the function f (x)= − cos x +12sin 2x
(a) Explain how you can tell that f is periodic with period 2π
(b) Find and classify all the critical points of f on the interval [0, 2π ] Do the trigonometric “algebra” on your own, then check your answers using a graphing
calculator (Hint: You’ll get a cos 2x that you’ll need to rewrite.)
For Problems 3 through 6, graph f on the interval [0, 2π ] labeling the x-coordinates
of all local extrema.
3 f (x)= cos x +√3 sin x
4 f (x)= cos x − sin x
5 f (x)= cos 2x − 2 cos x
6 f (x)= exsin x
7 Use a tangent line approximation to approximate the following In each case, use concavity to determine whether the approximation is larger or smaller than the actual value Then compare your results with the approximations given by a calculator or computer
(a) sin 0.2
(b) sin 0.1
(c) sin 0.01
(d) sin(−0.1)
8 Creme Fraiche and Caveat are battling their way to the finish line in the last leg of horseracing’s Triple Crown At the finish line, 30 feet away from the track itself, is a camera that is focused on the leading horse who is moving down the stretch at a rate of
46 feet per second At what rate is the camera rotating when the lead horse is 50 feet from the finish line?
9 Verify that sec x has local minima at x= 2πk and local maxima at x = π + 2πk (k an integer) by identifying its critical points and using the second derivative test for maxima and minima
10 Verify that tan x has points of inflection at x= πk, k an integer, by showing that the sign of its second derivative changes at these points
Trang 10700 CHAPTER 21 Differentiation of Trigonometric Functions
11 Let f (x)= 3 cos x + 2 sin x
(a) What is the period of f ? (b) What are the maximum and minimum values of f ?
12 If we ignore air resistance, a baseball thrown from shoulder level at an angle of θ radians with the ground and at an initial velocity of v0meters per second will be at shoulder level again when it is v
2 sin(2θ )
g meters away g is the acceleration due to gravity (9.8 m/sec2)
(a) Express the maximum distance the baseball can travel (from shoulder level to shoulder level) in terms of the initial velocity
(b) The fastest baseball pitchers can throw about 100 miles per hour How far would
such a ball travel if thrown at the optimal angle? (Note: 1 mile= 5280 feet and 1 meter≈ 3.28 feet.)(*)
13 A policewoman is standing 80 feet away from a long, straight fence when she notices someone running along it She points her flashlight at him and keeps it on him as he runs When the distance between her and the runner is 100 feet he is running at 9 feet per second At this moment, at what rate is she turning the flashlight to keep him illuminated? Include units in your answer
14 A sewage gutter is to be constructed from a piece of sheet metal 8 feet long and 4 feet wide by folding up a 1-foot strip on each side Denote by θ the angle between the sides and the vertical, as shown in the figure below What angle θ will result in a sewage gutter of maximum volume?
15 A lookout tower is located 0.5 kilometers from a line of warehouses A searchlight on the tower is rotating at a rate of 6 revolutions per minute How fast is the beam of light moving along the wall of warehouses when it passes by a window located 1 kilometer from the tower?
16 Graph f (x)= 2cos x (a) Is the function periodic? If so, what is its period?
(b) What is its maximum value? Its minimum value? Give exact answers
17 Let f (x)= − cos x and g(x) = sin x
(a) What is the maximum distance between these two curves on the interval [−π
4,3π4]? (b) What is the point of intersection of the tangent lines to these curves at the points from part (a) where the curves are farthest apart? Does this answer surprise you? Explain