Grafico 3. Uso de dispositivos de ayuda en carga y descarga de animales Uruguay 2002/
7.1 Caracterización del transporte de los bovinos
Fig. 10-1 10.2 Give the equations relating degree measure and radian measure of angles.
I 2-rr radians is the same as 360 degrees. Hence, 1 radian = 180/Tr degrees, and 1 degree = 77/180 radians. So, if an angle has a measure of D degrees and/? radians, then D = (180/7r).R and R = (77/180)D.
10.3 Give the radian measure of angles of 30°, 45°, 60°, 90°, 120°, 135°, 180°, 270°, and 360°.
I We use the formula R = (?r/180)D. Hence 30° = 77/6 radians, 45° = 77/4 radians, 60° = 77/3 radians, 90° = 77/2 radians, 120° = 27T/3 radians, 135° = 377/4 radians, 180° = 77 radians, 270° = 377/2 radians, 360° = 277 radians.
10.8
TRIGONOMETRIC FUNCTIONS AND THEIR DERIVATIVES
Fig. 10-3
Refer to Fig. 10-3. Place an arrow OA of unit length so that its initial point O is the origin of a coordinate system and its endpoint A is (1,0). Rotate OA about the point O through an angle with radian measure 0. Let OB be the final position of the arrow after the rotation. Then cos 6 is defined to be the ^-coordinate of B, and sin 0 is defined to be the ^-coordinate of B.
10.9 State the values of cos 0 and sin 0 for 0 = 0, 77/6, ir/4, ir/3, ir!2, -IT, 3ir/2, 2ir, 9ir/4.
10.10
sin 6 and cos (6 V2/2.
Evaluate: (a)cos(-ir/6) (b) sin (-7T/6) (c) cos(27r/3) (d) sin (2ir/3)
(a) In general, cos (-0) = cos ft Hence, cos (-ir/6) = cos (77/6) = V5/2. (*) In general, sin(-0)= -sin ft Hence, sin(-ir/6) = -sin (ir/6) = -|. (c) 2ir/3 = ir/2 + ir/6. We use the identity cos (0 + ir/2) = -sin ft Thus, cos(2ir/3)= -sin (-rr/6) = -\. (d) We use the identity sin (0 + ir/2) = cos ft Thus, sin(27r/3) = cos(7r/6) = V3/2.
10.11 Sketch the graph of the cosine and sine functions.
We use the values calculated in Problem 10.9 to draw Fig. 10-4.
10.12 Sketch the graph of y = cos 3*.
Because cos 3(* + 2tr/3) = cos (3>x + 2ir) = cos 3x, the function is of period p = 2ir/3. Hence, the length of each wave is 277/3. The number/of waves over an interval of length 2ir is 3. (In general, this number /, called the frequency of the function, is given by the equation /= 2ir/p.) Thus, the graph is as indicated in
Fig. 10-5.
10.13 Sketch the graph of y = 1.5 sin 4*.
The period p = ir/2. (In general, p = 2ir/b, where b is the coefficient of x.) The coefficient 1.5 is the amplitude, the greatest height above the x-axis reached by points of the graph. Thus, the graph looks like Fig.
10-6.
Give the definition of sin 0 and cos ft Fig. 10-2
63
Notice that 9ir/4 = 27r+ ir/4, and the sine and cosine functions have a period of 2ir, that is, sin(fl + 2ir) = and cos (97T/4) = cos (Tr/4) = 1- 277-) = cos «. Hence, sin(97r/4) = sin(7r/4) = V2/2
e
0
7T-/6 7T/4 IT/3
it 12
IT
37T/2 2lT
sin 6 0 1/2 V2/2 V3/2 1 0 -1 0
cos 0 1 V5/2 V2/2 1/2 0 -1 0
1
CHAPTER 10
Fig. 10-4
Fig. 10-5
Fig. 10-6 64
TRIGONOMETRIC FUNCTIONS AND THEIR DERIVATIVES 65
10.14 Calculate
10.15 Calculate
10.16 Calculate
10.17 Using the A-definition, calculate
Thus,
I By the identity sin (u + v) = sin ucos v + cos wsin v, sin (x + Ax) = sin x cos (A*) + cos x sin (Ax). Hence, sin (x + Ax) - sin ;c = sin x[cos (Ax) — 1] + cos x sin (Ax), and
Here, we have used (Problem 10.16).
10.18 Calculate (cos x) from the known derivative of sin x
10.19 Calculate
sin 3x is a composite function of 3x and the sine function. By the chain rule and the fact that
10.20 Calculate
Hence, by the chain rule,
10.21 Find
10.22 Find an equation of the tangent line to the graph of y = sin2 x at the point where x = ir/3.
The slope of the tangent line is the derivative y'. By the chain rule, since sin2 x = (sinx)2, y'=2(sinx)-and
When * = 7r/3, sinx = V5/2
At the point where x = if 13, y = (V5/2)2 = i. So a point-slope equation of the tangent line is y — \ = and
[chain By the identity cos x = sin
rule] = sin x • (-1)
cos2 x = (cos x)2. -2 sin x cos x = -sin 2x.
By the chain rule,
(sin x) = 2 sin x cos x. c o s x = i . So / = 2 - V 3 / 2 - i = V § / 2 . (V3/2)(Ar--n-/3).
66 CHAPTER 10
10.23 Find an equation of the normal line to the curve y = 1 + cos x at the point
The slope of the tangent line is the derivative y'. But, y' = -sin x = -sin (ir/3) = -V3/2. Hence, the slope of the normal line is the negative reciprocal 2A/3. So a point-slope equation of the normal line is y - \ = (2/V3)(x - IT/3) = (2V3/3)jc - 27rV3/9.
10.24 Derive the formula
Remember that tan x = sin AT/COS x and sec x = 1 /cos x. By the quotient rule,
10.25 Find an equation of the tangent line to the curve y = tan2 x at the point (7r/3,3).
Note that tan (ir/3) = sin(7j-/3)/cos (w/3) = (V3/2)/i = V3, and sec(ir/3) = l/cos(ir/3) = 1/| =2. By the chain rule, / = 2(tanx)- -7- (tan*) = 2(tan*)(sec2 *). Thus, when x = ir/3, y' =2V5-4 = 8V5, so the slope of the tangent line is 8V3. Hence, a point-slope equation of the tangent line is y — 3 = 8V3(x - ir/3).
10.26 Derive the formula
By the identity cot x = tan (IT12 - x) and the chain rule,
10.27 Show that
By the chain rule,
10.28 Find an equation of the normal line to the curve y = 3 sec2 x at the point (ir/6,4)
10.29 Find Dx
By the chain rule, y' = 3[2 sec x • -r- (sec x)] = 3(2 sec x • sec x • tan x) = 6 sec2 x tan x. So the slope of the tangent line is y' = 6(f)(V3/3) = 8V3/3. Hence, the slope of the normal line is the negative reciprocal -V3/8. Thus, a point-slope equation of the normal line is y - 4 = -(V3/8)(x - 77/6).
Recall that D,(csc;c) = -esc x cot x. Hence, by the chain rule,
10.30 Evaluate
10.31 Evaluate
Hence,
TRIGONOMETRIC FUNCTIONS AND THEIR DERIVATIVES 67
Hence,
10.32 Show that the curve y = xsin* is tangent to the line y = x whenever x = (4n + l)(ir/2), where n is any integer.
When x = (4n +l)(ir/2)= TT/2 + 2irn, sin * = sin ir/2 = 1, cos x = cos ir/2 = 0, and xsinx = x.
Thus, at such points, the curve y = *sin* intersects y = x. For y = *sinjc, y' = x • Dx(sm x) + sin x-Df(x) = x cos A; + sin x. Thus, at the given points, y' = x • 0 + 1 = 1. Hence, the slope of the tangent line to the curve y = jcsinx at those points is 1. But the slope of the line y = x is also 1, and, therefore, y = * is the taneent line.
10.33 At what values of x does the graph of y = sec x have a horizontal tangent?
I A line is horizontal when and only when its slope is 0. The slope of the tangent line is y' = D^sec x) = sec ;t tan*. Hence, we must solve sec*tan* = 0. Since sec x = 1 /cos x, sec* is never 0. Hence, tan* = 0. But, since tan x = sin* /cos-*, tan* = 0 is equivalent to sin* = 0. The latter occurs when and only when x = nir for some integer n.
10.34 For what values of x are the tangent lines to the graphs of y = sin x and y = cos x perpendicular?
I The tangent line to the graph of y = sin x has slope D^(sin x) = cos x, and the tangent line to the graph of >> = cosx has slope D^(cos x) = -sin x. Hence, the condition for perpendicularity is that cos x • (-sin x) = -1, which is equivalent to cos x sin x = 1. Since 2 cos x sin x — sin 2x, this is equivalent to sin 2x = 2, which is impossible, because |sin jr| :£ 1 for all x. Hence, there are no values of x which satisfy the property.
10.35 Find the angle at which the curve y = 3 sin 3x crosses the x-axis.
I The curve crosses the x-axis when y = | sin 3x = 0, which is equivalent to sin 3x = 0, and thence to 3x = ntr, where n is an arbitrary integer. Thus, x = mr/3. The slope of the tangent line is y' = 3 cos 3x • 3 = cos 3x = cos (mr) = ±1. The lines with slope ±1 make an angle of ±45° with the x-axis.
In Problems 10.36 to 10.43, calculate the derivative of the given function.
10.36 x sin x
Dx(x sin x) = x • D^(sin x) + Dx(x) • sin x = x cos x + sin x.
10.37 x2 cos 2x.
Dx(x2 cos 2x) = x2 • Dx(cos 2x) + 2x • cos 2x = ;c2(-sin 2x) • Dx(2x) + 2x cos 2x = -2x2 sin 2x + 2x cos 2x.
10.38
10.39
10.40 2 tan (x/2)-5.
10.41 tan x - sec x.
Dx(tan x - sec x) = sec x - sec x tan x = (sec x)(sec x - tan x).
Dx(2 tan (x/2) - 5) = 2 sec2 (x/2) • D,(x/2) = sec2 (x/2).
D,[sin3(5x + 4)] = 3 sin2 (5x + 4) • Dx(5x + 4) = 3 sin2 (5x + 4) • (5) = 15 sin2 (5x + 4).
sin3 (5x + 4).
10.43 esc (3* - 5).
10.44
10.45 For what value of A does 3 sin Ax have a period of 2?
10.46 Find the angle of intersection of the lines 3!,: y = x - 3 and 3!2: y = -5x + 4.
10.47 Find the angle of intersection of the tangent lines to the curves xy = 1 and y = x3 at the common point (1,1).
68 CHAPTER 10 10.42 cot2 x.
Dx[csc (3x - 5)] = [-esc (3x - 5) cot (3* - 5)] • D,(3x - 5) = -esc (3x - 5) cot (3* - 5) • (3)
= -3 csc (3*-5) cot (3*-5).
Evaluate
Remember that and use the definition of the derivative.
The angle 0, that .$?, makes with the Jt-axis has a tangent that is equal to the slope of the line. The angle 02 that .S?, makes with the *-axis has a tangent equal to the slope of &2. Thus tan 0, = 1 and tan 02 = -5. The angle 0 between ^ and <£2 is 02 - 0}. So, tan 6 = tan (02 - 0^ =
Reference to a table of tangents reveals that 0 = 56°.
Fig. 10-7
Let 0l be the angle between the horizontal and the tangent line to y = x3, and let 02 be the angle between the horizontal and the tangent line to xy = 1. Now, tan 0, is the slope of the tangent line to y = jc3, which is the derivative of x3 evaluated at (1,1), that is, 3x2 evaluated at x = 1 or 3. So, tan 0, = 3. Likewise, since the derivative of 1/JC is —(1/Jt2), which, when evaluated at x = 1, is —1, we have tan 02 = —1. Hence,
A table of tangents yields 02 — 0, » 63°.
10.48 Evaluate
Thus, the desired limit is f . D^cot2 x) = 2 cot x • D, (cot x) = 2 cot x (-esc2 *) = -2 cot x esc2 *.
= [D,(cos x)](ir/3) = -sin (w/3) = -V5/2.
The period p = 2ir/A. Thus, 2 = 2ir/X, 2>l = 27r, yl = IT.
But, and