EXAMPLE CHEMICAL INVENTORY
Scott Specialty Gases
Scott Specialty Gases
1. Leta be the xcoordinate of Q. Since the derivative ofy : l x2 is y : 2x, the slope at Q is 2a. But since
the triangle is equilateral, AO/OC = /l, so the slope at Q is _,\/ Therefore, we must have that 2a = x/
z; a = 9. Thus, the point Q
El CHAPTER3 DERIVATlVES
3. HP (x) :2 A + B (x a) + C (x (1)2, then P/ (x) = B + 2C (x a) and P (x) : 2C. Applying the conditions
(i), (ii), and (iii). we get
P (a) = f (a): A = f (a)
P' (a) = f/ (a): B = f (a)
P (a) = f (a): 2C = f (a) => C = %.f (a)
D CHAPTERS DERIVATIVES
(a) f(x) 24x tanx => f(x)=4sec2x :> f (x) = Zsecx (secxtanx): -Zsec2xtanx.
(b) We can see that our answers are reasonable, since the graph of f / is 0
where f has a horizontal tangent, and the graph of f is positive
156 CHAPTER 3 DERIVATIVES
We are given that d: = 4 ft/s and d? = 5 ft/S. z2 = (x +y)2 + 5002 :>
dz dx dy
2Zalt (x+y)(dt+ dt
x = (4 ft/s) (20 min) (60 s/min) : 4800 ft andy = 5-15-60 = 4500 =>
= 4/(4800 + 4500)2 + 5002, so
dz x + y dx dy) 4800 + 450
160 E CHAPTER3 DERIVATIVES
33. We are given that d:_ _ 300 km/h. By the Law of Cosines,
y2 =xz+12(1)xcoleO=x2+12x(%) =x2+x+ 1,30
dy dx dx dy_ 2x +1 dx
1 2=2 Aft 1 =1=
ydt xdt + dt :> dt 2y dt er minute, x 60 5
:> y=V52+5+ =/31 :>
dy 2(5)+1 1650
174 U CHAPTER3 DERiVATIVES
22. wk 4w] + x Fihl- : x m K/x l + .
, V J) y y y 2m
23. y z (x Vanni/"5 : y = % (x tanx)_4/5 (tanx +x seczx)
24. y : sin (cosx) :9 y : cos (005x) ( sinx) : sinx cos (cosx)
25-x3=y(y+1)=y2+y => 2X
SECTION 3.7 IMPLICIT DIFFERENTIATION 141
3x2 6x + 2
33.a b :_ => =lat
() ()y 2(2y3_3y2_y+1) y
(0, l) and y = at (o, 2).
Equations of the tangent lines are y = x + l
and y = %x + 2.
(c)y'=0 => 3x26x+2=0 =>
x : l :I: %\/
There are eight points with
53 CHAPTER3 DERIVATIVES
Differentiating the rst given equation implicitly with respect to x and using the Chain Rule, we obtain
ftfgtxn = x :> f/ (g (x)g (x) = 1 => g (x) = . Using the second given equation to expand
f (g (26)
l . .
LABORATORY PROJECT TAYLOR POLYNOMIALS C] 169
6. T (x) : f (a) + f (a) (x a) + fga) (x - )2 + . . . + fUZ'W) (x a)". To compute the coefcients in this
equation we need to calculate the derivatives of f at 0: '
f(x) = cosx f() : cosO :1
166 CHAPTER 3 DERIVATIVES
dF 4kR3 dR dR
44. F = kR4 : dF = 4kR3 dR. 7? = 7131 : 4 (R) z) the relative change in F is about 4 times the
relative change in R, So a 5% increase in the radius corresponds to a 20% increase in blood ow.
45. (a) dazdCdX=0dx=0
61ABWI48-104 3 JUNE 2008
3.5.1. Upon receipt of hazardous materials, receiving personnel will inspect the material and
review the MSDS. Personnel will examine containers to ensure materials are labeled or marked
properly, displaying the identity of the
61ABWI48-104 3 JUNE 2008
states the personal protective measures fire fighters should use. It is important that this section be
reviewed prior to using the chemical.
A126.96.36.199. Section V provides reactivity data. This section simply describes what can b
302AWI36-2005 5 JANUARY 2007
4.2.3. Board members will assess each candidate using the whole person concept. Factors such
as duty performance, job responsibility, experience, supervisory and leadership ability, professional competence, education, AFOQT
LABORATORY PROJECT TAYLOR POLYNOMIALS 161
47. (a) Thegraph showsthatf(l) :2, so L(x) =f(1)+f(1)(x l) = 5+2(x ~ 1) :2x +3.
f(0.9) % L (0.9) = 4.8 and f(1.l) a: L (1.1) : 5.2.
(b) From the graph, we see that f (x) is positive and decreasing. This mean
172 CHAPTER 3 DERlVATIVES
7. The graph of a has tangent lines with positive slope for x < 0 and negative slope for x > 0, and the values of c t
this pattern, so c must be the graph of the derivative of the function for a. The graph of c has horizon
SECTION 3.10 LINEAR APPROXIMATIONS AND DIFFERENTIALS D 161
36. The hour hand of a clock goes around once every 12 hours or, in radians per
hour, 2712 = 6- rad /h. The minute hand goes around once an hour, or at the rate
of 27x rad/h. So the angle 0 betwee
182 El PROBLEMSPLUS
5. We use mathematical induction. Let S, be the statement that
d n (sin4 x + cos4x) : 4" cos (4x + mr/2). S] is
d . . . . .
d (s1n4x + e034 x) = 4sm3x cosx 4cos x smx = 4smx cosx (smzx coszx)
= 4sinx cosx cos 2x
SECTION 3.9 RELATED RATES 157
16. Let D denote the distance from the origin (0, 0) to the point on the curve y = J)?
D = \/(x-O)2+(yO)2 =cfw_:+()2 = x2+x 2)
6119 1 2 _1/2 2x+1 dx W' dx
_=_ 2 1 Wthd :3 h :4.
dt 2(x +X) (x+) t 22mm wenx
SECT|0N3.8 HIGHER DERIVATIVES D 149
38.Dsinx:cosx 2 Dzsinxzsinx :> D3sinxzcosx 2) D4sinx:sinx
The derivatives ofsinx occur in a cycle offour. Since 99 : 4 (24) + 3, we have I)99 sillx = D3 sin x = cosx.
39. In general, Df (2x) = 2 f (2x), D2 f (2x) = 4f (
SECTION 3.8 HIGHER DERIVATIVES 147
(x2+l)~x(2x) _ lx2
22-<a)f<x>=x2+1 3 "2 (x2+1)2 (x2+1)2 2
m) 2 W z M211) = 2x. L ~3)
(x2 +1)4 (x2 +1)3 (962 + 03
We can see that our answers are plausible, since f has horizontal
tangents where f (x) = 0, and
144 U CHAPTER3 DERIVATIVES
48_y:ax3 : y=3ax2andx2+3y2=b => 2x+6yy=0 =5
/ i x 1 so the curves are ortho onal
y 3y _ 3ax3 3ax2 g '
49. y : 0 => x2 x (0) + 02 = 3 4:) x = i. So the graph ofthe ellipse crosses the x-axis at the points
(i, 0). Using implic
CHAPTEB3 REVIEW [3 171
IIl-I-I-Il-Illl-I-I-I-Imwmmmm1 TRUE_EALSE(1U|Z NWm$m_-.-_-_
1. False. See the warning after Theorem 3.2.4.
2. True. This is the Sum Rule.
3. False. See the warning before the Product Rule.
4. True. This is the Chain Rule.
5. True by
SECTION 3.8 HlGHER DEHlVATlVES D 151
d d d d d . . . .
a (t) = dlt) = a? 21,:- = v (2) E3 Then dlt) is the rate of change ofthe velocny With respect to tlme (1n other
words, the acceleration) whereas do/ds is the rate of ch
SECTION 3.8 HIGHER DERIVATIVES 145
x2 + 4y2 = 5 => 2x + 4 mm = 0 z; y = 41. Now let 11 be the height ofthe lamp, and let (a, b) be
the point of tangency of the line passing through the points (3, h) and (5, 0). This line has slope
(h - 0) /