2.3-4. Given the position s = f (t) of a body moving on a coordinate line for a t b,
with s in meters and t in seconds.
(a) Find the body's displacement and average velocity for the given time interval.
(b) Find the body's speed and acceleration at the endpoints of the interval.
(c) When during the interval does the body change direction (if ever)?
s = (t =4) ; t + t , 0 t 2
4
Sol :
3
2
(a) s = s(2) ; s(0) = 0 ; 0 = 0 m, vav = s=t = 0 m/sec
(b) v(t) = s0(t) = t ; 3t + 2t
) jv(0)j = 0 m/sec and jv(2)j = 0 m/sec,
a(t) = v0(t) = 3t ; 6t + 2
) a(0) = 2 m/sec and a(2) = 2 m/sec .
(c) Let v(t) = t ; 3t + 2t = 0
) t(t ; 1)(t ; 2) = 0
) t = 0; 1; 2
) v = t(t ; 1)(t ; 2) is positive for 0 < t < 1 and v is negative for 1 < t < 2
) the body changes direction at t = 1.
3
2
2
2
3
2
2
2.3-8. At time t 0, the velocity of a body moving along the s-axis is
v = t ; 4t + 3
2
(a) Find the body's acceleration each time the velocity is zero.
(b) When is the body moving forward? moving backward?
(c) When is the body's velocity increasing? decreasing?
Sol :
v(t) = t ; 4t + 3 ) a(t) = v0(t) = 2t ; 4
(a) Let v(t) = t ; 4t + 3 = 0 ) t = 1 or 3
) a(1) = ;2 m/sec and a(3) = 2 m/sec .
(b) Let v(t) = t ; 4t + 3 = (t ; 1)(t ; 3) > 0 ) the body is moving forward for
0 < t < 1 and t > 3;
v(t) = (t ; 1)(t ; 3) < 0 ) the body is moving backward for 1 < t < 3.
(c) Velocity increasing ) a(t) = 2t ; 4 > 0 ) t > 2;
velocity decreasing ) a(t) = 2t ; 4 < 0 ) 0 < t < 2.
2
2
2
2
2
2.3-19. When a model rocket is launched, the propellant burns for a few seconds, acceler-
ating the rocket upward. After burnout, the rocket coasts upward for a while and
then begins to fall. A small explosive charge pops out a parachute shortly after the
rocket starts down. The parachute slows the rocket to keep it from breaking when
it lands.
The gure here shows velocity data from the ight of the model rocket. Use the
data to answer the following.
1
(a)
(b)
(c)
(d)
(e)
(f)
(g)
How fast was the rocket climbing when the engine stopped?
For how many seconds did the engine burn?
When did the rocket reach its highest point? What was its velocity then?
When did the parachute pop out? How fast was the rocket falling then?
How long did the rocket fall before the parachute opened?
When was the rocket's acceleration greatest?
When was the acceleration constant? What was its value then (to the nearest
integer)?
(a) 190 ft/sec
(b) 2 sec
(c) at 8 sec, 0 ft/sec
(d) at 10+2(12-10)/5=10.8 sec, 90 ft/sec
(e) 10.8-8=2.8 sec
(f) at 2 sec
(g) from 2 to 10.8 sec, a = v : : ;;v ;32 ft/sec
2.3-24. The graphs in Fig. 2.31 show the position s, the velocity v = ds=dt, and the
acceleration a = d s=dt of a body moving along the coordinate line as functions of
time t. Which graph is which? Give reasons for your answers.
C = position, B = velocity, and A = acceleration.
Curve C cannot be the derivative of either A and B because C has only negative
values while A and B have some positive slopes. So C represents position. Curve C
has no positive slope, so its derivative, the velocity, must be B. That leaves A for
acceleration. Indeed, A is negative where B has negative slopes and positive where
B has positive slopes.
Sol :
(10 8)
10 8
2
(2)
2
2
2
Sol :
2.3-26. Marginal revenue.
Sol :
Suppose the revenue from selling x custommade oce desk is
r(x) = 2000(1 ; x +1 1 ) dollars:
(a) Find the marginal revenue when x desks are produced.
(b) Use the function r0(x) to estimate the increase in revenue that will result from
increasing production from 5 desks a week to 6 desks a week.
(c) Find the limit of r0(x) as x ! 1. How would you interpret this number?
(a)
r(x) = 2000(1 ; x +1 1 ) ) r0(x) = (x2000
+ 1)
2
r0(x) is the marginal revenue.
2
(b) r0(5) = 2000=(5 + 1) = 2000=36 55.56 dollars.
(c)
0 (x) = lim 2000 = 0:
lim
r
x!1
x!1 (x + 1)
The increase in revenue as the number of items increases without bound will
approach zero.
2
2
2.3-30. The volume V = (4=3)r of a spherical balloon changes with the radius.
3
(a) At what rate does the volume change with respect to the radius when r = 2
ft?
(b) By approximately how much does the volume increase when the radius changes
from 2 to 2.2 ft?
(a)
Sol :
dV j = 4(2) = 16 (ft =ft):
V = 34 r ) dV
=
4
r
)
dr
dr r
(b) By (a), dVdr = 16 when r = 2, so that when r changes by 1 unit, we expect V to
change approximately by 16. Therefore when r changes by 0.2 units V changes
approximately by (16)(0:2) = 3:2 10:05 ft . Note that V (2:2) ; V (2) 11:09 ft .
3
2
3
2
=2
3
3
2.4-30.
+ tan x ) =?
lim
sin
(
x!
tan x ; 2 sec x
0
Sol :
+ tan x ) = sin + tan 0 = sin ( ; ) = ;1
lim
sin
(
x!
tan x ; 2 sec x
tan 0 ; 2 sec 0
2
0
2.4-40.
lim
6x (cot x)(csc 2x) =?
!
2
x
0
Sol :
6x cos x = lim(3 cos x x 2x ) = 3
lim
6
x
(cot
x
)(csc
2
x
)
=
lim
x!
x! sin x sin 2x
x!
sin x sin 2x
2
2
0
2.4-46.
0
0
sin 5x =?
lim
x! sin 4x
0
3
Sol :
sin 5x = lim( sin 5x 4x 5 ) = 5 lim( sin 5x 4x ) = 5
lim
x! sin 4x
x! sin 4x 5x 4
4 x! 5x sin 4x 4
2.4-54. y = 2x + sin x
0
Sol :
0
0
y = 2x + sin x ) y0 = 2 + cos x
Howizontal tangent occurs where 2 + cos x = 0 ) cos x = ;2
But there are no x-values for which cos x = ;2
So answer is : No!
The graph of y = 2x + sin x:
y
15
10
y=2x
y=2x+sinx
5
x
0
O
−5
−10
−15
−8
−6
−4
−2
0
2
4
6
8
; 4) and (b) the horizontal
2.4-60. Find an equation for (a) the tangent to the curve at P ( p
4
tangent to the curve at Q. Where the graph is y = 1 + 2 csc x + cot x
Sol :
p
p
p2 cos
y = 1 + 2 csc x + cot x ) y0 = ; 2 csc x cot x ; csc x = ;( x )( xx )
(a) If x = , then y0 = ;4 ; the tangent line is y = ;4x + + 4
p
(b) Let y0 = 0 ) 2 cos x + 1 = 0; 0 < x < ) x = radians. When x = ,
y = 2.
Therefore y = 2 is the horizontal tangent.
p
The local graph of y = 1 + 2 csc x + cot x and two tangents y = ;4x + + 4,
y = 2:
1
sin
2
+1
sin
4
3
4
4
3
4
y
12
10
y = 1 + 1.414cscx + cotx
8
6
4
2
y=2
0
x
O
−2
−1
−0.5
y = −4x + 7.1416
0
0.5
2.4-64. Is there a value of b that will make
1
1.5
2
2.5
3
3.5
x+b
x<0
g(x) = cos
x
x0
continuous at x=0 ? dierentiable at x=0 ? Give reasons for your answers.
Sol :
(a) limx! ; g(x) = limx! ; (x + b) = b and limx! + g(x) = limx! + cos x = 1:
g is continuous at x = 0 , limx! ; g(x) = limx! + g(x) , b = 1
(b) the left-hand derivative is dxd (x + b) jx = 1 but the right-hand derivative is
d
(cos x) jx = ; sin 0 = 0 ,
dx
so g is not dierentiable at x = 0 for any value of b (including b = 1) .
The graph of g(x) for b = ;1; 0; 1; 2; 3:
0
0
0
0
0
=0
=0
5
0
y
3
y = x + b, x < 0
2
1
y = cosx, x > 0
0
x
b=3
−1
b=2
−2
b=1
−3
b=0
−4
b = −1
−5
−4
O
−3
−2
−1
0
1
p
2.5-36. r = sec tan ( )
p
r = (sec ) tan , then
p p
1
dr
2
3
1
1
Sol :
1 )+(sec p)(sec 1 )(; 1 )
p
=
tan
(
)(sec
tan
)(
d
2 p
p p
= ; 1 sec sec ( 1 ) + p1 tan 1 sec tan 2 p
p tan tan ( ) sec ( )
p
= (sec )[
; ]
2 p
p
2.5-48. y = 4 sin ( 1 + t)
p
p
Sol : y = 4 sin (
1 + t), then
2
2
2
2
1
1
2
2
dy = 4 cos (q1 + pt) d (q1 + pt)
dt
dt
q
p
p
= 4 cos ( 1 + t) p 1 p dtd (1 + t)
2 1+ t
p
p
= 2pcos ( p1 + pt)
1+ t2 t
p
p
cos
(
1
+
= p p t)
t+t t
6
4
2.5-62. Find dy=dt when x = 1 if y = x + 7x ; 5 and dx=dt = 1=3 .
2
Sol :
dy j = d x + 7 dx
dt x
dt
dt
2
=1
x=1
dx
= 2x dx
+
7
dt dt
x=1
= 2 13 +7 13 = 3
2.5-64. Find dy=dx if y = xp= by using the Chain Rule with y as a composite of
(a) y = up and u = x
(b) y = u and u = x .
3 2
3
3
Sol :
dy
y = x 32 ) dx
= x 12
3
2
p
dy
(a) y = u ) du
= 3u ; u = x ) du
= px i:e:
dx
p
p
3( x) px = x as expected.
p
dy
(b) y = u ) du
= pu ; u = x ) du
= 3x
dx
1
p 3 3x = x 2 again as expected .
x
3
1
2
3
2
1
2
2
2
1
3
2
1
2
2
2
dy
dx
dy du
= du
dx = 3u px =
i:e:
1
2
2
dy
dx
= pu 3x =
1
2
2
3
2
2.5-72. A particle moves along the x-axis with velocity dx=dt = f (x). Show that the
particle's acceleration is f (x)f 0(x).
Sol :
Let
dx
dt
= f (x) , then
dv dx = dv f (x) = d ( dx ) f (x) = d (f (x)) f (x) = f 0 (x)f (x)
=
a = dv
dt dx dt dx
dx dt
dx
as required.
2.5-73. Temperature and the period of a pendulum.
For oscillations of small amplitude (short swings), we may safely model the relationship between the period T and the length L of a simple pendulum with the
equation
s
T = 2 Lg ;
where g is the constant acceleration of gravity at the pendulum's location. If we
measure g in centimeters per second squared, we measure L in centimeters and T in
seconds. If the pendulum is made of metal, its length will vary with temperature,
either incresing or decreasing at a rate that is roughly proportional to L. In symbols,
with u being temperature and k the proportionslity constant
dL = kL:
du
Assuming this to be the case, show that the rate at which the period changes with
respect to temperature is kT=2.
7
s
Sol :
1 1
q = q = p
T = 2 Lg ) dT
=
2
dL
gL
2 Lg g g Lg
Then by Chian Rule,
s
dT = dT dL = p kL = k 2 L = kT
du dL du
2
g 2
gL
as required.
2.5-76. Suppose u = g(x) is dierentiable at x = ;5 , y = f (u) is dierentiable at u = g(;5)
, and (f g)0(;5) is negative. What , if anything, can be said about the values of
g0(;5) and f 0(g(;5)) ?
(f g)0(;5) < 0 ) f 0(g(;5)) g0(;5) < 0 ) f 0(g(;5)) and g0(;5) are both nonzero
Sol :
and have opposite signs .
i:e: either [f 0(g(;5)) > 0 and g0(;5) < 0] or [f 0(g(;5)) < 0 and g0(;5) > 0] .
2.6-16. Find the rst derivative of the function: g(x) = 2(2x; = + 1); = .
1 2
1 3
g(x) = 2(2x; = + 1); = , then
d (2x; = + 1) = ; 2 (2x; 21 +1) ; 43 [(2 ;1 )x ; 12 ; ]
g0(x) = ; 31 2(2x; 21 +1) ; 13 ; dx
3
2
= 2 (2x; 12 + 1); 43 x; 32
3
2.6-57. Find the two points where the curve x + xy + y = 7 crosses the x-axis, and show
that the tangents to the curve at these points are parallel. What is the common
slope of these tangents?
Solve
x + xy + y = 7
y=0
p
p
p
) x = 7 ) x = 7 ) (; 7; 0) and ( 7; 0) are the points where the curve
crosses the x-axis.
dy + 2y dy = 0 ) (x + 2y) dy = ;2x ; y
x + xy + y = 7 ) 2x + y + x dx
dx
dx
dy = ; 2x + y ) m = ; 2x + y
) dx
x + 2y
x + 2y
p
p
) the slope at (; 7; 0) is m = ;2 and the slope at ( 7; 0) is also m = ;2.
Since the slopes are the same, the corresponding tangents must be parallel.
1 2
Sol :
1 3
(
1 2
1)
(
2
Sol :
2
2
2
2
2
2
8
)
(
1)
The graph of x + xy + y = 7 and the tangents:
2
2
y
4
3
The tangent line passing through x0
2
2
x +xy+y =7
2
1
x1
x
0
x2
O
−1
−2
The tangent line passing through x2
−3
−4
−4
−3
−2
−1
0
1
2
3
4
2.6-66. Find the normals to the curve xy +2x ; y = 0 that are parallel to the line 2x + y = 0.
dy + 2 ; dy = 0 ) dy = y + 2
xy + 2x ; y = 0 ) y + x dx
dx
dx 1 ; x
Sol :
dy
= ;2. In order to be parallel, the normal
The slope of the line 2x + y = 0 is dx
lines must also have slope of ;2. Since a normal is perpendicular to a tangent, the
slope of the tangent is 1/2. Then
y + 2 = 1 ) 2y + 4 = 1 ; x ) x = ;3 ; 2y:
1;x 2
Substituding in the original equation:
y(;3 ; 2y) + 2(;3 ; 2y) ; y = 0
) y + 4y + 3 = 0
) y = ;1 or y = ;3.
If y = ;1; then x = ;1: m = ;2 ) y + 1 = ;2(x + 1) ) y = ;2x ; 3.
If y = ;3; then x = 3: m = ;2 ) y + 3 = ;2(x ; 3) ) y = ;2x + 3.
2
The graph of xy + 2y ; x = 0:
9
y
250
x=1
200
150
100
50
xy+2x−y=0
0
x
O
xy+2x−y=0
−50
−100
−150
−200
−250
−1
−0.5
0
0.5
1
1.5
2
2.5
3
2.6-67. Show that if it is possible to draw these three normals from the point (a; 0) to the
parabola x = y shown here, then a must be greater than 1/2. One of the normals
is the x-axis. For what value of a are the other two normals perpendicular?
2
dy = 1 ) dy = 1 :
y = x ) 2y dx
dx 2y
Sol :
2
So the slope of the normal is ;2y. If a normal is drawn from (a; 0) to (x ; y ) on
the curve, its slope satises
y ; 0 = ;2y ) y = ;2y (x ; a) ) a = x + 1 :
x ;a
2
Since x 0 on the curve, we must have that a 1=2.p
By symmetry, the two points on the
parabola
are (x ; x ) and (x ; ;px ), so the
p
p
two slopes of the normals are ;2 x andp2 x . p
For the normals to be perpendicular, (;2 x )(2 x ) = ;1 ) x = 1=4
) ( 14 ; 21 ); a = x + 21 = 34 :
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
0
2.6-70. Find dy=dx (treating y as a function of x) and dx=dy (treating x as a function of y)
for x + y = sin y. How do dy=dx and dx=dy seem to be related? Can you explain
the relationship geometrically in terms of the graphs?
3
2
2
10
Sol :
dy = (2 sin y)(cos y) dy ) dy =
3x
x +y = sin y ) 3x +2y dx
dx dx 2 sin y cos y ; 2y ;
3
2
2
2
2
dx = 2 sin y cos y ; 2y ;
x +y = sin y ) 3x dx
+2
y
=
(2
sin
y
)(cos
y
)
)
dy
dy
3x
3
2
2
2
2
thus
dx = 1 :
dy
dy dx
The two dierent treatments view the graphs as functions symmetric across the line
y = x so their slopes are reciprocals of each other at the corresponding points.
The graph of x + y = sin y:
3
2
2
y
4
3
3
2
2
x +y =sin y
2
1
0
x
O
−1
−2
−3
−4
−10
−8
−6
−4
−2
0
2
2.7-9. The area A of a triangle with sides of lengths a and b enclosing an angle of measure
is
A = 12 ab sin :
(a) How is dA=dt related to d=dt if a and b are constant?
(b) How is dA=dt related to d=dt and da=dt if only b is constant?
(c) How is dA=dt related to d=dt, da=dt, and db=dt if none of a; b, and are
constant?
11
Sol :
(a)
1 ab cos d
A = 12 ab sin ) dA
=
dt 2
dt
(b)
= 1 ab cos d + 1 b sin da
A = 12 ab sin ) dA
dt 2
dt 2
dt
(c)
1 ab cos d + 1 b sin da + 1 a sin db
A = 12 ab sin ) dA
=
dt 2
dt 2
dt 2
dt
2.7-11. Changing dimensions in a rectangle.
Sol :
The length l of a rectangle is decreasing at the rate of 2 cm/sec while the width w
is increasing at the rate of 2 cm/sec. When l=12 cm and w=5 cm, nd the rates
of change of (a) the area, (b) the perimeter, and (c) the lengths of the diagonals of
the rectangle. Which of these quantities are decreasing, and which are increasing?
Given dtdl = ;2 cm/sec, dw
= 2 cm/sec, l=12 cm and w=5 cm.
dt
(a)
A = lw ) dA
= l dw +w dl ) dA = 12(2)+5(;2) = 14cm =sec;
dt
dt dt dt
increasing.
(b)
P = 2l+2w ) dP
= 2 dl +2 dw ) 2(;2)+2(2) = 0cm=sec;
dt
dt dt
constant.
(c)
p
= 1 (w +l ); = (2w dw +2l dl )
D = w + l = (w +l ) = ) dD
dt 2
dt dt
+ l dtdl (5)(2) + (12)(;2)
w dw
dD
dt
) dt = p
= p
= ; 14
13 cm=sec;
25 + 144
w +l
decreasing.
2
2
2
2
2
2
2 1 2
2
1 2
2
2.7-20. A growing raindrop. Suppose that a drop of mist is a perfect sphere and that,
through condensation, the drop picks up moisture at a rate proportional to its
surface area. Show that under these circumstances the drop's radius increases at a
constant rate.
12
Sol :
V = 34 r ; S = 4r ; dV
dt = kS = 4kr
dr = 4kr ) dr = k
=
4
r
) dV
dt
dt
dt
k is a constant; therefore, the radius is increasing at a constant rate.
3
2
2
2
2
2.7-30. A moving shadow.
Sol :
A man 6 ft tall walks at the rate of 5 ft/sec toward a streetlight that is 16 ft above
the ground. At what rate is the tip of his shadow moving? At what rate is the
length of his shadow changing when he is 10 ft from the base of the light?
s= the length of the shadow, x= the distance of the man from the streetlight.
) s = x and dxdt = ;5.
(a) If I = the distance of the tip of the shadow from the streetlight, then I = s + x
ds + dx ) j dI j = j ds + dx j = j 3 dx + dx j = 8 j;5j
) dI
=
dt dt dt
dt
dt dt
5 dt dt 5
= 8 ft/sec, the speed the tip of the shadow is moving along the ground.
(b)
ds = 3 dx = 3 (;5) = ;3 ft=sec;
dt 5 dt 5
so the length of the shadow is decreasing at a rate of 3 ft/sec.
3
5
2.7-33. A melting ice layer.
A spherical iron ball 8 in. in diameter is coated with a layer of ice of uniform
thickness. If the ice melts at the rate of 10 in /min, how fast is the thickness of
the ice decreasing when it is 2 in. thick? How fast is the outer surface area of ice
decreasing?
The radius of the spherical iron ball is 8/2=4.
dr
=
4
r
V = 34 r ; 43 (4) ) dV
dt
dt
When the thickness of the ice decreases to 2 in. thick, the total radius of iron ball
and ice is 6, then
dr j = 10 j = 5 in=min
dt r
4r r
72
dr
S = 4r ) dS
=
8
r
dt
dt
dr j = 5 ) dS j = 48( 5 ) = 10 in =min
dt r
72 dt r
72
3
3
Sol :
3
=6
3
2
2
=6
2
=6
2
=6
13