Solutions to Assignment #10 –MATH 1401
Kawai/Engau
(Page 1) Write carefully:
(i) Sect. 4.7. As x ! 1; which function grows faster: 2x or xx ? Show your reasoning?
Plan: Compare 2x to xx as x ! 1:
2x
lim x = lim
x!1 x
x!1
2
x
x
=???
So the part inside the parentheses goes to zero. Thus, we have 01 : This is NOT an
indeterminate form.
01 ! 0
So xx grows much faster than 2x :
(ii) Sect. 4.7, Exercise #18 (p. 290). This is 0=0:
4
4 sec2 (4z)
4 (1)
tan (4z)
= lim
=
=
2
z!0 7 sec (7z)
z!0 tan (7z)
7 (1)
7
lim
(iii) Sect. 4.7, Exercise #26 (p. 290). This is 0=0; and we don’t want to to use the di¤erence
of two cubes formula (and thus, not use L’Hôpital).
lim
x!2
p
3
3x + 2
x 2
2
=
=
1
3
lim
(3x + 2)
1
=
1
2
81=3
(3)
= lim
1
x!2
82=3
2=3
x!2
=
1
(3x + 2)2=3
1
4
We note that the original limit was the limit de…nition for f 0 (2) where f (x) = (3x + 2)1=3 :
(Page 2) Write carefully:
(i) Sect. 4.7, Exercise #34 (p. 290).
Plan: We’re probably better o¤ without the square root in the denominator.
r
p
p
p
1 x
sin (x) 1 x
sin (x) 1 x x
p
p p
= lim
= lim
lim sin (x)
x
x
x x
x!0+
x!0+
x!0+
p
p
p
sin (x) 1 x x
sin (x)
= lim
= lim
lim
1
x
x
x!0+
x!0+
x!0+
p
x x
By L’Hôpital, the …rst limit is zero (we also showed this with geometry and the Squeeze
Theorem).
p
p
(1)
1 0 0 = 0
1
(ii) Sect. 4.7, Exercise #44 (p. 290).
everything and then do L’Hôpital.
L =
ln (L) =
lim
x!1
1+
Suppose the limit is L: We can take the ln of
ln(x)
1
x
ln(x)
1
1+
x
lim ln
x!1
!
= lim ln (x) ln 1 +
x!1
This is 1 0: We will need to rewrite this as a fraction.
0
1
1
B
x2 C
B
C
@
1 A
1
1+
ln 1 +
x
x
= lim
ln (L) = lim
x!1
x!1
1
1
1
1
2
ln (x)
x
ln (x)
=
lim
x!1
1
x (x + 1)
1
x ln2 (x)
x!1
We still need to use L’Hôpital again.
0
B 2 ln (x)
ln (L) = lim B
x!1 @
1
1
1
2
x
C
B
x2
C
B
A
@
1
1+
x2
x
= lim
x!1
1
x ln2 (x)
0
x ln2 (x)
1
1
x (x + 1)
= lim
1
1
x C
C = 2 lim
A
x!1
One more time!
= lim
x!1
ln (x)
x
1
1
B x C
C = 2 (0) = 0
ln (L) = 2 lim B
x!1 @ 1 A
0
ln (L) = 0 ) L = e0 = 1
That was very long.
The faster way is this: Rewrite ln (x) as:
ln (x)
x
ln (x) = x
:
Now we have:
lim
x!1
1
1+
x
x (ln(x)=x)
= lim
x!1
1
1+
x
x
(ln(x)=x)
We can evaluate the limits in the base and in the exponent separately.
lim
x!1
1+
1
x
x
2
lim ln(x)=x
x!1
1
x
= e0 = 1: X
ln2 (x)
x+1
:
(iii) Sect. 4.7, Exercise #64 (p. 291). The historical example...
lim
p
2a3 x
x!a
p
a4
1=4
1=3
1=2
x4
a5=3 x1=3
a1=4 x3=4
a
x!a
a a3
a4
2a3 x
= lim
(ax3 )1=4
a
The numerator becomes
a
1=3
a a2 x
x4
= 0 and the denominator becomes
= 0: L’Hôpital.
L =
lim
1
2a3 x
2
x!a
2a3
1=2
x4
a3
lim
x!a
a1=4
4x3
3
x
4
a1=4
2a3 x
=
1=2
x4
2=3
1=4
2x3
3
x
4
1
x
3
a5=3
1
x
3
a5=3
2=3
1=4
Multiply top and bottom by 12 and then try direction substitution of x = a:
2a3 x
L =
=
=
a3
lim
x!a
lim
a1=4
12 a3 2x3
p
2a3 x x4
12 a3 2a3
p
2a3 a a4
a3
a2
9
2x3
3
x
4
1=4
3
(12)
(12)
!
2=3
!
12
=
p
4a
=
2=3
1=4
4a5=3 a
9a1=4 a 1=4
!
1
x
3
a5=3
4a5=3 x 2=3
9a1=4 x
x!a
12
=
1=2
x4
12a 4a
16a
=
( 9)
9
a3
a4
9
!
4a
(Page 3) Write carefully:
(i) Sect. 4.8, Exercises #22 & #24 (p. 301).
(#22) Power Rule for antiderivatives:
Z
xn dx =
xn+1
+ C; n 6= 1
n+1
Z
Z
Z
1
dx
1
x dx =
dx =
= ln jxj + C:
x
x
Find the antiderivative (family of functions).
Z
Z
Z
Z
1
5
2
2
+ 4t dt = 5
dt + 4 t dt = 5 t
t2
t2
= 5
t 1
( 1)
t3
3
+4
+C =
2
dt + 4
5
1
t
Z
t2 dt
4
+ t3 + C
3
5 4 3
+ t +C
t
3
=
(#24) Distribute …rst.
Z
Z
3
5m 12m
10m dm = 5
4
12m
m5
5
= 60
2
10m
50
dm = 60
m3
3
Z
4
m dm
+ C = 12m5
50
2
Z
tan (kx)
+C
k
sec2 (2v) dv = 2
tan (2v)
2
+ C = tan (2v) + C
(#32) Plan: Split it into two fractions. It’s still a bit tricky, though...
Z
Z
Z
sin ( ) 1
sin ( )
1
d
=
d
d
cos2 ( )
cos2 ( )
cos2 ( )
Z
Z
sin ( )
1
=
d
sec2 ( ) d
cos ( )
cos ( )
Z
Z
=
tan ( ) sec ( ) d
sec2 ( ) d
=
sec ( )
4
tan ( ) + C
m2 dm
50 3
m +C
3
(ii) Sect. 4.8, Exercises #30 & #32 (p. 301).
(#30) Trig. antiderivatives.
Z
sec2 (kx) dx =
Z
(Page 4) Write carefully:
(i) Sect. 4.8, Exercise #42 (p. 301).
With an initial condition, we can determine the value of C:
0
1
p
6
Z
p
4
x
+
B
xC
C dx; F (1) = 4:
F (x) = B
@
A
2
x
Plan: Power Rule. Split.
Z
Z
Z
6x 1=2
4x1=2
dx +
dx = 4 x
x2
x2
0
3=2
dx + 6
1
Z
x
5=2
dx
1
0
3=2 C
B x 1=2 C
B
C + 6B x
C+C
= 4B
@
@
1 A
3 A
2
2
=
1=2
8x
8
p
x
F (x) =
2
3
+ (6)
4
x3=2
x
3=2
+C
+ C:
When x = 1; we have F = 4:
8
p
1
F (1) = 4 =
4
13=2
+C )4=
12 + C ) C = 16:
The particular solution is
8
p
x
F (x) =
4
x3=2
+ 16
(ii) Sect. 4.8, Exercises #47 & #48 (p. 301).
(#47) f 0 (u) = 4 cos (u)
f (u) =
4 sin (2u)
Z
(4 cos (u)
= 4 sin (u)
4
4 sin (2u)) du
cos (2u)
2
+ C = 4 sin (u) + 2 cos (2u) + C
The initial value is f ( =6) = 0:
0 = 4 sin
C =
+ 2 cos 2
6
6
+C =4
1
2
+2
3:
The particular solution is
f (u) = 4 sin (u) + 2 cos (2u)
5
3:
1
2
+C =0
(#48) p0 (t) = 10e
t
p (t) = 10
Z
e t dt = 10
e t
( 1)
+C =
10e
t
+C
The intial value is p (0) = 100:
100 =
0
10e
+ C ) 100 =
10 + C ) C = 110:
The particular solution is
p (t) = 10e
t
+ 100 :
(Page 5) Write carefully.
(i) Sect. 4.8, Exercise #65abcd (p. 302).
The vertical position function is
4:9t2 + v0 t + s0 ;
s (t) =
where v0 is the initial velocity (up is positive) and s0 is the initial height.
In this case, since the balloon was rising at the time the rope on the payload was cut,
the payload had an initial velocity of +10 m/s.
The initial height was s0 = 400 m. Thus, the position function is
4:9t2 + 10t + 400:
s (t) =
(a) Find the velocity function.
s0 (t) = v (t) =
9:8t + 10 m/s
(b) The position function is given above.
(c) When does the payload reach its highest point?
Plan: Set s0 (t) = 0: When v (t) = 0; the velocity will go from positive to negative.
v (t) =
9:8t + 10 = 0 ) t =
10
seconds.
9:8
This is approx. 1.02 seconds . Why is this a local maximum?
Check the second derivative!
a (t) = v 0 (t) = s00 (t) =
9:8 m/sec2 :
The second derivative is always negative, so the curve is always concave DOWN.
Here is the graph of s (t) :
y
400
300
200
100
0
0
2. 5
5
7. 5
10
x
6
(d) When will the object hit the ground?
Plan: Set s (t) = 0: Quadratic Formula.
4:9t2 + 10t + 400 = 0
Hint: Multiply through by ( 1) …rst. We often make errors when a is a negative
number.
4:9t2
10t
400 = 0
q
( 10)2
( 10)
t =
4 (4:9) ( 400)
=
10
p
7940
9:8
2 (4:9)
p
10 + 7940 :
= 10:1 seconds.
We only want the positive solution: t =
9:8
We note that the velocity at impact is
!
!
p
p
p
10 + 7940
10 + 7940
v
= 9:8
+ 10 =
7940 m/sec.
9:8
9:8
This is approx. ( 89) m/sec. (Downward, of course!)
(ii) Sect. 4.8, Exercise #80 (p. 302). Mass on a spring.
Plan: Begin with acceleration. Integrate to …nd the velocity function. Integrate again
to the …nd position function.
a (t) = sin ( t) ; v (0) = 3
The velocity function is
Z
v (t) =
cos ( t)
sin ( t) dt =
cos ( (0))
v (0) = 3 =
C1 = 3 +
1
+ C1 =
+ C1
1
+ C1
:
Thus, the particular solution is
v (t) =
cos ( t)
1
+ 3+
:
Integrate again to …nd the position function. Remember that
s (t) =
=
=
Z
v (t) dt =
1
1
Z
Z
cos ( t)
cos ( t) dt +
sin ( t)
7
Z
+ 3+
+ 3+
3+
1
1
dt
t + C2
3+
1
1
dt
is a constant.
The initial value is
s (0) = 0 =
1
sin ( (0))
+ 3+
1
(0) + C2
C2 = 0:
The particular solution is
s (t) =
1
sin ( t)
+ 3+
1
t:
Here is the graph.
s(t)
25
20
15
It’s rather subtle. We have a sine wave added to a line.
The amplitude of the sine wave is small 1= 2 , so it’s barely
noticeable on this graph.
10
5
0
0
1
2
3
4
5
6
t
8