Document 6603414

1. Suppose a hot air balloon is rising vertically upward, and it is tracked by a scout standing 500 ft apart from
the liftoff point. At the moment the scout’s elevation angle is π/4, the angle is increasing at the rate of 0.14
rad/min. How fast is the balloon rising at that moment?
Let y be the distance of the balloon from the ground and θ be the elevation angle of the scout, then y = 500 tan θ.
dy
dθ
Differentiating with respect to t,
= 500 sec2 θ .
dt
dt
√
dy dθ 2 π
At the moment concerned, the rate of change of y is
= 500 sec
·
= 500( 2)2 · 0.14 = 140.
dt
4 dt
π
π
θ= 4
θ= 4
The balloon is rising at a rate of 140 ft/min at that moment.
2. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the
building at a speed of 1.6 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m
from the building?
Let L be the length of the shadow of the man on the wall, x be the distance of the man from the spotlight. By
considering similar triangles,
L
2
=
12
x
24
.
L=
x
dL
−24 dx
= 2 ·
.
dt
x
dt
When the man is 4 m from the building, he is 8 m from the spotlight, so the rate of change of L is
3
24 dx 24 8
dL =
−
·
=− · =− .
dt x=8
82 dt x=8
64 5
5
3
The length of his shadow is decreasing at m/s.
5
Differentiating with respect to t,
3. A 13-ft ladder is leaning against a wall when its base starts to slide away. By the time the base is 12 ft from the
wall, the base is moving at the rate of 5 ft/sec.
(a) How fast is the top of the ladder sliding down the wall?
Let x√be the distance
√ of the base of the ladder from the wall, y be the height of the top of the ladder, then
y = 132 − x2 = 169 − x2 .
dy
1
−x
dx
dx
Differentiating with respect to t,
= √
(−2x) ·
=√
·
.
dt
dt
2 169 − x2
169 − x2 dt
dy −12
dx −12
At this moment, the rate of change of y is
=√
·
=
· 5 = −12.
2
dt
dt
5
169 − 12
x=12
x=12
The top of the ladder is sliding down at 12 ft/sec.
(b) At what rate is the area of the triangle formed by the ladder, wall, and the ground changing?
1
xy.
2
1
dy
dx
dA
Differentiating with respect to t,
=
x
+y
.
dt
2
dt
dt
At this
the rate
moment,
of change of
A is
dA dy 119
1
dx 1
(12)
.
=
+
5
= (12(−12) + 5(5)) = −
dt x=12
2
dt x=12
dt x=12
2
2
119 2
119 2
The area is changing at −
ft /sec. (Or decreasing at
ft /sec)
2
2
The are of the triangle is A =
1
(c) At what rate is the acute angle θ between the ladder and the ground changing?
x y
y
. Or sin−1 , tan−1
13
13
x
dθ
1 dx
−1
Differentiating with respect to t,
·
=p
·
.
x
2
dt
1 − ( 13 ) 13 dt
dθ −1
1 dx 13 1
q
At this moment, the rate of change of θ is
=
·
=− ·
·
· 5 = −1.
dt x=12
dt x=12
5 13
2 13
1 − ( 12
)
13
The angle θ is cos−1
The angle θ is changing at −1 rad/sec. (Or decreasing at 1 rad/sec)
4. The minute hand on a watch is 8 mm long and the hour hand is 4 mm long. How fast is the distance between
the tips of the hands changing at one o’clock? (Recall the law of cosines c2 = a2 + b2 − 2ab cos γ)
Let L be the distance between the two tips, θ be the angle between the two hands.
p
√
√
By the law of cosine, L = 82 + 42 − 2(4)(8) cos θ = 80 − 64 cos θ = 4 5 − 4 cos θ.
2π
π
At one o’clock, θ =
= .
12
6
The minute hand rotates clockwise in a uniform angular speed that rotates a circle in 602 seconds, the angle θ
decreases as the minute hand rotates clockwise. At the same time, the hour hand rotates clockwise in a uniform
angular speed a circle in (12 · 602 ) seconds, where the angle θ increases as the hour hand rotates clockwise, so
2π
2π
−11π
dθ
=− 2 +
=
(measured in rad/s).
2
dt
60
12 · 60
21600
4
dθ
8 sin θ
dθ
dL
= √
· (4 sin θ) ·
=√
· .
Differentiating with respect to t,
dt
dt
2 5 − 4 cos θ
5 − 4 cos θ dt
At one o’clock,
8 sin π6
8( 12 )
dL
−11π
−11π
11π
q
p
=p
·
=
=
·
√ .
√
π
dt
21600
5 − 4 cos 6 21600
5400 5 − 2 3
5 − 4( 23 )
The distance between the tips is decreasing at
11π
p
√ mm/s.
5400 5 − 2 3
5. Let f (x) = tan−1 x + tan−1 x1 .
(a) Identify the domain of f .
f is defined when x 6= 0. The domain of f is (−∞, 0) ∪ (0, ∞).
(b) Show that tan−1 x is an odd function.
Let u = tan−1 (−x), then
tan u = −x
− tan u = x
tan(−u) = x
−u = tan−1 x
u = − tan−1 x.
Since tan−1 (−x) = − tan−1 x for any arbitrary x, tan−1 x is an odd function.
(c) Show that f is an odd function.
−1
f (−x) = tan
(−x) + tan
−1
1
−x
= − tan−1 x − tan−1
2
1
= −f (x), therefore f (x) is an odd function.
x
(d) Find f (x) for each x in the domain of f .
If x > 0, let u = tan−1 x, then tan u = x.
Consider a right angled triangle with an angle u, where the opposite side has length x and adjacent side
has length 1.
π
1
1
The other angle is v = − u, which satisfies tan v = , that is, v = tan−1 .
2
x
x
π
−1
−1 1
Therefore f (x) = tan x + tan
=u+v = .
x
2
π
If x < 0, then −x > 0. Since f is odd, f (x) = f [−(−x)] = −f (−x) = − .
2
(
π
,
x>0
Combining, f (x) = 2 π
.
−2, x < 0
6. Let f (x) =
ex − e−x
, it is known that f is a one-to-one function.
2
(a) Find f (ln 2).
2 − 12
eln 2 − e− ln 2
3
=
= .
2
2
4
3
1
(b) Find (f −1 )0 ( ), by (f −1 )0 (x) = 0 −1
.
4
f [f (x)]
f (ln 2) =
2+
ex + e−x
eln 2 + e− ln 2
and f 0 (ln 2) =
=
2
2
2
1
1
4
−1 0 3
(f ) ( ) = 0 −1 3 = 0
= .
4
f (ln 2)
5
f [f ( 4 )]
f 0 (x) =
1
2
=
5
.
4
(c) Find f −1 (x).
Let y =
ex − e−x
, then
2
2y = ex − e−x
2yex = e2x − 1
e2x − 2yex − 1 = 0.
Regarding this equation as a quadratic equation in ex , by the quadratic formula,
p
2y ± 4y 2 + 4
x
e =
p2
= y ± y 2 + 1.
Since
p
y2 + 1 >
p
y 2 = |y|, y −
Therefore f −1 (x) = ln(x +
√
p
y 2 + 1 < y − |y| ≤ 0, while ex > 0, we reject the root y −
p
ex = y + y 2 + 1
p
x = ln(y + y 2 + 1).
x2 + 1).
3
p
y 2 + 1. Now
(d) Differentiate the function obtained in (c) to verifty the result in (b).
1
√
·
x + x2 + 1
1
3
=
(f −1 )0 ( ) = q
4
( 34 )2 + 1
(f −1 )0 (x) =
2x
1+ √
2 x2 + 1
=
√
1
2 x2 + 1 + 2x
√
√
=√
.
2(x + x2 + 1) x2 + 1
x2 + 1
4
.
5
7. Find the values of a such that the two curves y = ax and y = x are tangent to each other.
Suppose that y = ax and y = x are tangent at x = p.
Then the two function ax and x have the same value and same derivative at x = p.
ax and x have the same value at p implies ap = p.
ax and x have the same derivative at p implies ap ln a = 1.
(
Now we have the two equations in two unknowns a and p:
From the first equation, p ln a = ln p, so ln a =
ap
=p
.
p
a ln a = 1
ln p
.
p
ln p
ln p
into the second equation, we have p ·
= 1, which solves p = e.
p
p
Hence from the first equation,
Put ap = p and ln a =
ae = e
e ln a = 1
1
ln a =
e
a = e1/e .
4