– Homework 6 – tran – (52970)
This print-out should have 16 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
001
Estimate the value of 171/4 using differentials.
1. 171/4 ≈
67
32
2. 171/4 ≈
33
16
10.0 points
Find the linearization of
f (x) = √
1
2+x
3. 171/4 ≈ 2
at x = 0.
1 1. L(x) = √ 1 −
2
1 2. L(x) = √ 1 +
2
1 x
4
1 x
4
63
32
5. 171/4 ≈
65
32
1. Max error ≈ ±8.0584 sq.ins
10.0 points
Find the differential, dy, of
2
y = f (x) = tan(4x ) .
2. Max error ≈ ±8.0344 sq.ins
3. Max error ≈ ±8.0664 sq.ins
4. Max error ≈ ±8.0504 sq.ins
1. dy = 8x sec2 (4x) + dx
5. Max error ≈ ±8.0424 sq.ins
2
2. dy = 8x sec (4x )
005
2
2
10.0 points
2
3. dy = 4 sec (4x ) tan(4x )
Find the linearization of
4. dy = 4 sec2 (4x2 ) tan(4x2 ) dx
5. dy = 8x sec2 (4x2 ) dx
6. dy = 4 sec2 (4x2 ) tan(4x2 ) + dx
003
10.0 points
The radius of a circle is estimated to be
16 inches, with a maximum error in measurement of ±0.08 inches. Use differentials to
estimate the maximum error in calculating
the area of the circle using this estimate.
1
1
6. L(x) = √ − x
2 2
2
4. 171/4 ≈
004
1
1
3. L(x) = √ + x
2 2
1 1
1− x
4. L(x) =
2
2
1 1
1+ x
5. L(x) =
2
4
002
1
10.0 points
f (x) = 6 cos x
at x = −π/4.
1. L(x) =
√ 3
2 3 − π + 3x
4
3
2. L(x) = 3 − π − 3x
4
– Homework 6 – tran – (52970)
3. L(x) =
√ 3
2 3 + π − 3x
4
3
4. L(x) = 3 − π + 3x
4
3
5. L(x) = 3 + π − 3x
4
√ 3
6. L(x) = 2 3 + π + 3x
4
006
10.0 points
Find the derivative of
f (t) = 2 sinh(t) − cosh(t) .
1. f ′ (t) = et + 3e−t
2. f ′ (t) = 3et − e−t
1 t 3 −t
e − e
2
2
3. f ′ (t) =
4. f ′ (t) = et − 3e−t
5. f ′ (t) =
3 t 1 −t
e − e
2
2
6. f ′ (t) =
1 t 3 −t
e + e
2
2
007
10.0 points
Find the derivative of
f (x) =
1. f ′ (x) =
cosh(x)
.
3 + 4 cosh(x)
3 cosh(x) + 4
(3 + 4 cosh(x))2
2. f ′ (x) =
3 sinh(x)
3 + 4 cosh(x)
3. f ′ (x) =
3 sinh(x) − 4
3 + 4 cosh(x)
4. f ′ (x) =
3 cosh(x) − 4
3 + 4 cosh(x)
5. f ′ (x) =
3 sinh(x)
(3 + 4 cosh(x))2
008
2
10.0 points
Find the rate at which the volume of a
sphere is changing with respect to its radius r
when r = 4 cm.
1. rate = 24 cm3 /cm
2. rate = 4 cm3 /cm
3. rate = 48 cm3 /cm
4. rate = 16π cm3 /cm
5. rate = 64π cm3 /cm
009 10.0 points
A mass attached to a vertical spring has position function given by
y(t) = C cos(ωt),
where C is the amplitude of its oscillations
and ω is a constant.
Find the velocity and acceleration as functions of time.
1. v(t) = C sin(ωt), a(t) = −C cos ωt
2.
v(t)
=
a(t) = −C ω 2 cos(ωt)
C ω sin(ωt),
3. v(t) =
−C ω 2 cos(ωt)
− C ω sin(ωt),
a(t)
=
4. v(t) =
−C ω 2 cos(ωt)
− C ω cos(ωt),
a(t)
=
5. v(t) = − C sin(ωt), a(t) = −C cos(ωt)
010
10.0 points
If the half-life of a certain radioactive substance is 1800 years, estimate how many years
– Homework 6 – tran – (52970)
must elapse before only 65% of the radioactive
substance remains.
2. speed =
15
ft/sec
2
1. # years ≈ 1018
3. speed =
29
ft/sec
4
2. # years ≈ 1418
4. speed =
31
ft/sec
4
3. # years ≈ 1218
5. speed = 8 ft/sec
4. # years ≈ 1318
5. # years ≈ 1118
011
013
dA
= 0.08 A.
dt
Katy deposits $100 in such a savings account.
Find out how much money will be in her
account after 7 years, leaving the answer in
exponential form.
1. A(7) = $100 e5.6
2. A(7) = $100 e0.56
3. A(7) = $100 e−0.56
10.0 points
Determine the value of dy/dt at x = 3 when
y = x2 − 3x
10.0 points
The amount, $A, in a Wells Fargo savings
account satisfies the differential equation
3
and dx/dt = 2.
dy = 14
1.
dt x=3
dy 2.
= 8
dt x=3
dy 3.
= 6
dt x=3
dy = 10
4.
dt x=3
dy 5.
= 12
dt x=3
014
10.0 points
A point is moving on the graph of
4. A(7) = $100 e−5.6
5. A(7) = $100 e56
012
3x3 + 2y 3 = xy.
When the point is at
P =
10.0 points
A 5 foot ladder is leaning against a wall. If
the foot of the ladder is sliding away from the
wall at a rate of 10 ft/sec, at what speed is
the top of the ladder falling when the foot of
the ladder is 3 feet away from the base of the
wall?
1. speed = 7 ft/sec
1 1 ,
,
5 5
its y-coordinate is increasing at a speed of 8
units per second.
What is the speed of the x-coordinate at
that time and in which direction is the xcoordinate moving?
1. speed =
5
units/sec, decreasing x
4