– 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.
002
1
10.0 points
Find the differential, dy, of
y = f (x) = tan(2x2 ) .
001
10.0 points
Find the linearization of
f (x) = √
1. dy = 4x sec2 (2x) + dx
1
6+x
2. dy = 4x sec2 (2x2 )
at x = 0.
1 1 1. L(x) = √ 1 − x correct
12
6
1
1
2. L(x) = √ − x
6 6
3. dy = 2 sec2 (2x2 ) tan(2x2 ) dx
4. dy = 2 sec2 (2x2 ) tan(2x2 )
5. dy = 2 sec2 (2x2 ) tan(2x2 ) + dx
1
1
3. L(x) = √ + x
6 6
1 1
1− x
4. L(x) =
6
6
1 1
1+ x
5. L(x) =
6
12
1 1 6. L(x) = √ 1 + x
12
6
6. dy = 4x sec2 (2x2 ) dx correct
Explanation:
The differential, dy, of
y = f (x)
is given by
dy = f ′ (x) dx .
Explanation:
The linearization of f is the function
L(x) = f (0) + f ′ (0)x .
But for the function
1
f (x) = √
= (6 + x)−1/2 ,
6+x
the Chain Rule ensures that
On the other hand,
d
tan x = sec2 x .
dx
Thus by the Chain Rule,
f ′ (x) = 4x sec2 (2x2 ) .
Consequently,
1
f ′ (x) = − (6 + x)−3/2 .
2
dy = 4x sec2 (2x2 ) dx .
Consequently,
1
f (0) = √ ,
6
f ′ (0) = −
1
√ ,
12 6
and so
keywords: differential, trig function, tan function, Chain Rule,
003
1 1 L(x) = √ 1 − x .
12
6
10.0 points
Estimate the value of 151/4 using differentials.
– Homework 6 – tran – (52970)
31
16
1. 151/4 ≈
5. Max error ≈ ±2.0026 sq.ins
Explanation:
The area of a circle having radius x is given
by A = πx2 . By differentials, therefore,
2. 151/4 ≈ 2
3. 151/4 ≈
65
32
4. 151/4 ≈
63
correct
32
5. 151/4 ≈
33
16
∆A ≈ ∆x
dA
= 2πx∆x.
dx
At x = 4 and ∆x = ±0.08, therefore,
Max error ≈ ±2.0106 sq.ins.
Explanation:
Set f (x) = x1/4 . Then
005
df
1
.
=
dx
4x3/4
f (a + ∆x) − f (a)
df ∆x
≈
.
∆x =
dx x = a
4a3/4
Thus, with a = 16 and ∆x = −1,
151/4 − 2 ≈ −
1
.
32
Consequently,
151/4 ≈
63
.
32
10.0 points
The radius of a circle is estimated to be
4 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. Max error ≈ ±2.0346 sq.ins
2. Max error ≈ ±2.0266 sq.ins
3. Max error ≈ ±2.0106 sq.ins correct
4. Max error ≈ ±2.0186 sq.ins
10.0 points
Find the linearization of
f (x) = 2 cos x
By differentials, therefore, we see that
004
2
at x = −π/4.
1
1. L(x) = 1 − π − x
4
√ 1
2. L(x) = 2 1 + π + x correct
4
√ 1
3. L(x) = 2 1 − π + x
4
1
4. L(x) = 1 − π + x
4
1
5. L(x) = 1 + π − x
4
√
1
6. L(x) = 2 1 + π − x
4
Explanation:
The linearization of f at x = a is the function
L(x) = f (a) + f ′ (a)(x − a) .
But for the function
f (x) = 2 cos x ,
we see that
f ′ (x) = −2 sin x .
– Homework 6 – tran – (52970)
Consequently,
π
√
2
f′ −
= √ = 1 2,
1
5
4
2
f ′ (t) = et + e−t .
2
2
In this case
π
√
2
f −
= √ = 1 2,
4
2
and so
√
L(x) =
2+
√ π √ 1
2 x+
= 2 1+ π+x .
4
4
keywords:
006
10.0 points
Find the derivative of
007
f (x) =
1. f ′ (t) = et − 5e−t
1 t 5 −t
e − e
2
2
t
′
−t
3. f (t) = 5e − e
4. f ′ (t) =
5 t 1 −t
e − e
2
2
5. f ′ (t) = et + 5e−t
6. f ′ (t) =
1 t 5 −t
e + e correct
2
2
Explanation:
Since
(cosh(t))′ = sinh(t) ,
(sinh(t))′ = cosh(t) ,
cosh(x)
.
2 + 3 cosh(x)
1. f ′ (x) =
2 cosh(x) − 3
2 + 3 cosh(x)
2. f ′ (x) =
2 sinh(x) − 3
2 + 3 cosh(x)
3. f ′ (x) =
2 sinh(x)
correct
(2 + 3 cosh(x))2
4. f ′ (x) =
2 sinh(x)
2 + 3 cosh(x)
5. f ′ (x) =
2 cosh(x) + 3
(2 + 3 cosh(x))2
Explanation:
Since
(cosh(x))′ = sinh(x) ,
we see that
f ′ (t) = 3 cosh(t) − 2 sinh(t) .
On the other hand,
cosh(t) =
1 t
(e + e−t ) ,
2
sinh(t) =
1 t
(e − e−t ) .
2
while
3
f (t) = (et + e−t )−(et − e−t )
2
t 3
−t 3
= e
− 1 +e
+1 .
2
2
(sinh(x))′ = cosh(x) ,
the quotient rule shows that
f ′ (x)
=
sinh(x)(2 + 3 cosh(x)) − cosh(x)(3 sinh(x))
.
(2 + 3 cosh(x))2
Consequently,
Thus
′
10.0 points
Find the derivative of
f (t) = 3 sinh(t) − 2 cosh(t) .
2. f ′ (t) =
3
f ′ (x) =
008
2 sinh(x)
(2 + 3 cosh(x))2
10.0 points
.
– Homework 6 – tran – (52970)
Find the rate at which the surface area of
a cube is changing with respect to its side
length x when x = 2 cm.
1. rate = 27 cm2 /cm
4
5. v(t) = A ω sin(ωt), a(t) = −A ω 2 cos(ωt)
Explanation:
2. rate = 4 cm2 /cm
y(t) = A cos(ωt)
3. rate = 36π cm2 /cm
4. rate = 24 cm2 /cm correct
v(t) = y ′ (t) = − A ω sin(ωt)
5. rate = 16π cm2 /cm
a(t) = v ′ (t) = −A ω 2 cos(ωt)
Explanation:
For a cube with side length x its
010
surface area = 6 x2 .
Now the rate at which the surface area is
changing with respect to x is the derivative of
surface area with respect to x. Thus
rate =
d
(surface area) = 12 x .
dx
2. # years ≈ 4129
.
3. # years ≈ 4229 correct
009 10.0 points
A mass attached to a vertical spring has position function given by
y(t) = A cos(ωt),
where A is the amplitude of its oscillations
and ω is a constant.
Find the velocity and acceleration as functions of time.
1. v(t) = A sin(ωt), a(t) = −A cos ωt
2. v(t) =
−A ω 2 cos(ωt)
− A ω cos(ωt),
a(t)
=
4. # years ≈ 4329
5. # years ≈ 4429
Explanation:
The radioactivity, R(t), remaining after t
years is given by
R(t) = R(0)e−kt
where R(0) is the initial amount of radioactivity and k is the decay constant. Since the
half-life is 3200 years,
R(3200) = R(0)e−3200k =
3. v(t) = − A sin(ωt), a(t) = −A cos(ωt)
4. v(t) = − A ω sin(ωt),
−A ω 2 cos(ωt) correct
If the half-life of a certain radioactive substance is 3200 years, estimate how many years
must elapse before only 40% of the radioactive
substance remains.
1. # years ≈ 4029
Consequently, when x = 2 cm.,
rate = 24 cm2 /cm
10.0 points
a(t)
1
R(0),
2
so
=
−k =
1
1
ln ,
3200 2
i.e., k =
1
ln 2 .
3200
– Homework 6 – tran – (52970)
Thus the number, n, of years which have to
elapse before only 40% of the radioactivity
remains is given by
R(n) = R(0)e−kn =
40
R(0).
100
Taking logs of both sides we see that
1 40 1 100 n = − ln
= ln
.
k
100
k
40
Hence
# years ≈ 4229 .
011
10.0 points
The amount, $A, in a Wells Fargo savings
account satisfies the differential equation
dA
= 0.07 A.
dt
Katy deposits $100 in such a savings account.
Find out how much money will be in her
account after 6 years, leaving the answer in
exponential form.
−4.2
1. A(6) = $100 e
2. A(6) = $100 e−0.42
3. A(6) = $100 e42
In turn, after exponentiation this becomes
A(t) = eC e0.07 t
with C an arbitrary constant. The constant C
is determined by the initial condition A(0) =
100 for
A(0) = 100
5. A(6) = $100 e
correct
Explanation:
After integration the differential equation
dA
= 0.07A
dt
becomes
1
dA =
A
Z
eC = 100.
A(t) = 100 e0.07 t .
At t = 6, therefore,
A(6) = $100 e0.42 .
012
10.0 points
A 15 foot ladder is leaning against a wall.
If the foot of the ladder is sliding away from
the wall at a rate of 12 ft/sec, at what speed
is the top of the ladder falling when the foot
of the ladder is 9 feet away from the base of
the wall?
1. speed =
35
ft/sec
4
2. speed =
19
ft/sec
2
3. speed =
39
ft/sec
4
4. speed = 9 ft/sec correct
5. speed =
Z
=⇒
Consequently, the amount in the account after
t years is given by
4. A(6) = $100 e4.2
0.42
5
37
ft/sec
4
0.07 dt.
Thus
ln A = 0.07 t + C.
Explanation:
Let y be the height of the ladder when the
foot of the ladder is x feet from the base of
the wall as shown in figure