MAT 122 Homework 6 Solutions
Section 3.2, Problem 10 —
d
dy
=
(5 · 2x − 5x + 4)
dx
dx
d
d
d
=
(5 · 2x ) − (5x) + (4)
dx
dx
dx
d x
d
= 5 (2 ) − 5 (x) + 0
dx
dx
= 5 ln(2) · 2x − 5
Section 3.2, Problem 12 —
d 0.7t dy
=
e
= 0.7 · e0.7t
dt
dt
Section 3.2, Problem 28 —
f 0 (t) =
=
=
=
=
d
Aet + B ln t
dt
d
d
(Aet ) + (B ln t)
dt
dt
d
d t
A · (e ) + B · (ln t)
dt
dt
1
t
Ae + B ·
t
B
Aet +
t
Section 3.2, Problem 34 — The function f (t) = 6.8e0.012t describes the world population, in
billions, as a function of t, the time in years since 2009.
Part a: f (0) = 6.8e0.012·0 = 6.8 · 1 = 6.8. This means that the world population in the year 2009
is 6.8 billion.
Part b: We first find the derivative, which is
d
6.8e0.012t
dt
d
=
6.8e0.012t
dt
d 0.012t = 6.8 ·
e
dt
= 6.8 · 0.012 · e0.012t
= 0.0816 · e0.012t
f 0 (t) =
Next, we evaluate the derivative at t = 0 to find
f 0 (0) = 0.0816 · e0.012·0 = 0.0816
1
This means that the world population is growing at an (instantaneous) rate of 0.0816 billion people
per year in the year 2009 (or 81.6 million people per year).
Part c: f (10) = 6.8e0.012·10 = 6.8e.12 = 7.66698, so the world population in 2019 is estimated to
be 7.7 billion people.
Part d: Using the expression we found earlier for the derivative, we have that
f 0 (10) = 0.0816 · e0.012·10 = 0.0816 · e0.12 = 0.0920
This means that the world population is (estimated to be) growing at an (instantaneous) rate of
0.092 billion people per year in the year 2019 (or 92 million people per year).
Section 3.4, Problem 10 —
d 3
dy
=
(t − 7t2 + 1)et
dt
dt
d
d 3
=
(t − 7t2 + 1) · et + (t3 − 7t2 + 1) · (et )
dt
dt
2
t
3
2
t
= (3t − 14t)e + (t − 7t + 1)e
= (3t2 − 14t + t3 − 7t2 + 1)et
= (t3 − 4t2 − 14t + 1)et
Section 3.4, Problem 16 —
d √ −z d 1/2 −z ze
=
z e
dz
dz
d 1/2
d
=
(z ) · e−z + z 1/2 · (e−z )
dz
dz
1 −1/2 −z
1/2
z
· e + z · −e−z
=
2
√
1
= √ z − ze−z
2 ze
f 0 (z) =
Section 3.4, Problem 24 —
d
3z
dw
=
dz
dz 1 + 2z
d
d
(1 + 2z) dz
(3z) − (3z) dz
(1 + 2z)
=
(1 + 2z)2
(1 + 2z) · 3 − (3z) · 2
=
(1 + 2z)2
3 + 6z − 6z
=
(1 + 2z)2
3
=
(1 + 2z)2
2
Section 3.4, Problem 34 —
f (x) = (3x + 8)(2x − 5)
d
[(3x + 8)(2x − 5)]
f 0 (x) =
dx
d
d
=
(3x + 8) · (2x − 5) + (3x + 8) · (2x − 5)
dx
dx
= 3 · (2x − 5) + (3x + 8) · 2
= 6x − 15 + 6x + 16
= 12x + 1
d
f 00 (x) =
(12x + 1)
dx
= 12
Section 3.4, Problem 36 — To find the equation of the tangent line at x = 0, we need to
find the point the line will pass through and its slope.
2x − 5
−5
The point will have coordinates (0, f (0)). We have f (x) =
, so f (0) =
= −5. Note
x+1
1
that this gives that the y-intercept of the line is at −5.
The slope is given by the value of the derivative at 0, that is, f 0 (0). We first find f 0 (x). Using
the quotient rule,
d 2x − 5
0
f (x) =
dx x + 1
d
d
(x + 1) dx
(2x − 5) − (2x − 5) dx
(x + 1)
=
2
(x + 1)
(x + 1) · 2 − (2x − 5) · 1
=
(x + 1)2
2x + 2 − 2x + 5
=
(x + 1)2
7
=
(x + 1)2
7
Evaluating at x = 0, we have that f 0 (0) = (0+1)
2 = 7. Therefore the tangent line has slope 7.
The equation of a line with slope 7 and y-intercept −5 is y = 7x − 5.
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