Math 1A: Homework 4 Solutions
July 16
1. Find the derivatives of the following functions.
q
3
(a) f (x) = xx5 +3
.
+5
We have
x3 + 3
x5 + 5
−1/2
d x3 + 3
dx x5 + 5
1/2 5
1 x5 + 5
(x + 5)(3x2 ) − (x3 + 3)(5x4 )
=
2 x3 + 3
(x5 + 5)2
1/2 1 x5 + 5
(3x7 + 15x2 ) − (5x7 + 15x4 )
=
2 x3 + 3
(x5 + 5)2
1/2 1 x5 + 5
−2x7 − 15x4 + 15x2
=
.
2 x3 + 3
(x5 + 5)2
1
f (x) =
2
0
(b) f (x) = csc(x3 + 1).
We have
d 3
(x + 1)
dx
= −3x2 csc(x3 + 1) cot(x3 + 1).
f 0 (x) = − csc(x3 + 1) cot(x3 + 1)
x
(c) f (x) = ee .
We have
d x
(e )
dx
x
x
= ee ex = ee +x .
x
f 0 (x) = ee
(d) f (x) = ln | sec(x) + tan(x)|.
1
We have
1
d
(sec(x) + tan(x))
sec(x) + tan(x) dx
sec(x) tan(x) + sec2 (x)
=
sec(x) + tan(x)
sec(x)(tan(x) + sec(x))
=
sec(x) + tan(x)
= sec(x).
f 0 (x) =
(e) f (x) = sin2 (ecos(4x) ).
We have
d cos(4x)
(e
)
dx
d
= 2 sin(ecos(4x) ) cos(ecos(4x) )(ecos(4x) ) (cos(4x))
dx
cos(4x)
cos(4x)
cos(4x)
= 2 sin(e
) cos(e
)(e
)(−4 sin(4x))
cos(4x)
cos(4x) cos(4x)
= −8 sin(4x) sin(e
) cos(e
)e
.
f 0 (x) = 2 sin(ecos(4x) ) cos(ecos(4x) )
(f) f (x) = ln
ex +e−x
1+x2
.
We have
d ex + e−x
f (x) = ex +e−x dx
1 + x2
1+x2
(1 + x2 )(ex − e−x ) − (ex + e−x )(2x)
1 + x2
= x
e + e−x
(1 + x2 )2
ex − e−x
2x
= x
−
.
−x
e +e
1 + x2
0
2. Find
dy
dx
1
by implicit differentiation.
(a) x2/3 + y 2/3 = 4.
Differentiate throughout with respect to x to get
2 −1/3 2 −1/3 dy
x
+ y
= 0
3
3
dx
dy
y −1/3
= −x−1/3
dx
y 1/3
dy
= −
.
dx
x
2
(b)
√
y = 2tan(x) + y 2 .
Differentiate throughout with respect to x to get
1 dy
d
dy
= ln(2)2tan(x) (tan(x)) + 2y
√
2 y dx
dx
dx
dy
1
= ln(2)2tan(x) sec2 (x)
√ − 2y
2 y
dx
dy
ln(2)2tan(x) sec2 (x)
.
=
1
dx
√
− 2y
2 y
√
(c) xe4
y
+ cos2 (y) = 0.
Differentiate throughout with respect to x to get
√
dy
d √
(4 y) + e4 y (1) + 2 cos(y)(− sin(y))
= 0
dx dx
√
√
4 dy
dy
xe4 y
+ e4 y − 2 cos(y) sin(y)
= 0
√
2 y dx
dx
√
√
dy 2xe4 y
− 2 cos(y) sin(y)
= −e4 y
√
dx
y
√
xe4
y
√
e4 y
dy
=
dx
2 cos(y) sin(y) −
(d) y 2 (y 2 − 4) = x2 (x2 − 5).
Differentiate throughout with respect to x to get
y 2 (2y)
dy
dy
+ (y 2 − 4)(2y)
dx
dx
dy 2
2y
y + y2 − 4
dx
dy
dx
dy
dx
(e) y = (ln(x))ln(x) .
3
= x2 (2x) + (x2 − 5)(2x)
= (2x)(x2 + x2 − 5)
2x(2x2 − 5)
=
2y(2y 2 − 4)
x(2x2 − 5)
=
.
y(2y 2 − 4)
√
4
2xe
√
y
y
.
We have
ln(y)
ln(y)
1 dy
y dx
1 dy
y dx
dy
dx
dy
dx
= ln[(ln(x))ln(x) ]
= ln(x) ln(ln(x))
1
1 1
=
ln(ln(x)) + ln(x)
.
x
ln(x) x
ln(ln(x)) 1
=
+
x
x
y
=
(ln(ln(x)) + 1)
x
(ln(x))ln(x)
=
(ln(ln(x)) + 1).
x
3. A particle moves with position function
s(t) = t4 − 4t3 − 20t2 + 20t
t ≥ 0.
(a) At what times does the particle have a velocity greater than 20 m/s?
The velocity function is
v(t) = s0 (t) = 4t3 − 12t2 − 40t + 20.
We then have
⇔
⇔
⇔
⇔
⇔
v(t) > 20
4t3 − 12t2 − 40t + 20 > 20
4t3 − 12t2 − 40t > 0
4t(t2 − 3t − 10) > 0
4t(t + 2)(t − 5) > 0
−2 < t < 0 or t > 5.
Since we are only considering t ≥ 0, the only values of t for which the particle has
a velocity greater than 20 are t > 5.
(b) At what times is the acceleration negative? Does this necessarily mean that the
particle is slowing down?
The acceleration function is
a(t) = v 0 (t) = 12t2 − 24t − 40.
We then have
a(t) < 0
⇔ 12t2 − 24t − 40 < 0
⇔ 3t2 − 6t − 10 < 0
⇔ (t − α) (t − β) < 0
4
√
√
where α = 6− 6 156 ≈ −1.081 and β = 6+ 6 156 ≈ 3.082. Thus, a(t) < 0 ⇔ −1.081 <
t < 3.082. Since t ≥ 0, this shrinks to 0 ≤ t < 3.082.
Negative acceleration means that the velocity is decreasing. This however does not
necessarily mean that the speed is decreasing: a negative velocity could decrease
to become more negative but the speed would increase. Put another way, the
particle would slow down if and only if the signs of its velocity and acceleration
values at any time are different.
4. The population of a town grows at a rate proportional to the population present at
time t. The initial population of 500 increases by 15% in 10 years. What will be the
population in 30 years?
Let P (t) be the population of the town at time t (in years). We then have for some
constant r
P 0 (t) = rP (t) ⇔ P (t) = Aert
for some constant A.
We are given that P (0) = 500 and P (10) = 500 × 115% = 500(1.15). We then have
500 = Ae0 ⇒ A = 500.
We also have
500(1.15) = 500e10r ⇒ e10r = 1.15.
Observe then that
P (30) = 500e30r = 500(e10r )3 = 500(1.15)3 ≈ 760.44.
(Note that we could have found the value of r but this is much easier.)
5. A dead body was found within a closed room of a house where the temperature was
a constant 25◦ C. At the time of discovery the core temperature of the body was
determined to be 32◦ C. One hour later a second measurement showed that the core
temperature of the body was 28◦ C. Assume that the core temperature of the body at
the beginning was 37◦ C and that the cooling process obeys Newton’s Law of Cooling.
Determine how many hours elapsed before the body was found.
Let B(t) be the temperature of the body (in ◦ C) t hours after death. From Newton’s
Law of Cooling, we have for some constant k
B 0 (t) = k(B(t) − 25).
Let F (t) = B(t) − 25. We then have
F 0 (t) =
=
=
⇒ F (t) =
B(t) − 25 =
B(t) =
B 0 (t)
k(B(t) − 25)
kF (t)
Aekt (A is some constant)
Aekt
25 + Aekt .
5
Let T be the time that elapses before the body is found. We are given that B(0) = 37,
B(T ) = 32 and B(T + 1) = 28. We then have
37 = 25 + Ae0 ⇒ A = 12.
We also have
32 = 25 + 12ekT ⇒ ekT =
7
.
12
Also,
28 = 25 + 12ek(T +1) ⇒ ekT +k =
Note then that
3
.
12
ekT +k
3
3/12
⇒ ek = .
=
kT
e
7/12
7
Finally, note that
ln(7/12)
7
= ekT = (ek )T = (3/7)T ⇒ T =
≈ 0.636 = 38.16 mins.
12
ln(3/7)
6. Find the linearizations of the following functions about the given points.
(a) f (x) = ln(1 + 8x) about x = 0.
Note that f (0) = ln(1) = 0 and f 0 (x) =
8
1+8x
so f 0 (0) = 8. We therefore have
L(x) = f (0) + f 0 (0)(x − 0) = 8x.
(b) f (x) = tan(2x) about x = − π8 .
√
Note that f (− π8 ) = tan(− π4 ) = −1 and f 0 (x) = 2 sec2 (2x) so f 0 (− π8 ) = 2( 2)2 =
4. We therefore have
π
π
π π
L(x) = f −
+ f0 −
x − (− ) = −1 + 4 x +
.
8
8
8
8
(c) f (x) = ex cos(x) about x = 0.
Note that f (0) = e0 cos(0) = 1 and f 0 (x) = ex cos(x) − ex sin(x) so f 0 (0) =
e0 cos(0) − e0 sin(0) = 1. We therefore have
L(x) = f (0) + f 0 (0)(x − 0) = 1 + x.
(d) f (x) = ln(sin(x)) about x = π4 .
√
and f 0 (x) =
Note that f ( π4 ) = ln(sin(π/4)) = ln(1/ 2) = − ln(2)
2
cot(x) so f 0 (π/4) = 1. We therefore have
L(x) = f (π/4) + f 0 (π/4)(x − π/4) = −
6
1
sin(x)
ln(2)
+ (x − π/4).
2
cos(x) =