math 1a: homework 4 solutions
TRANSCRIPT
Math 1A: Homework 4 Solutions
July 16
1. Find the derivatives of the following functions.
(a) f(x) =√
x3+3x5+5
.
We have
f ′(x) =1
2
(x3 + 3
x5 + 5
)−1/2d
dx
(x3 + 3
x5 + 5
)=
1
2
(x5 + 5
x3 + 3
)1/2((x5 + 5)(3x2)− (x3 + 3)(5x4)
(x5 + 5)2
)=
1
2
(x5 + 5
x3 + 3
)1/2((3x7 + 15x2)− (5x7 + 15x4)
(x5 + 5)2
)=
1
2
(x5 + 5
x3 + 3
)1/2(−2x7 − 15x4 + 15x2
(x5 + 5)2
).
(b) f(x) = csc(x3 + 1).
We have
f ′(x) = − csc(x3 + 1) cot(x3 + 1)d
dx(x3 + 1)
= −3x2 csc(x3 + 1) cot(x3 + 1).
(c) f(x) = eex.
We have
f ′(x) = eex d
dx(ex)
= eex
ex = eex+x.
(d) f(x) = ln | sec(x) + tan(x)|.
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We have
f ′(x) =1
sec(x) + tan(x)
d
dx(sec(x) + tan(x))
=sec(x) tan(x) + sec2(x)
sec(x) + tan(x)
=sec(x)(tan(x) + sec(x))
sec(x) + tan(x)
= sec(x).
(e) f(x) = sin2(ecos(4x)).
We have
f ′(x) = 2 sin(ecos(4x)) cos(ecos(4x))d
dx(ecos(4x))
= 2 sin(ecos(4x)) cos(ecos(4x))(ecos(4x))d
dx(cos(4x))
= 2 sin(ecos(4x)) cos(ecos(4x))(ecos(4x))(−4 sin(4x))
= −8 sin(4x) sin(ecos(4x)) cos(ecos(4x))ecos(4x).
(f) f(x) = ln(ex+e−x
1+x2
).
We have
f ′(x) =1(
ex+e−x
1+x2
) ddx
(ex + e−x
1 + x2
)=
1 + x2
ex + e−x
((1 + x2)(ex − e−x)− (ex + e−x)(2x)
(1 + x2)2
)=
ex − e−x
ex + e−x− 2x
1 + x2.
2. Find dydx
by implicit differentiation.
(a) x2/3 + y2/3 = 4.
Differentiate throughout with respect to x to get
2
3x−1/3 +
2
3y−1/3
dy
dx= 0
y−1/3dy
dx= −x−1/3
dy
dx= −
(yx
)1/3.
2
(b)√y = 2tan(x) + y2.
Differentiate throughout with respect to x to get
1
2√y
dy
dx= ln(2)2tan(x) d
dx(tan(x)) + 2y
dy
dx(1
2√y− 2y
)dy
dx= ln(2)2tan(x) sec2(x)
dy
dx=
ln(2)2tan(x) sec2(x)(1
2√y− 2y
) .
(c) xe4√y + cos2(y) = 0.
Differentiate throughout with respect to x to get
xe4√y d
dx(4√y) + e4
√y(1) + 2 cos(y)(− sin(y))
dy
dx= 0
xe4√y
(4
2√y
dy
dx
)+ e4
√y − 2 cos(y) sin(y)
dy
dx= 0
dy
dx
(2xe4
√y
√y− 2 cos(y) sin(y)
)= −e4
√y
dy
dx=
e4√y
2 cos(y) sin(y)− 2xe4√y
√y
.
(d) y2(y2 − 4) = x2(x2 − 5).
Differentiate throughout with respect to x to get
y2(2y)dy
dx+ (y2 − 4)(2y)
dy
dx= x2(2x) + (x2 − 5)(2x)
2ydy
dx
(y2 + y2 − 4
)= (2x)(x2 + x2 − 5)
dy
dx=
2x(2x2 − 5)
2y(2y2 − 4)
dy
dx=
x(2x2 − 5)
y(2y2 − 4).
(e) y = (ln(x))ln(x).
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We have
ln(y) = ln[(ln(x))ln(x)]
ln(y) = ln(x) ln(ln(x))
1
y
dy
dx=
1
xln(ln(x)) + ln(x)
1
ln(x).1
x
1
y
dy
dx=
ln(ln(x))
x+
1
xdy
dx=
y
x(ln(ln(x)) + 1)
dy
dx=
(ln(x))ln(x)
x(ln(ln(x)) + 1).
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) = s′(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 hasa velocity greater than 20 are t > 5.
(b) At what times is the acceleration negative? Does this necessarily mean that theparticle is slowing down?
The acceleration function is
a(t) = v′(t) = 12t2 − 24t− 40.
We then have
a(t) < 0
⇔ 12t2 − 24t− 40 < 0
⇔ 3t2 − 6t− 10 < 0
⇔ (t− α) (t− β) < 0
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where α = 6−√156
6≈ −1.081 and β = 6+
√156
6≈ 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 notnecessarily mean that the speed is decreasing: a negative velocity could decreaseto become more negative but the speed would increase. Put another way, theparticle would slow down if and only if the signs of its velocity and accelerationvalues at any time are different.
4. The population of a town grows at a rate proportional to the population present attime t. The initial population of 500 increases by 15% in 10 years. What will be thepopulation in 30 years?
Let P (t) be the population of the town at time t (in years). We then have for someconstant r
P ′(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 have500(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 wasa constant 25◦ C. At the time of discovery the core temperature of the body wasdetermined to be 32◦ C. One hour later a second measurement showed that the coretemperature of the body was 28◦ C. Assume that the core temperature of the body atthe 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’sLaw of Cooling, we have for some constant k
B′(t) = k(B(t)− 25).
Let F (t) = B(t)− 25. We then have
F ′(t) = B′(t)
= k(B(t)− 25)
= kF (t)
⇒ F (t) = Aekt (A is some constant)
B(t)− 25 = Aekt
B(t) = 25 + Aekt.
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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 =3
12.
Note then thatekT+k
ekT=
3/12
7/12⇒ ek =
3
7.
Finally, note that
7
12= ekT = (ek)T = (3/7)T ⇒ T =
ln(7/12)
ln(3/7)≈ 0.636 = 38.16 mins.
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 ′(x) = 81+8x
so f ′(0) = 8. We therefore have
L(x) = f(0) + f ′(0)(x− 0) = 8x.
(b) f(x) = tan(2x) about x = −π8.
Note that f(−π8) = tan(−π
4) = −1 and f ′(x) = 2 sec2(2x) so f ′(−π
8) = 2(
√2)2 =
4. We therefore have
L(x) = f(−π
8
)+ f ′
(−π
8
)(x− (−π
8))
= −1 + 4(x+
π
8
).
(c) f(x) = ex cos(x) about x = 0.
Note that f(0) = e0 cos(0) = 1 and f ′(x) = ex cos(x) − ex sin(x) so f ′(0) =e0 cos(0)− e0 sin(0) = 1. We therefore have
L(x) = f(0) + f ′(0)(x− 0) = 1 + x.
(d) f(x) = ln(sin(x)) about x = π4.
Note that f(π4) = ln(sin(π/4)) = ln(1/
√2) = − ln(2)
2and f ′(x) = 1
sin(x)cos(x) =
cot(x) so f ′(π/4) = 1. We therefore have
L(x) = f(π/4) + f ′(π/4)(x− π/4) = − ln(2)
2+ (x− π/4).
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