1. (a) (i)
lim (sin x + cos x) = sin
x→π/2
π
π
+ cos = 1 + 0 = 1.
2
2
(ii) Since x2 + x − 12 = (x − 3)(x + 4) (the zeros are 3 and −4) we have
(x − 3)(x + 4)
x2 + x − 12
= lim
= lim (x + 4) = 7.
x→3
x→3
x→3
x−3
x−3
lim
The same result could also be obtained with l’Hospital’s rule.
(iii) By l’Hospital’s rule:
x
1
lim xex = lim −x = lim
= 0.
x→−∞ e
x→−∞ −e−x
| {z }
Type [ −∞
∞ ]
x→−∞
(iv)
x−ln x
lim (x − ln x) = lim ln e
x→∞
x→∞
By l’Hospiatal’s rule
= lim ln
x→∞
ex
eln x
= lim ln
x→∞
ex
x
.
ex
ex
= lim
= ∞,
x→∞ x
x→∞ 1
lim
so we get
lim (x − ln x) = lim ln
x→∞
x→∞
ex
x
=
x
u= ex
x→∞⇒u→∞
= lim ln u = ∞.
u→∞
(b) Since c is an interior point of [a, b] and f is continuous on [a, b], by
definition of continuity
lim f (x) = f (c).
x→c
(c)
lim (H ◦ f )(x) = lim H(f (x)) = lim H(H(x) + x).
x→0−
x→0−
x→0−
−
For x < 0 (which we are considering since x → 0 ) we have H(x)+x = 0+x < 0
so H(H(x) + x) = 0. Consequently
lim H(H(x) + x) = 0.
x→0−
2. (a) (i)
√
√
d
d
d √
cos x sin x =
cos x
sin x + cos x ·
sinx =
dx
dx
dx
√
√
1
d
cos2 x
− sin x sin x + cos x √
·
sin x = − sin x sin x + √
.
2 sin x dx
2 sin x
(ii) In case we’ve forgotten the derivative of tanh we can remind ourselves
like this:
d
d sinh x
cosh2 x − sinh2 x
1
tanh x =
=
=
,
2
dx
dx cosh x
cosh x
cosh2 x
1
so
tanh x · 2x − x2 cosh1 2 x
d x2
=
=
dx tanh x
tanh2 x
2x sinh x cosh x−x2
cosh2 x
2
tanh x
=
2x sinh x cosh x − x2
.
sinh2 x
(iii)
d x
d ln 10x
d x ln 10
10 =
e
e
=
= ex ln 10 ln 10 = 10x ln 10.
dx
dx
dx
(iv) Using implicit differentiation with respect to x, we have
4x − 2y
so
dy
= 0,
dx
dy
4x
2x
=
=
.
dx
2y
y
(b) With y = y(x) we have
y(x) = ax3 + bx2 + cx + d,
y 0 (x) = 3ax2 + 2bx + c,
y 00 (x) = 6ax + 2b.
The requirement that y has a local extrema at (−2, 6) and (2, 0) give us y(−2) =
6, y(2) = 0, y 0 (−2) = 0 and y 0 (2) = 0, and we get the system of equations
−8a + 4b − 2c + d = 6
8a + 4b + 2c + d = 0
12a − 4b + c = 0
12a + 4b + c = 0.
Solving this system yields
a=
3
,
16
9
c=− ,
4
b = 0,
d = 3,
so
3 3 9
x − x + 3.
16
4
0
We can verify that y(−2) = 6, y(2) = 0, y (−2) = 0, y 0 (2) = 0 and also that
y 00 (−2) = −9/4 < 0 and y 00 (2) = 9/4 > 0 so (−2, 6) is a local maximum and
(2, 0) is a local minimum as required.
(c) Suppose that f (x) is even, so f (−x) = f (x). We want to show that f 0 (x)
is odd, so we want to show that −f 0 (−x) = f 0 (x). But, using that f (−x) = f (x)
and f (−(x − h)) = f (x − h) (since f is even), we have
y(x) =
f (−x + h) − f (−x)
f (−(x − h)) − f (−x)
= − lim
=
h→0
h
h
f (x − h) − f (x) u=−h f (x + u) − f (x)
− lim
= h→0⇒u→0 = − lim
= f 0 (x),
u→0
h→0
h
−u
|
{z
}
− f 0 (−x) = − lim
h→0
→−f 0 (x)
2
and we have shown that f 0 is odd.
3. (a) (i)
π/2
Z
u=cos x du=− sin x dx
x=0⇒u=1
x=π/2⇒u=0
3
sin x cos x dx =
0
1
0
u4 1
1
u du = − = 0 − (− ) = .
4
4
4
Z
R
0
Z
3
=−
1
(ii)
Z
∞
xe−x/3 dx = lim
R→∞
0
xe−x/3 dx.
0
But
↑
R
Z
0
z }| {
R Z
−x/3
−x/3 x
e
dx
=
x
·
(−3e
)
−
|{z}
0
R
−3e−x/3 dx =
0
↓
R
R
− 3xe−x/3 − 9e−x/3 = −3Re−R/3 − 9(e−R/3 − 1) → 9,
0
Hence
as R → ∞.
0
Z
∞
xe−x/3 dx = 9.
0
(iii)
Z
h
i
1
2
,u≥0 =
√
dx = x+2=u
dx=2udu
x− x+2
Z
2u du
=
u2 − 2 − u
Z
2u du
(u − 2)(u + 1)
Using a partial fraction decomposition we get
2u
A
B
A(u + 1) + B(u − 2)
(A + B)u + A − 2B
=
+
=
=
.
(u − 2)(u + 1)
u−2 u+1
(u − 2)(u + 1)
(u − 2)(u + 1)
Equating coefficients gives the system
A+B =2
A − 2B = 0,
and solving this gives A = 4/3, B = 2/3, so
Z
Z
Z
4
du
2
du
2u du
=
+
=
(u − 2)(u + 1)
3
u−2 3
u+1
√
√
4
2
4
2
ln |u − 2| + ln |u + 1| + C = ln | x + 2 − 2| + ln( x + 2 + 1) + C.
3
3
3
3
(b) (i) For a > 0,
Z y=x
Z x
Z y=x
y=a tan t
dy
(a/ cos2 t)dt
2
dy=(a/
cos
t)dt
p
,
=
=
sec t dt =
=
√
a/ cos t
(a/ cos t)= y 2 +a2
y 2 + a2
0
y=0
y=0
y=x
x
1
1p
y 2
2
ln + tan t = ln y + a + =
cos t
a
a y=0
0
r
!
r
x2
2
√
x
0 x
x
+ 1 + − ln 0 + 1 + = ln
+1+
= g(x).
ln a2
a
a
a2
a
3
(ii)
x
x
⇔ = sinh y.
a
a
Implicit differentiation with resprct to x of the right hand equation gives
y = sinh−1
1
dy
= cosh y ,
a
dx
so
1
dy
=
.
dx
a cosh y
Using the identity
cosh2 y − sinh2 y = 1,
we get
q
cosh y = ± 1 + sinh2 y,
but using that cosh y = (ex + e−x )/2 > 0 and sinh y = x/a, a > 0, we get
r
x2
1p 2
a + x2 ,
cosh y = 1 + 2 =
a
a
and
f 0 (x) =
dy
1
1
=
=√
.
2
dx
a cosh y
a + x2
(c) From the equations
x
Z
0
dy
p
y2
+ a2
and
f 0 (x) = √
= g(x)
1
,
a2 + x2
the fundamental theorem gives us that
g 0 (x) = √
1
= f 0 (x).
+ a2
x2
Since f and g have the same derivative, they can differ only by a constant, so
f (x) = g(x) + C,
(1)
for some constant C. We can also evaluate g(0) = . . . = 0 and since sinh 0 = 0
we get f (0) = sinh−1 0 = 0 which, substituted into (1) gives us
0=0+C
, so C = 0 and we have
f (x) = g(x),
or in other words
−1
sinh
x
= ln
a
r
4
x2
x
+1+
a2
a
!
,
or simply (with a = 1)
sinh−1 x = ln
p
x2 + 1 + x ,
4. We are looking for the arc length L of the curve y = cosh x for −1 ≤ x ≤ 1.
This is given by
Z
s
1
1+
L=
−1
Z
dy
dx
2
Z
1
dx =
1
Z
p
2
1 + sinh x dx =
−1
Z
Z
cosh x dx = 2
−1
−1
p
cosh2 x dx =
−1
1
| cosh x| dx =
1
0
1
1
cosh x dx = sinh x =
0
sinh 1 − sinh 0 = sinh 1.
In the above calculation we used that cosh x =p(ex + e−x )/2 is an even function,
√
and sinh 0 = (e0 − e−0 )/2 = 0. Also sinh 1 = cosh2 1 − 1 = a2 − 1, so
p
L = a2 − 1.
5. From day 0 to day 10 the difference in concentration of nitrogen dioxide is
given by
Z 1
dN
N (10) − N (0) =
0
dt,
dt
0
by the fundamental theorem. We get
Z
0
10
dN
dt =
dt
Z
10
Z
10
kP dt =
0
10−9 k(1 − e−t/20 ) dt =
0
10
10−9 k(t+20e−t/20 ) = 10−9 ·0.05(10+20e−1/2 −20) ≈ 1.0653·10−11
(kg/m3 ).
0
In 1011 m3 of air, this corresponds to an increase of ≈ 1 kg.
6. (i) From the figure we see that
|XZ| =
p
A2 + s2
and
|ZY | =
p
B 2 + (D − s)2 .
(ii) Since time equals distance/velocity, we get that the time t1 required to
travel the distance |XZ| through the air is
√
A2 + s2
t1 =
,
u
and the time t2 required to travel the distance |ZY | through the glass is
p
B 2 + (D − s)2
t2 =
v
5
and the total time t = t1 + t2 is
p
√
B 2 + (D − s)2
A2 + s2
t=
+
.
u
v
(iii) Since Fermat’s principle states that the path taken is the one that minimizes t, we want to minimize this expression. Since t depends on s we calculate
dt/ds and find when that derivative is zero. We have
dt
1
1
2s + p
2(D − s)(−1) =
= √
ds
2u A2 + s2
2v B 2 + (D − s)2
D−s
s
√
− p
.
2
2
2
u A +s
v B − (D − s)2
The critical points occur when dt/ds = 0 which gives us
s
D−s
= p
2
2
2
u A +s
v B − (D − s)2
√
which is the desired equation.
To verify that this is really a minimum becomes slightly messy in this case,
so I guess that was not part of the problem.
Finally, rearranging the equation above, we have
p
s B 2 − (D − s)2
u
√
=
,
v
(D − s) A2 + s2
and looking at the figure we see that
sin a =
sin b =
so
s
s
=√
,
|XZ|
A2 + s2
D−s
D−s
=p
,
2
|ZY |
B + (D − s)2
s
√
sin a
A2 +s2
=
D−s
√
sin b
2
B +(D−s)2
p
s B 2 − (D − s)2
u
√
= .
=
2
2
v
(D − s) A + s
6