1. No category

Math 113/114–Solutions to Final Exam Sample Problems
1. Find y 0 .
(a) y 0 = (−1 − 3x2 ) sec2 (x2 − 1) + (1 − x − x3 ) · 2 sec2 (x2 − 1) tan(x2 − 1)(2x).
1
3 sec2 3x
(b) ln y = ln x ln(tan(3x)) =⇒ y1 y 0 = ln(tan(3x)) + ln x ·
x
tan 3x
3 ln x sec2 3x
0
ln x ln(tan 3x)
∴ y = (tan 3x)
+
x
tan 3x
√
√
1
(c) y 0 = 3(tan2 x2 − sin x)(sec2 x2 − sin x) · 21 (x2 − sin x)− 2 (2x − cos x)
(d) ln y = 3 ln(sin x) + 12 ln(3x5 + 1) − x3 ln e − 5 ln(2 − 5x2 )
1
3 cos x 1 15x4
5(−10x)
=⇒ y 0 =
+
− 3x2 −
y
sin√x
2 (3x5 + 1)
2 − 5x2
3
4
5+1
sin
x
5x
15x
50x
2
∴ y 0 = x3
3
cot
x
+
−
3x
+
2(3x5 + 1)
2 − 5x2
e (2 − 5x2 )5
√
√
√
1
(e) y 0 = [sec2 (x2 − sin 1 − x2 )4 ](4(x2 − sin 1 − x3 )3 ){2x − (cos 1 − x3 ) 12 (1 − x3 )− 2 (−3x2 )}
(f) Diff. implicitly, [cos(x − y 2 )](1 − 2yy 0 ) = 2x − y + xy 0 =⇒
2x − y − cos(x − y 2 )
(−2y cos(x − y 2 ) + x)y 0 = 2x − y − cos(x − y 2 ) =⇒ y 0 =
x − 2y cos(x − y 2 )
(g) We use logarithmic differentiation: taking natural log, we get ln y = (tan x) ln(sec x)
1
1
sec x tan x =⇒ y 0 = (sec x)tan x [(sec2 x) ln(sec x) + tan2 x]
=⇒ y 0 = (sec2 x) ln(sec x) + tan x
y
sec x
(h) Diff. implicitly. exy (y + xy 0 ) − 2yy 0 =
yexy + xexy y 0 − 2yy 0 =
1
x
1
x
=⇒
=⇒ (xexy − 2y)y 0 =
1
x
− yexy =⇒ y 0 =
1
x
− yexy
xexy − 2y
2. Evaluate the limits.
(a)
lim (x −
x→∞
p
√
√
x2 + x + 1)(x + x2 + x + 1)
√
x→∞
x + x2 + x + 1
2
2
x(−1 − x1 )
x −x −x−1
√
q
= lim
= lim
x→∞
x→∞ x + x2 + x + 1
x + x2 (1 + 1 +
x2 + x + 1) = lim
(x −
x
x(−1 − x1 )
q
x→∞
x 1 + 1 + x1 +
= lim
1
x2
−1
1
= 1 + 1 = −2
1
x2 )
√
( x2 = x, x > 0).
(b)
(1 − cos x)(1 + cos x)
1 − cos2 x
1 − cos x
= lim
= lim
x→0 x sin 2x(1 + cos x)
x→0 x sin 2x(1 + cos x)
x→0 x sin 2x
lim
1 sin x
1
1
1
1
1
sin2 x
= lim ·
·
·
= ·1·1· = .
x→0 2
x→0 x2 sin x cos x(1 + cos x)
x
cos x 1 + cos x
2
2
4
= lim
(c)
lim
x→0
sin 3x
sin 3x 3x
2x
3
3
sin 3x
= lim
cos 2x = lim
·
·
· cos 2x = 1 · · 1 · 1 = .
x→0 3x
tan 2x x→0 sin 2x
2x sin 2x
2
2
1
2
(d)
lim
√
√
√
2x − x
2x − x)( 2x + x)
2x − x2
√
√
= lim
= lim
x→2 (x − 2)( 2x + x)
x→2 (x − 2)( 2x + x)
x−2
x(2 − x)
1
x
2
√
√
= lim
=− .
= lim
=
x→2 −(2 − x)( 2x + x)
x→2 −( 2x + x)
−(2 + 2)
2
x→2
(e)
lim √
x→−∞
(since
√
2x
2x
= lim q
2
x + 1 x→−∞ x2 (1 +
1
x2 )
2x
q
x→−∞
−x 1 +
= lim
x2 = −x, x < 0).
1
x2
=
2
= −2,
−1
(f)
x3 ( x13 − x42 + 1)
x( x13 − x42 + 1)
1 − 4x + x3
=
lim
=
lim
= −∞,
x→−∞ x2 (3 − 1 + 42 )
x→−∞ 3x2 − x + 4
x→−∞
3 − x1 + x42
x
x
lim
since 1/xr → 0 as x → −∞.
3. Evaluate the integrals.
(a)
1 − x2 − 3x2
Z
x
dx =
2
3
Z
4
1
3 7
1
12 7
2
(x− 3 − 4x 3 ) dx = 3x 3 − 4 · x 3 + C = 3x 3 − x 3 + C.
7
7
(b)
Z
1
x sin(1 − 2x ) dx = −
4
Z
2
sin u du =
1
1
cos u + C = cos(1 − 2x2 ) + C
4
4
(c) Let u = 1 + cos 2x, du = −2 sin 2xdx. x = 0 =⇒ u = 2, x =
Z
π
3
0
sin 2x
1
√
dx = −
2
1 + cos 2x
Z
1
2
2
1
du
√ =
u
2
Z
2
u
− 21
1
2
(d)
Z
2
0
|3x − 1| dx =
Z
1
3
0
−(3x − 1) dx +
Z
2
1
3
π
3
(u = 1 − 2x2 , du = −4xdx).
=⇒ u = 12 .
2
√
1
1
1
2−1
1
du = · 2u 2 = 2 − √ = √ = √ .
2
1
2
2
2
2
1 2
2
3x2 3
3x
(3x − 1) dx = x −
+
− x
2 0
2
1
3
1 3 1 3
3 1 1
13
= − · + ·4−2− · + =
.
3 2 9 2
2 9 3
3
(e)
Z
1
0
2
|2x + x − 1| dx =
Z
1
0
|(2x − 1)(x + 1)| dx = −
Z
0
1
2
2
(2x + x − 1) dx +
Z
1
1
2
(2x2 + x − 1) dx
21 3
1
3
x
x
x2
x2
=− 2 +
− x + 2 +
− x
3
2
3
2
1
0
2
1 1
1
1 1
1
7
3
1
2
7
+ −
+ 12 − 1 −
+ −
+ +
= .
=−
+
=
12 8 2
3
12 8 2
24 6 24
4
3
(f) Let u = 2 + cos2 θ, du = −2 cos θ sin θ dθ.
Z
Z
1 u3
1
1
u2 du = − ·
sin θ cos θ(2 + cos2 θ)2 dθ = −
+ C = − (2 + cos2 θ)3 + C.
2
2 3
6
(g) Let u = tan 7x, du = 7 sec2 7x dx. Then
Z π4 tan 7x
Z π4
e
dx
=
sec2 7x etan 7x dx
cos2 7x
0
0
Z
π
7π
1 −1 u
e du
x = 0 =⇒ u = tan 0 = 0; x =
=⇒ u = tan
= −1
=
7 0
4
4
0
Z 0
1 1
1
1
1
eu du = − eu = − (e0 − e−1 ) = −
1−
.
=−
7
7
7
7
e
−1
(h) Note that f (x) =
−1
−x3 sin4 x
x3 sin4 x
is
an
odd
function
since
f
(−x)
=
= −f (x). So
1 + x4
1 + x4
Z π2 3 4
x sin x
dx = 0.
1 + x4
−π
2
(i) Let u = x3 + 1, du = 3x2 dx.
Z
Z
Z
p
p
√
1
3
2
5
3
3
(u − 1) u du
x x + 1 dx = x x + 1 x dx =
3
Z
5
3
1
3
1 2 5
2 3
2
1
2 3
2
2
2
2
u − u +C =
(x + 1) 2 − (x3 + 1) 2 + C.
=
(u − u ) du =
3
3 5
3
15
9
(j) Let u = 2 − x, du = −dx.
Z
Z
Z
√
√
3
1
x 2 − x dx = − (2 − u) u du = − (2u 2 − u 2 ) du
4 3
3
5
2 5
4
2
= − u 2 − u 2 + C = − (2 − x) 2 + (2 − x) 2 + C.
3
5
3
5
(k) Let u = 3 + 7 tan 5x. So du = 35 sec2 5x dx. Then
Z
Z
sec2 5x
1
du
1
1
dx =
=
ln |u| + C =
ln |3 + 7 tan 5x| + C.
3 + 7 tan 5x
35
u
35
35
(l)
Z
(1 + ecos x ) sin x dx =
Z
(sin x + ecos x sin x) dx = − cos x −
Z
eu du (u = cos x, du = − sin x dx)
= − cos x − eu + C = − cos x − ecos x + C.
(m) First let u = π ln x, du = πx dx. Then
Z e
Z
1 π
cos2 (π ln x) sin(π ln x)
cos2 u sin u du
dx =
x
π 0
1
Z
1 −1 2
=−
t dt (t = cos u, dt = − sin u du and cos 0 = 1, cos π = −1)
π 1
1
Z
Z
2
2 1 2
2 t3 2
1 1 2
=
t dt =
t dt =
−0=
.
=
π −1
π 0
π 3 0
3π
3π
4
4.
(a) f (x) =
Z
2
tan x
p
1 + t4 dt.
Z tan x p
Z up
d
d
du
1 + t4 dt = −
1 + t4 dt ·
dx 2
du 2
dx
tan x
p
p
F.T.C.
2
2
4
4
= − 1 + u sec x = − 1 + tan x sec x.
f 0 (x) =
(b) f (x) =
Z
d
dx
0
2
Z
p
1 + t4 dt = −
cos t2 dt.
cot2 (1−tan x)
f 0 (x) =
d
dx
Z
0
cot2 (1−tan x)
d
=−
du
F.T.C.
=
Z
u
0
cos t2 dt = −
du
cos t2 dt ·
dx
d
dx
Z
cot2 (1−tan x)
cos t2 dt
0
(where u = cot2 (1 − tan x)
−(cos u2 )2(cot(1 − tan x))(− csc2 (1 − tan x))(− sec2 x)
= −2 cos(cot4 (1 − tan x)) cot(1 − tan x)(csc2 (1 − tan x)) sec2 x.
5. We first graph the function and the region.
y
y=x 2−2x
1
x
−1
A=
Z
1
−1
2
|x − 2x| dx =
Z
0
−1
2
(x − 2x) dx −
Z
1
0
−1
1
1
1
=−
+ 1 − + 1 = + 1 − + 1 = 2.
3
3
3
3
0
1
3
x3
x
2
2
(x − 2x) dx = −
−x −
−x 3
3
−1
0
2
6. We wish to maximize the volume of the cylinder, i.e., V = πr 2 h.
r
h
20−h
20 cm
r
h
5 cm
5
From the diagram we have
20
20 − h
=
=⇒ 20 − h = 4r =⇒ h = 20 − 4r.
r
5
5
Then
V = πr2 (20 − 4r) = 4πr 2 (5 − r) = 4π(5r 2 − r3 ),
0≤r≤5
Differentiating,
dV
= 4π(10r − 3r 2 ) = 4πr(10 − 3r).
dr
So the critical points are r = 0, 10/3. We will use the Extreme Value Theorem to verify:
2 10
10
100 5
10
2000π
= 4π
5−
= 4π ·
V (0) = 0, V (5) = 0, V
· =
.
3
3
3
9 3
27
So the maximum volume of the cylinder is
2000π
cm3 .
27
7.
y
y=16−x2
y
x
x
We wish to maximize the area A = 2xy. Then
dA
= 2(16 − 3x2 ).
dx
√
To find the critical points let dA/dx = 0 so that x2 = 16/3 =⇒ x = 4/ 3. Let’s use the first derivative
test to verify:
A = 2xy = 2x(16 − x2 ) = 2(16x − x3 ) =⇒
4
x < √ =⇒
3
4
x > √ =⇒
3
√
So√the area is a maximum at x = 4/ 3. Then
8/ 3 by 32/3.
dA
> 0 =⇒ A increasing
dx
dA
< 0 =⇒ A decreasing.
dx
y = 16 − (16/3) = 32/3. The dimensions of the rectangle are
√
8. We wish to maximize the area A = 4xy = 4x r2 − x2 .
y
x 2 +y2 =r2
2x
2y
x
p
dA
4x 2
1
4(r2 − 2x2 )
1
= 4 r 2 − x2 +
(r − x2 )− 2 (−2x) = 4(r2 − x2 )− 2 (r2 − x2 − x2 ) = √
.
dx
2
r 2 − x2
√
√
Setting dA/dx = 0,
√ we get x = r/ 2. Let’s use the first derivative test to verify:
√ if 0 < x < r/ 2, then
dA/dx > 0√and if r/ 2 < x, then dA/dx < 0. √So A is increasing on the left r/ 2 and decreasing on the
right of r/ 2, so A is a maximum when x = r/ 2.
6
Then y =
q
r2 −
r2
2
=
q
r2
2
=
√r .
2
√
√
So it is a square with 2r/ 2 by 2r/ 2.
9. The statement of the Mean Value Theorem: If a function f is continuous on [a, b] and differentiable on
(a, b), then there is a number c ∈ (a, b) such that f 0 (c) = [f (b) − f (a)]/(b − a), a 6= b.
Then since 1 ≤ f 0 (x) ≤ 4 for all x ∈ (2, 5), f is differentiable on (2, 5), then f is continuous on [2, 5]. Then
(2)
. But 1 ≤ f 0 (c) ≤ 4.
by the Mean Value Theorem there exists a number c ∈ (2, 5) such that f 0 (c) = f (5)−f
3
(2)
So 1 ≤ f (5)−f
≤ 4 =⇒ 3 ≤ f (5) − f (2) ≤ 12.
3
10. Step 1: Let f (x) = 3x + 2 cos x + 5. The function is continuous everywhere since it is the sum of
everywhere-continuous polynomial and sine functions. We note that
f (−π) = 3(−π) + 2 cos(−π) + 5 = −3π − 2 + 5 = −3π + 3 < 0,
f (0) = 0 + 2 cos 0 + 5 = 7 > 0,
so that f (−π) < 0 < f (0). Then by the Intermediate Value Theorem, there is a number c ∈ (−π, 0) such
that f (c) = 0, i.e., there is a root x = c of the equation.
We now show that there is exactly one real root. Assume that there are two roots, a and b such that
a < b. Then f (a) = 0 = f (b). By Step 1, f is continuous on [a, b]. f 0 (x) = 3 − 2 cos x so f is differentiable
on (a, b). Then by Rolle’s Theorem, there is a number d ∈ (a, b) such that f 0 (d) = 0, i.e. 3 − 2 cos d = 0. But
this equation has no solution since −1 ≤ cos d ≤ 1 and hence f 0 (d) > 0. This is a contradiction to Rolle’s
Thm. Therefore, there cannot be two roots of the given equation.
11. f (x) = 4 cos x on [0, π2 ]. Using x∗i = π6 , π4 , π3 , π2 , Note that the width ∆x is not the same for all the
π
π
subintervals. ∆x1 = π6 , ∆x2 = 12
, ∆x3 = 12
, and ∆x4 = π6 .
y=4cos x
π π π π
6 4 3 2
So
ππ
π π π π ππ
+f
+f
6
6
12 √
3
12
2
6
√4
π
π
π π
π π π π 2π
3 π
2 π1
=
4 cos
+
4 cos
+
4 cos
+
4 cos
=
·
+ ·
+
+0
6
6
12
4
12
3
6
2
3
2
3 2
32
√
√
√
√ !
2
2 3
π
1+2 3+ 2
π.
=
π+
π+ =
6
6
6
6
A ≈ f (x∗1 )∆x1 + f (x∗2 )∆x2 + f (x∗3 )∆x3 + f (x∗4 )∆x4 = f
12. f (x) = 2 + x2 on [1, 3]. x∗i = xi = 1 + i∆x = 1 +
y
i2
n
where ∆x =
f(x)=2+x2
2
1
+f
3
x
2
n.
7
Z
3
1
2 n
n
n X
i2 2
i2
2X
2X
∗
f 1+
f (xi )∆x = lim
2+ 1+
= lim
(2 + x ) dx = lim
n→∞
n→∞ n
n n n→∞ n i=1
n
i=1
i=1
n
n
n
n
2 X
4i 4i2
4 X 2
2X
4X
= lim
3+
i+ 2
+ 2 = lim
3+
i
n→∞ n
n→∞ n
n
n
n i=1
n i=1
i=1
i=1
4 n(n + 1)
4 n(n + 1)(2n + 1)
2
3n + ·
+ 2·
= lim
n→∞ n
n
2
n
6
4(n + 1) 4(n + 1)(2n + 1)
= lim 6 +
+
n→∞
n
3n · n
1
4
1
1
4
38
= lim 6 + 4 1 +
+
1+
2+
= 6 + 4 + (2) =
.
n→∞
n
3
n
n
3
3
2
13.
(a) y =
x
.
(x − 1)2
(x − 1)[x − 1 − 2x]
−1 − x
−(x + 1)
(x − 1)2 − x · 2(x − 1)
=
=
=
(x − 1)4
(x − 1)4
(x − 1)3
(x − 1)3
−(x − 1)3 + (x + 1) · 3(x − 1)2
(x − 1)2 [−(x − 1) + 3(x + 1)]
y 00 =
=
6
(x − 1)
(x − 1)6
−x + 1 + 3x + 3
2x + 4
2(x + 2)
=
=
=
.
(x − 1)4
(x − 1)4
(x − 1)4
y0 =
√
(b) y = x 9 − x2 .
9 − 2x2
x
1
1
9 − x2 + (9 − x2 )− 2 (−2x) = (9 − x2 )− 2 (9 − x2 − x2 ) = √
,
2
9 − x2
√
1
−4x 9 − x2 − (9 − 2x2 ) 12 (9 − x2 )− 2 (−2x)
y 00 =
9 − x2
1
−x(36 − 4x2 − 9 + 2x2 )
x(2x2 − 27)
−x(9 − x2 )− 2 [4(9 − x2 ) − (9 − 2x2 )]
=
=
.
=
3
3
2
9−x
(9 − x2 ) 2
(9 − x2 ) 2
y0 =
p
√
14. y = x 9 − x2 . So the domain is [−3, 3] and the function is continuous there. So by the Extreme Value
Thm, there is an absolute max. and absolute min. on [−3, 3]. Using the result of #12(b), the critical
points are x = ± √32 , ±3.
At x = −3,
at x = 3,
y = 0,
y = 0,
r
3
9
9
9
= ,
9− = √
2
2
2 2
r
3
3
9
9
at x = − √ , y = − √
9− =− .
2
2
2
2
3
at x = √ ,
2
3
y=√
2
r
So the absolute min. value is − 92 at x = − √32 and absolute max. value is
9
2
at x =
√3 .
2
8
15. The domain of f is x 6= 1 or (−∞, 1) ∪ (1, ∞). Using the results of 12(a), the critical point is x = −1.
−(x + 1) (x − 1)3
(−∞, −1)
(−1, 1)
+
−
(1, ∞)
−
−
−
+
y0
y
−
+
decreasing
increasing
−
decreasing
So y is decreasing on (−∞, −1) ∪ (1, ∞) and increasing on (−1, 1). A local min. at x = −1, y = −1/4.
No local maximum. From y 00 above possible inflection point is at x = −2.
(−∞, −2)
(−2, 1)
x + 2 (x − 1)4
(1, ∞)
y 00
y
−
+
+
+
−
+
conc. down
conc. up
+
+
+
conc. up.
Hence, the inflection points occur at x = −2, y = −2/9.
We now look for asymptotes.
lim
x→±∞
x
1
0
x
x2
x
=
lim
= 0.
=
lim
=
x−1
1
2
2
2
x→±∞
x→±∞
(x − 1)
(1 − 0)2
( x )
(1 − x )
So y = 0 is a horizontal asymptote.
x
= ∞ (since x → 1 and (x − 1)2 → 0 pos.)
(x − 1)2
x
= ∞ (since x → 1 and (x − 1)2 → 0 pos.).
lim+
x→1 (x − 1)2
lim
x→1−
So x = 1 is a vertical asymptote.
y
−2
−1
x
0
(−2,−2/9)
1
(−1,−1/4)
x=1
16. See the figure below. Then we are given that dV /dt = 2 cm3 /min, dr/dt = 1 cm/min. We wish to find
dh/dt when r = 5 and V = 60.
r
h
dV
dr
2 dh
V = πr h
=⇒
= π 2rh + r
dt
dt
dt
dV
dr
dh
1 dV
dr
dh
2
=
− 2πrh
=⇒
= 2
− 2πrh
=⇒
πr
dt
dt
dt
dt
r π dt
dt
2
9
Now when r = 5 and V = 60, h = 12/5π, then
2 − 24
1
22
dh
12
2 − 2π(5)
(1) =
=
=−
.
dt
25π
5π
25π
25π
So the height is decreasing at a rate of
22
25π
cm/min.
17. See the figure below. Let x be the length of the shadow cast by the building and θ the angle of elevation
of the sun.
400
θ
x
We are given
dθ
dt
= −0.25 rad/h and wish to find
dx
dt
when θ = π/6.
dθ
400 dx
400
=⇒ sec2 θ
=− 2
x
dt
x dt
dx
x2
dθ
∴
=−
sec2 θ .
dt
400
dt
tan θ =
We will now find x when θ = π/6,
tan
√
400
400
400
π
=
=⇒ x =
π = √1 = 400 3.
6
x
tan 6
3
Then
4002 · 3
dx
π
=−
sec2 (−0.25) = 400(3)
dt
400
6
2
√
3
2 1
= 400 ft/h.
4