1. Calculate the following. Be sure to show all of your work.
x3 − 8
x→2 x − 2
(a) lim
x3 − 8
(x − 2)(x2 + 2x + 4)
= lim
= lim (x2 +2x+4) = 22 +2(2)+4 = 12
x→2 x − 2
x→2
x→2
x−2
lim
sin (x2 )
x→0
πx
(b) lim
sin(x2 )
sin(x2 ) x2
sin(x2 )
x2
= lim
·
=
lim
·
lim
= 1·0 = 0
x→0
x→0
x→0 πx
πx
x2
πx x→0 x2
lim
√
(c)
lim
x→+∞
4x2 + 1
x+3
√
lim
x→+∞
√
4x2 + 1
4
=
=2
x+3
1
(d) lim f (x),
x→0
where f (x) =
The limit does not exist.
1
1, x rational
0, x irrational
2. Calculate the following. Be sure to show all of your work.
Z
sin x
x2 tdt. Find F 0 (x).
(a) F (x) =
0
2
sin x
Z
F (x) = x ·
tdt
0
sin x
Z
0
tdt + x2 · (sin x · cos x)
F (x) = 2x ·
0
Z
x
(cos t − 1) dt
0
(b) lim
x
Z
x→0
x
(cos t − 1)dt. Then F (0) = 0. Furthermore,
Let F (x) =
0
Z
x
(cos t − 1)dt
0
lim
x→0
= lim
x
x→0
F (x) − F (0)
= F 0 (0)
x−0
On the other hand, F 0 (x) = cos x − 1. Therefore
Z x
(cos t − 1)dt
0
lim
=0
x→0
x
d2
(c) 2
dt
t2
1 + t2
2 d
t
2t · (1 + t2 ) − t2 · (2t)
2t
=
=
2
2
2
dt 1 + t
(1 + t )
(1 + t2 )2
d2
t2
d
2t
=
2
dt 1 + t2
dt (1 + t2 )2
2 · (1 + t2 )2 − 2t · 2(1 + t2 ) · (2t)
=
(1 + t2 )4
2 − 6t2
=
(1 + t2 )3
2
π/2
Z
sin x cos3 xdx
(d)
0
Let u = cos x. Then du = − sin xdx. When x = 0, u = 1.
When x = π/2, u = 0. Thus
Z
π/2
3
Z
3
Z
u · (−1)du =
sin x cos xdx =
0
0
0
1
3
1
u3 · du =
u4 1
1
=
0
4
4
3. A population of frogs in a swamp is found to grow at a rate proportional to the square root of the population size. The initial population
P is 400 frogs, and 5 years later there are 900 of them. Write the differential equation for the frog population P (t) with the two corresponding
conditions, and find the particular solution which incorporates both
conditions.
p
d
P (t) = k · P (t)
dt
where k the proportional constant.
Solve the differential equation by the separation of variable method:
dP (t)
p
= kdt
P (t)
then find the antiderivative on both sides (treat P (t) as another varidP (t)
dy
able y, so p
=√ )
y
P (t)
p
2 P (t) = kt + C
Therefore P (t) =
kt + C
2
2
.
From the two conditions: P (0) = 400 and P (5) = 900, we have
C
2
2
= 400 and
5k + C
2
2
= 900
Therefore C = 40 and k = 4. Finally,
P (t) =
4t + 40
2
4
2
= (2t + 20)2
4. Find the dimensions of the rectangle with maximum area inscribed in
a circle of unit radius. Be sure to show all your work, and verify your
result with the second derivative test.
Let x be the width of the rectangle and y be the length. Since it is
inscribed in a unit circle, we have x2 + y 2 = d2 , where
√ d denotes the
diameter of the circle. In this case, d = 2. Thus, y = 4 − x2 .
Let A denote the area of the rectangel. Then
p
A = x · y = x · 4 − x2
Thus A is a function of x. By taking derivatives, we have
p
p
d
1
A(x) = 4 − x2 +x· (4−x2 )−1/2 ·(−2x) = 4 − x2 −x2 (4−x2 )−1/2
dx
2
and
d2
1
1
A(x) = (4 − x2 )−1/2 (−2x) − 2x(4 − x2 )−1/2 − x2 (− )(4 − x2 )−3/2 (−2x)
2
dx
2
2
= −3x(4 − x2 )−1/2 − x3 (4 − x2 )−3/2
√
d
d2
A(x) = 0. Moreover,
A(x) < 0 for all x > 0.
dx
dx2
Thus at x = 0, A(x) achieve its maximum value 2.
When x =
2,
5
1
. Indicate if f has any
1 + x2
symmetry, and find where f is increasing, decreasing, concave up and
concave down. Identify all local maxima, local minima, and inflection points. Does f have a global maximum? Does f have a global
mimimum?
5. Give a sophisticated graph of f (x) =
1
1
1. Since f (−x) =
=
= f (x), we see that f (x) is
1 + (−x)2
1 + x2
even.
−2x
2. f 0 (x) =
. For x > 0, f 0 (x) < 0 and f (x) is descreasing;
(1 + x2 )2
for x < 0, f 0 (x) > 0 and f (x) is increasing;
√
√
6x4 + 4x2 − 2
3
3 00
00
3. f (x) =
. For −
< x <
, f (x) < 0 and
2
4
(1 + x )
3√
3
√
3
3 00
f (x) is concave down; for x < −
or x >
, f (x) > 0 and
3
3
f (x) is concave up.
√
√
√
3 00
3 3
3 3
, f (x) = 0. Thus −
,
and
,
are
4. When x = ±
3
3 4
3 4
inflextion points.
5. At x = 0, f (x) achieves its local maximum value 1. There is no
local minimum value.
6. f (x) achieves its global maximu at x = 0. f (x) does not have a
global minimum value.
6. For a given spring, the force required to keep it stretched x feet is
given by F = 9x pounds. How much work in ft·lb is done in stretching
the spring 1 foot from its unstretched equilibrium state? Be sure to
show all your work.
The work needed is given by
Z 1
Z
W =
F (x)dx =
0
0
6
1
9xdx = 9
9
x2 1
=
0
2
2
7. Consider the region under the graph of y(x) = sin x, over the x−axis
between x = 0 and x = π. Revolve this region around the x−axis, and
find the volume of the resulting solid.
It’s better to use the method of slabs in this case:
Z π
Z π
1 − cos(2x)
2
π(sin x) dx = π
V =
dx
2
0
0
sin(2x) π
π
x−
=
0
2
2
π
sin(2π)
=
π−
−0
2
2
π2
=
2
8. Find the area of the finite region bounded by the graphs of y = −x + 2
and y = x2 . Sketch the region.
By solving −x+2 = x2 , we see that the parabola and the line intersect
at (−2, 4) and (1, 1). Therefore the area is given by
Z
1
Z
1
[(−x + 2) − x ]dx = −
(x2 + x − 2)dx
−2
−2
3
x
x2
1
=−
+
− 2x −2
3
2
3
1
12
(−2)3 (−2)2
=−
+
− 2(1) −
+
− 2(−2)
3
2
3
2
9
=
2
A=
2
7
9. Consider a circle of radius 2 centered at the origin, parameterized by
(x(t), y(t)) = (2 cos t, 2 sin t). Find the circumference of this circle by
calculating the arc length of the curve parameterized by (x(t), y(t))
with t ∈ [0, 2π].
x0 (t) = −2 sin t and y 0 (t) = 2 cos t. Thus
Z
2π
L=
Z
p
0
2
0
2
[x (t)] + [y (t)] dt =
p
(−2 sin t)2 + (2 cos t)2 dt
0
0
Z
2π
2π
2dt
=
0
= 4π
10. Use the Theorem of Pappus to find the volume of the doughnut, or
torus, obtained when the region bounded by the circle of radius 1, centered at the point (3, 0), is revolved around the y−axis. The equation
of the circle is (x − 3)2 + y 2 = 1. (Hint: where is the centroid of the
circular region?)
Since the region is bounded by a circle, its centroid is the center of
the circle. As the region is revolving around the y-axis, the trace of
the centroid traveled is a circle with radius 3. By Pappus’ Theorem
we have
V = C · R = (2π · 3) · (π · 12 ) = 6π 2
where C is the length of the trace traveled by the centroid and R is
the area of the region.
8