calculo diferencial

1. UNIVERSIDAD DEL QUINDÍO
FACULTY OF BASIC SCIENCES AND TECHNOLOGIES
PHYSICS PROGRAM
DIFFERENTIAL CALCULUS COURSE
Juan A. Álvarez A. - jaalvareza@uqvirtual.edu.co
Part 1. Limits and Their Properties
Exercise. 5) Find the limit L. Then use ε − δ denition to prove that the limit is L.
lim
x→1
(x + 4) = 5
For all ε 0, exists δ 0, such that, |x − 1| δ, implies that |x + 4 − 5| ε
|x − 1| δ ⇒ |x + 4 − 5| ε
⇒ |x − 1| ε
ε = δ
Exercise. 16) Find the limit (if it exists)
lim
t→3
t2
− 9
t − 3
= lim
t→3
$$$$(t − 3)(t + 3)
$$$$(t − 3)
= lim
t→3
t + 3 = 6
Exercise. 36) Find the limit (if it exists). If the limit doesn't exist, explain why.
lim
x→4
x − 1
Remark. By the nature of the greatest integer function, we calculate the limit by their lateral limits.
lim
x→4−
x − 1 = 2 ; lim
x→4+
x − 1 = 3
The function is not continuous since its lateral limits are not equal, therefore there isn t limit.
Exercise. 44) Determine the intervals on which the function is continuous.
f (x) =
3x2
− x − 2
x − 1
Intervals (−∞, 1) ∪ (1, ∞)
Remark. The discontinuity is removable, since g (x) = 3x2
+ 2 , is a function that has the same behavior that f (x), except at
x = 1.
1

2. UNIVERSIDAD DEL QUINDÍO 2
Exercise. 72) Find the one-sided limit (if it exists).
lim
x→0+
sec x
x
= lim
x→0+
1
x cos x
+∞
Part 2. Dierentiation
Exercise. 33) (Vibrating string) When a guitar string is plucked, it vibrates with a frequency of F = 200
√
T, where F is
measured in vibrations per second and the tension is measured in pounds. Find the rates of change of when (a) T = 4 and (b)
T = 9.
F´(T) =
200
2
(T)− 1
2 =
100
√
T
(a)
F´(4) =
100
√
4
= 50
(b)
F´(9) =
100
√
9
=
100
3
Exercise. 42) nd the derivative of the function g (x) = (x3
+ 7x)(x + 3).
g´ (x) = (3x2
+ 7)(x + 3) + (x3
+ 7x)
= 3x3
+ 9x2
+ 7x + 21 + x3
+ 7x
= 4x3
+ 9x2
+ 14x + 21
Exercise. 66) nd the second derivative of the function h (t) = 10 cos t − 15 sin t.
h´ (t) = −10 sin t − 15 cos t
= −5(2 sin t + 3 cos t)
h´´ (t) = −5(2 cos t − 3 sin t)
= 15 sin t − 10 cos t)
Exercise. 105) nd dy/dx by implicit dierentiation.

3. UNIVERSIDAD DEL QUINDÍO 3
√
xy = x − 4y
x
1
2 y
1
2 = x − 4y
1
2
x− 1
2 y
1
2 dx +
1
2
x
1
2 y− 1
2 dy = dx − 4dy
√
x
2
√
y
dy + 4dy = dx −
√
y
2
√
x
dx
dy
√
x + 8
√
y
2
√
y
= dx
2
√
x −
√
y
2
√
x
dy
dx
=
¡2
√
y(2
√
x −
√
y)
¡2
√
x(
√
x + 8
√
y)
dy
dx
=
2
√
xy − y
x + 8
√
xy
Exercise. 115) (Moving Shadow) A sandbag is dropped from a balloon at a height of 60 meters when the angle of elevation to
the sun is 30
°
(see gure). Find the rate at which the shadow of the sandbag is traveling along the ground when the sandbag is
at a height of 35 meters. Hint: The position of the sandbag is given by s (t) = 60 − 4.9t2
.
s (t) = 60 − 4.9t2
35 = 60 − 4.9t2
5.1 = t2
2.26s = t
Be x the position of the shadow of the sandbag.

4. UNIVERSIDAD DEL QUINDÍO 4
x =
s (t)
tan 30
x =
60 − 4.9t2
tan 30
dx
dt
=
(−9.8t)(tan 30) − 0
tan2
30
dx
dt
=
(−9.8)(2.26)(tan 30)
tan2
30
dx
dt
= −3(22.148)(tan 30)
dx
dt
= −38.36
m
s
The rate of change is negative, because the distance decreases.
Part 3. Applications of dierentation
Exercise. 14) Determine whether the Mean Value Theorem can be applied to on the closed interval [a, b]. If the Mean Value
Theorem can be applied, nd all values of in the open interval (a, b) such that f (c) = f(b)−f(a)
b−a . If the Mean Value Theorem
cannot be applied, explain why not.
f (x) =
1
x
, [4, 1]
f (x) = −
1
x2
f (c) = −
f (4) − f1
4 − 1
f (c) = −
1
4
−
1
4
= −
1
c2
c = ±2

5. UNIVERSIDAD DEL QUINDÍO 5
Exercise. 39) Use the Second Derivative Test to nd all relative extrema. g (x) = 2x2
1 − x2
.
g (x) = 2x2
− 2x4
g (x) = 4x − 8x3
4x − 8x3
= 0
4x 1 − 2x2
= 0
x = 0
x =
√
2
2
x = −
√
2
2
g (x) = 4 − 24x2
g (0) = 4 − 24(0)2
g (0) = 4 0
g
√
2
2
= 4 − 24
√
2
2
2
g
√
2
2
= −8 0
g −
√
2
2
= 4 − 24 −
√
2
2
2
g −
√
2
2
= −8 0
By the Second Derivative Test, in x = 0 a relative minimum occurs, whose value is 0; in x =
√
2
2 n' x = −
√
2
2 relative maxima
occur, whose value is given in both points, by
1
2 .
Exercise. 56) Find the limit.
lim
x→−∞
x
2 sin x
=
limx→−∞ x
limx→−∞ 2 sin x
=
limx→−∞
x
x
limx→−∞
2 sin x
x
=
limx→−∞ 1
limx→−∞ 0±
= +∞, −∞
= Limit doesn t exists
Exercise. 45) (Section 3.7, Projectile Range) The range R of a projectile red with an initial velocity Vo at an angle θ with the
horizontal is R =
V
2
o sin 2θ
g , where g is the acceleration due to gravity. Find the angle θ such that the range is a maximum.

6. UNIVERSIDAD DEL QUINDÍO 6
R (θ) =
V
2
o sin 2θ
g
R (θ) =
2Vo
g
cos 2θ
Sea, R (θ) = 0
Entonces, 0 =
2Vo
g
cos 2θ
0 = cos 2θ
cos
−1
0
2
= θ
45°
= θ
Part 4. Integration
Exercise. 21) Use the properties of summation and Theorem 4.2 to evaluate the sum.
20
i=1
(i + 1)
2
=
20
i=1
i2
+ 2i + 1
=
20
i=1
i2
+ 2
20
i=1
i +
20
i=1
1
=
n(n + 1)(2n + 1)
6
+ n(n + 1) + n
=
(n2
+ n)(2n + 1)
6
+ n(n + 1) + n
=
2n3
+ 3n + n
6
+ n(n + 1) + n
=
2(20)3
+ 3(20) + (20)
6
+ (20)((20) + 1) + (20)
=
16000 + 60 + 20
6
+ 420 + 20
=
16080
6
+ 440
= 3120
Exercise. 28) Use the limit process to nd the area of the region between the graph of the function and the x-axis over the
given interval. Sketch the region.
y = x2
+ 3, [0, 2]
Drawn region,
a = 0 ; b = 2 ; x = 2
n ; xi = 2i
n

7. UNIVERSIDAD DEL QUINDÍO 7
Area = lim
n→∞
n
i=1
f (xi) x
n
i=1
f (xi) x =
n
i=1
2i
n
2
+ 3
2
n
=
n
i=1
4i2
n2
+ 3
2
n
=
n
i=1
8i2
n3
+
6
n
=
8
n3
n
i=1
i2
+
6
n
n
i=1
1
=
8
n3
n(n + 1)(2n + 1)
6
+
6
n
[n]
=
(4n + 4)(2n + 1)
3n2
+ 6
=
8n2
+ 12n + 4
3n2
+ 6
=
(8n2
+ 12n + 4) + 18n2
3n2
=
26n2
+ 12n + 4
3n2
lim
n→∞
n
i=1
f (xi) x = lim
n→∞
26n2
+ 12n + 4
3n2
= lim
n→∞
26n2
n2 + 12n
n2 + 4
n2
3n2
n2
= lim
n→∞
26 + 12
n + 4
n2
3
Area =
26
3
Exercise. 69) Find the indenite integral.
´
x 1 − 3x2 4
dx
u = 1 − 3x2
du = −6xdx
−
1
6
du = xdx
ˆ
x 1 − 3x2 4
dx = −
1
6
ˆ
u4
du
= −
1
6
u5
5
+ constant
= −
1 − 3x2 5
30
+ constant
Exercise. 69) Evaluate the denite integral. Use a graphing utility to verify your result.
´ 3
0
1√
1+x
dx
u = 1 + x
du = dx

8. UNIVERSIDAD DEL QUINDÍO 8
3ˆ
0
1
√
1 + x
dx =
4ˆ
1
1
√
u
du
=
4ˆ
1
u− 1
2 du
= 2 u
1
2
4
1
= 2
√
4 −
√
1
= 2