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111
S
3
×
r
-
Because, V =
осн
=
осн
hS
;
V
h
S
осн
2/510
1203
510
( units 2)
5104
===
17
.
( units)
Solve and check yourself using the answer:
1. Given two vectors a(11,10,2) and b(4,0,3). Find a unit vector c orthogonal to the vectors a and b and aimed so that the ordered triple of
vectors a, b, c was right.
Answer: x =
6
, y = -
55
1
5
, z =-
8
.
55
2. To find the volume of pyramids АВСD if А(3,4,5), В(1,2,1), С(-2,
-3,6), D(3,-6,-3). Answer: 42.
3. To find the volume of pyramids АВСD if А(-7,-5,6), В(-2,5,-3),
С(3,-2,4), D(1,2,2). Answer: 77/3
4. To find the volume of pyramids АВСD if А(1,3,1), В(-1,4,6),
С(-2,-3,4), D(3,4,-4). Answer: 3
5. To find the volume of pyramids АВСD if А(2,4,1), В(-3,-2,4),
С(3,5,-2), D(4,2,-3). Answer: 25/3.
The various types of equations.
Example 13. Find the equation of a straight line passing through the
point A(1, 2) perpendicular to the vector
n
(3, -1).
Prepare when A = 3 and B = -1 the equation of a straight line: 3x – y
+ C = 0. To find the coefficient C we substitute in the expression of the
coordinates of a given point A.
Received: 3 – 2 + C = 0, hence C = -1. So: 3х – у – 1 = 0.
Example 14. Find the equation of the line passing through the points
A(1, 2) and B(3, 4).
Applying the formula we get:
24
2
=-
-
12
-=-
xy
01
=+-
yx
)1(
-
xy
13
Example 15: Given the General equation of a straight line
12x – 5Y – 65 = 0.
You need to write different types of equations of the line
112
5
5125655
3
2;0
300
2
-
-
4
4
1
1
4
3
65
1
=-yхух
1
+
=
)13()12/65(
-
the equation of the line in pieces:
12
65
the equation of the line with an angular ratio: (divide by 5)
12
.13
-=-= xxy
the normal equation of a straight line:
13
5
05
=--=
;
= ух
m
1
-+
1
22
13
)5(12
12
13
cosj = 12/13; sinj = -5/13; p = 5.
Solve and check yourself using the answer:
1. Write the line equation passing through the point A(3,1) and in-
clined to the line 2x+3y-1 = 0 at angle of 450.
Answer: x-5y+2=0 and 5x + y – 16 = 0.
2. Write an equation of a line passing through the point A(-2, -3) and
the origin O (0,0).
Answer:
0
=
--
---
yxyx
=
-
.023;
=-
yx
3. Straight cuts along the coordinate axes is equal to the positive seg-
ments. Write an equation of a straight line, if the area of the triangle
yx
1
formed by these lines is equal to 8 см2. Answer:
=+
or х + у – 4 = 0.
4. Find the equation of a straight line with the guiding vector (1, -1)
and passing through the point A(1, 2). Answer: х + у – 3 = 0
5. Set the General equation of a straight line x – y + 1 = 0. Find the
ух
1
equation of the line in pieces. Answer:
=+-
, а = -1, b = 1.
6. To find the angular coefficient of the straight 2x – y + 3 = 0. An-
swer 2.
yx
1
7. The amount of segments, cut straight
=--
to the coordinate
axes is equal to...? Answer: -7
113
8. The angular coefficient of the straight 3x – 2y +3 = 0 is ….? An-
2
4
2
3
swer:
9. The amount of segments, cut straight
yx
to the coordinate
1
=+
axes is equal to...? Answer 6.
10. What is the angular coefficient of the straight line 4x – y = 0? An-
swer: 4.
Conditions of parallelism, perpendicularity direct, the distance
from a point to a straight line.
The angle between lines in the plane.
Definition. If f the two lines are y = k1x + b1, y = k2x + b2,, then the
acute angle between the straight lines will be determined by the formula:
kk
-
12
+
21
.
(tangent)
tg
=
a
1 kk
Recall that the direct set of General equations, parallel, if the coefficients of x and y are proportional to each other!
The distance from a point to a straight line.
Theorem. if given a point М(х0, у0), then the distance to the straight
line Ах + Ву + С =0 is defined as
CByAx
++
.
d
=
00
22
BA
+
Example. Show that the lines 3х – 5у + 7 = 0 and 10х + 6у – 3 = 0
are perpendicular.
Find: k1 = 3/5, k2 = -5/3, k1k2 = -1, therefore, straight lines are perpendicular.
Example. To find the angle between lines: y = -3x + 7; y = 2x + 1.
)3(2
--
1
k1 = -3; k
= 2 tgj =
2
=
; j = p/4.
2)3(1
--
Solve and check yourself using the answer:
1. Write the line equation passing through the point A(3,1) and in-
clined to a straight line 2x+3y-1 = 0 on the angle 45o.
Answer: x – 5y + 2 = 0 и 5x + y – 16 = 0.
114
2. At what value of the parameter t straight lines defined by the equa-
5
220
+-=
6310
++=
7
35
3210
-+=
25120
+-=
tions 3tx-8y+1 = 0 and (1+t)x-2ty = 0 are parallel ?
Answer: t1 = 2, t2 = -2/3.
3. Given the vertices of a triangle ABC: A (4; 3) , B (-3; -3) , C (2; 7).
The distance from the point A to the straight line BC is the number …?
Answer:
58
4. Find the distance between the parallel straight
xy
Answer:
5. Have a direct lines
mon point?
Answer: yes.
.
.
xy
and
xy
xy
and
a com-
115
Section 2. Basics of calculus
Topic 9. The limit of a function
at a point and infinity
Introduction. This workshop is devoted to the discussion of the following concepts: "epsilon-delta definition" of limits, the rule of Lapetal,
uncertainties, some basic theorems of limits.
One of the basic operations in the theory of functions is the operation
limiting process (the operation of finding the limit of variables). This
operation is based on the notion of limit of a function at a point or at infinity.
Let the function y = f(x) is dened in some neighborhood of the point
х0 and the point х0 can be either defined or not defined.
Definition . The number А is called the limit of a function y = f(x) at
the point х0 (or x seeking to х0), if for any, even an arbitrarily small number of Е > 0, there is a number δ>0 (depending on E, δ = δ(φ)) that for all
x not equal to х0 and satisfying the inequality | х – х0 | < δ.
The number is called the limit of a function y = f(x) в точке х0 (or
when x seeking to х0 ), if for any, even an arbitrarily small number, there
will be such number δ>0 (dependent of Е, δ = δ(φ)),that for all x not
equal to х0 and satisfying the inequality | х – х0 | < δ will run inequality
|f(x) – A| < E and is denoted by lim
The denition of limit of a function f(x) at the point х0 lies in the fact
that for all values of х sufficiently close to х0,, the values of the function
f(x) arbitrarily differ little from the number A (absolute) value.
The geometric meaning of the limit of a function at a point, i.e. equality А = lim
x-x0
f(x).
The limit is a constant value at any point is equal to this constant value.
Infinitesimals and their properties.
Consider the class of functions y = f(x) whose limit point х0 is equal to 0.
Definition. Function α(х) is an infinitesimal function or infinitely small
value at the point х0, if its limit at this point is equal to zero, lim
f(x) = A or f(x)→ A if х → х0.
x-x0
x-x0
α (х)= 0.
116
Similarly defines an infinitely small value at х → х0 + 0; х→ х0 – 0;
х→+∞; х→+∞.
Examples. α1(х) = х2 if х →0 (approaches zero) ,α2(х) = х – 2 if
х→2; α3 (х) = sin x if х→πk, k ∈ z ; α4(x) = if х→∞.
The ratio of two infinitesimal quantities is called the indefinite expression or uncertainty of the form .
Definition. Infinitesimals α(х) и β(х) in the point х0 is called infinitesimal of the same order of smallness, if the limit of the ratio of these
quantities at the point х0 is a constant different from zero
=А≠0.
Definition. Infinitesimals α(х) и β(х) are called equivalent, if the limit of their ratio is equal to one
= 1 and is denoted by α(х) ~ β(х).
Definition. Infinitesimals α(х) is called an infinitesimal of higher or-
der of smallness in comparison with β(х) if the limit of their ratio is equal
to the zero, =0 and is denoted by α(х) = 0[β(х)].
Definition. Infinitesimals α(х) is called an infinitesimal of higher or-
der of smallness in comparison with β(х) if the limit of their ratio is equal
to the infinity = ∞ and is denoted by α(х) = 0[β(х)].
Noticeable limits.
When evaluating of uncertainties (indeterminate forms) convenient to
use the following Noticeable limits.
1. The first noticeable limit. = 1. This formula allows to
reveal the uncertainty .
For example: to calculate the limit = , To apply the 1st
noticeable limit
= = = 1,5.
When evaluating of uncertainties under the sign of the limit of infinitely small quantities, you can replace them with equivalent. This sim-
117
plifies the disclosure of uncertainties . In the previous example
n
x
(
)
(
)
the function sin 6x is replaced with equivalent function 6x,
get = = = = = 1,5.
1
n
2. The second noticeable limit is written as lim(1+
1
x
lim(1+
)
=e
)
=e ,
=e (4.6); or , е ≈ 2,718281
The second noticeable limit is used for the evaluation of the uncertainty [1∞].
For example: 1) = [1∞] then = е2 .
2) = [1∞]= = = е2.
To calculate limits of functions.
2
15
+
а) Find
lim
x
x
5
¥®
.
327
++
xx
Solution. First of all, we find the limits of the numerator and denomi-
nator of the fraction. Functions
large. Therefore,
lim
x
¥®
152x
2
,
lim .
¥=+
x
Therefore, dealing with indeterminate form
and
15
+x
¥®
5
5
327
xx
¥
ü
ì
.
ý
í
¥
þ
î
are infinitely
327
++ xx
¥=++
For disclosure of this uncertainty extract in the numerator and in the
denominator х in the higher degrees as cofactor and reduce the raction
æ
5
ç
x
2
15
+
x
5
xx
¥
ü
ì
=
=
ý
í
¥
þ
î
327
++
ç
è
æ
5
ç
7
x
ç
è
ö
15
÷
+
÷
53
xx
ø
=
ö
32
++
54
xx
¥®¥®¥®
xxx
÷
÷
ø
7
15
+
53
0
xx
32
++
54
xx
.limlimlim 0
==
7
Answer. 0.
b) Find
lim
®
x
2
2
2
3214
-+
xx
.
86
+-
xx
118
0
()(
)
()(
)
2
ü
Solution. For evaluating of uncertainty
ì
in this case, it is neces-
ý
í
0
þ
î
sary to decompose the numerator and denominator by multipliers and
reduce the fraction by a common factor (find the roots of square trinomials using discriminant and apply theorem about the decomposition of a
square trinomials by multipliers).
2
-+
xx
2
2
+-
xx
0
3214
86
ü
ì
=
=
ý
í
0
þ
î
( )( )
162
+-
xx
=
42
--
xx
xxx
16
+
x
4
-
x
22
®®®
162
+
=
-
.limlimlim 9
-=
42
Answer. -9.
c) Find
x
lim
2
2
1
-®
3214
-+
xx
.
86
+-
xx
Solution. To calculate this limit, we will substitute the value under the
sign of the limit into the function,
2
=
2
( ) ( )
--×+=
15
8161
+-×--
3
-=
.
45
321141
-
Get
lim
x
2
2
1
-®
3214
-+
xx
86
+-
xx
Answer. -3.
2
x
d) Find
lim
x-+®
0
Solution. For evaluating of uncertainty
11
.
2
x
0
ü
ì
in this case, you need to
ý
í
0
þ
î
multiply the numerator and the denominator in the expression of the adjoint of the numerator, and then reduce the fraction by a common factor.
2
x
2
0
x
Answer.
æ
æ
011
-+
ü
ì
=
ý
í
0
þ
î
1
.
ç
è
=
0
ö
-+
æ
ç
è
22
xx
ç
÷
è
ø
++
xx
ö
1111
++
÷
ø
=
ö
÷
ø
®®®
xxx
x
æ
222022
xx
ç
è
=
ö
1111
++
÷
ø
1
.limlimlim
2
e) Find
lim
x
.
x
0®
kx
sin
Solution. For evaluating of uncertainty
select the first noticeable limit:
lim 1
®
A
0
ü
ì
in this case, you need to
ý
í
0
þ
î
sin
A
.
=
A
0
119
sin
p
p
2
p
p
212
2
kx
0
ü
lim kk
ì
=
ý
í
x
0
þ
î
lim
xx
®®
sin
kxk
kx
00
.
=×==
1
Answer. k
f) Find
lim
x
®
( )
1
-
1
x
tgx
.
2
Solution. For evaluating of uncertainty {0 ∙∞} in this case, the prod-
uct should be converted to the ratio, that is to reduce uncertainty {0 ∙∞}
to uncertainty
( ) { } ( )
1
xy
é
=
ê
ë
ì
or
ý
í
0
þ
î
x
p
tgx
2
,
1
-=
,
ù
lim
=
ú
yxy
10
+=®
û
ý
í
¥
þ
î
.
x
limlim
cos
10
-
p
2
sin
x
cos
y
( )
y
1
+
x
p
é
2
ê
x
p
2
=
lim
ë
cos
2
y
-
æ
y
+
ç
è
ù
lim,sin
xпри
11
=
===-=¥×=-
ú
xxx
û
lim
=
pp
ö
yyy
000
®®®
÷
22
ø
-
®®®
-
sin
x
1
-
111
cos
y
p
y
2
0
ü
ì
=
ý
í
x
p
0
þ
î
2
.
¥
ü
ì
0
ü
Isolate the first noticeable limit, that is, multiply the numerator and
=
the denominator by a
. Get
lim
®
y
-
y
2
0
-
sin
y
2
==
ppp
p
.
lim
x
2
.
p
x
x
-
1
ö
æ
÷
ç
x
¥®
è
.
+
1
ø
Answer.
i) Find
Solution. For evaluating of uncertainty {1∞} in this case, you need to
x
1
ö
select the second noticeable limit:
æ
+
1lim
ç
x
¥®
è
e
=
÷
x
.
ø
120
2
2
4
-
x
1
+
x
æ
x
1
-
x
ö
æ
ç
¥®
x
è
{ }
÷
1
+
x
ø
é
¥
æ
11
+==
ç
ê
x
è
ë
x
ù
1
-
x
ö
-
÷
ú
1
+
x
ø
û
2
-
é
11
+=
ê
1
+
xx
ë
ç
x
ç
ù
ú
û
ê
ç
1
+=
ê
ç
¥®¥®¥®
x
ê
ç
ê
ëé-
ç
è
ö
1
+
x
÷
ù
2
-
÷
ú
1
ú
1
+
x
ú
ú
2
û
lim
¥®
x
÷
÷
÷
÷
ø
Answer. e-2.
1
g) Find
®
x
( )
-
x
2
53
.lim
2
-
x
Solution. For evaluating of uncertainty {1∞} in this case, you need to
1
select the second noticeable limit:
1
( ) ( )( ) ( )
lim
x
®
53 eyy
x
2
Answer.
Find
®
x
2
x
-
3
e
25
é
=-
ê
ë
.
( )
-
x
,
2
xy
-=
,
02
yyx
®+=
1
.lim
53
2
-
x
Solution.. We will substitute the value
( )
a
1lim
®
a
0
ù
ú
0
y
û
=x
.
e=+
a
1
y
y
5
standing under the sign
31523
0
®®
of the limit into the function. Get
Answer.
25
1
( )
x
25
x
2
-®x
5
æ
353
ç
2
è
1
5
ö
2
-
5
-×=-
2
÷
ø
.
2
25
5
ö
æ
=
÷
ç
2
ø
è
.lim
=
4
2
-
x
2
-
1
+
x
.limlimlimlim
==
ee
31
×
3
.limlim
=+=-+=
3
y
×
Topic 10. Derivative and differential of a function
Introduction. This workshop is devoted to consideration of the basic
rules of differentiation, the derivative of the derivative and using derivatives in curve tracing.
Example .
Find the derivatives of the given functions
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