Sáng kiến kinh nghiệm Một số kỹ thuật truyền cảm hứng học môn Toán bằng tiếng anh cho học sinh lớp 10

 học bình thường cần phải học ở trường
G. Không thấy việc học toán bằng tiếng Anh là cần thiết
Câu hỏi 3: Theo em chương trình học toán bằng tiếng Anh nào sau đây là phù hợp?
A. Chương trình Toán của Việt Nam dịch ra tiếng Anh
B. Chương trình của các nước tiên tiến
C. Chương trình được soạn riêng mà tích hợp môn Toán, môn tiếng Anh và môn
chuyên
Câu hỏi 4: Thời gian học toán bằng tiếng Anh trong 1 tuần?
A. 1 tiết B. 2 tiết C. 3 tiết D. 4 tiết
Câu hỏi 5: Khó khăn khi học toán bằng tiếng Anh?
A. Trình độ tiếng Anh giáo viên chưa cao
B. Trình độ tiếng Anh học sinh chưa cao
C. Nội dung chương trình dạy và học toán bằng tiếng Anh chưa có
Câu hỏi 6: Khó khăn khi học từ vựng toán bằng tiếng Anh
A. Từ chuyên ngành chỉ gặp trong tài liệu toán nên khó nhớ
B. Một từ có nhiều nghĩa
C. Nhiều từ có cùng một nghĩa
D. Là một cụm từ để diễn đạt khái niệm toán học
E. Một khái niệm toán học có thể diễn đạt theo nhiều cách khác nhau
F. Nghĩa của từ vựng trong toán học khác với nghĩa thông thường.
Câu hỏi 7: Em có thích học Toán bằng tiếng Anh không?
A. Rất thích B. Thích C. Bình thường D. Không thích
Xin chân thành cảm ơn các em !
dPHỤC LỤC 2
LESSON PLAN 1
PARAMETRIC EQUATIONS OF A LINE
I. Lesson’s objectives.
At the end of the lesson, the students will be able to:
- Prove the parametric equations of a line.
- Identify the direction vector and the coordinates of
points of a line with given equation.
- Find the parametric equations of a line with given
conditions.
- Solve problems involving lines.
II. Subject Matter
- Reference: Geometry for High-school Textbook.
- Materials: Sheets of paper.
III. Procedure
T: teacher; S: Student; Q: questions; Ans: Answer ;
Contents Teacher and
Students’activities
1. Introduce the lesson
I. Direction vector of a line
u
r
Definition: A non –zero vector u
r
 is a direction
vector of a line d if the line containing n is
parallel or coincident to d.
Notes:
- A given line has infinitely direction vectors
which are collinear. (i.e: if u
r
 is a direction vector of
a line d then k. u
r
 is also direction vector of the line
(for any non –zero scalar k).
T: State the definition
of direction
vector of a line
Q: How many direction
vector does a line have?
What is the relation
between them?
- A non-zero vector u
r
 is a direction vector of
infinitely parallel lines.
- Given a point M and a direction vector u
r
 ,
there is one and only one line passing through M
and get u
r
 be direction vector.
Problem: Given the line d passing through a
point ( )0 0 0;M x y and get a
non-zero vector u =
r
(a; b) be a direction vector.
What is the condition for the point M(x; y) to lie on
the line d?
Solution: M dÎ if 0M M
uuuuuur
 and u
r
are collinear, it
means there exists a parameter t such that 0M M tu=
uuuuuur r
, then we have
( ) ( )0 0; ;x x y y t a b- - =
0
0
x x at
y y bt
= +ì
Û í = +î
II. Theorem
The point M(x; y) is on the line d if and only if
there exists a number t such that
( )0
0
x x at
t R
y y bt
= +ì
Û Îí = +î
The system above is called the parametric
equations ofthe line d, and t is called the parameter.
Note:
- Each point on the line
corresponds to exactly one value of t.
- By equating t values, we obtain the equation
0 0x x y y
a b
- -
=
We call this equation the
Cartesian equation of the line.
- The gradient of the line b
a
How many lines does a
vector be direction
vector of? Why?
S:Answer all
the questions
T:State the problem
S: Find the solutions.
Q:What is the relation
between 0M M
uuuuuur
 and u
r
if
M lies on the line d?
Which vector equality
can
we deduce?
T:State the parametric
equations of a line
officially.
Q.How do we calculate
the gradient of the line?
2. Examples
Example 1: Given a line din the form of T: Deliver the
parametric equations ( )2 3
4 5
x t
t R
y t
= +ì
Îí = - +î
 and a point
M( -3 ; 6)
a) State the coordinates of a direction vector of
dand two points on d. Explain?
b) Find the parametric equations of the line 1d
passing through the point M and being parallel to
the line d.
c) Convert to Cartesian form.
Answer:
a) A direction vector of d is u =
r
(3; 5)
To obtain the points on the line d, we substitute
t by a real number.
Let t = 1, we get the point (5;1)
Let t = 0, we get the point (2;-4).
b) Two parallel line have direction vectors in
common, then a direction vector of 1d
is (3; 5). Because the line 1d passes through the
point M then the parametric equations of the line 1d
is ( )3 3
6 5
x t
t R
y t
= - +ì
Îí = +î
c) The Cartesian form of the line 2 4
3 5
x y- +
=
Example 2: Find the parametric equations of
the line d passing through the points A(1; 2) and
B(2;1).
Answer:
The line d passing through the points A and B
then it has direction vector AB =
uuur
 (1; -3). Then the
parametric equations of the line d is
( )1
2 3
x t
t R
y t
= +ì
Îí = -î
worksheet
containing with
problems.
S: Find the solution to
the
problem
T:Is the parametric
equations form of a line
unique? Why?
3. Summary the lesson
- Review the terms learned during the lesson.
- Summary the knowledge focus
4. Homework
Exercise 1:Find the parametric equations of the line which passing through
 (3; -4) and is parallel to the line ( )1 2
2
x t
t R
y t
= +ì
Îí = -î
Exercise 2: (k; 4) lies on the line with parametric equations ( )1 2
1
x t
t R
y t
= -ì
Îí = +î
.
Find k.
Exercise 3: Given the line ( )2
1 2
x t
t R
y t
= +ì
Îí = -î
 and the point A(1; 1). Prove that
A doesn’t lie on the line and find the coordinates of the foot of perpendicular of
A on d.
Exercise 4: Yacht A moves according to ( )4
5 2
x t
t R
y t
= +ì
Îí = -î
Where the distance units are kilometers and the time units are hours. Yacht B
moves according to ( )1 2
8
x t
t R
y t
= +ì
Îí = - +î
a. Find the initial position of each yacht.
b. Show that the speed of each yacht is constant
and state the speeds.
c. If they start at 6:00am, find the time when
the yachts are closest to each other.
d. Prove that the paths of the yachts are right
angles to each other.
LESSON PLAN 2
EQUATION OF CIRCLE
I. Lesson’s objectives.
At the end of the lesson, the students will be able to:
- Recognize the equation of a circle (the standard form and the
general form).
- Prove the equation of a circle.
- Find the equation of a circle with given conditions.
- Solve problems involving circles.
II.Subject Matter
- Reference: Geometry for High-school Textbook.
- Materials: Sheets of paper, Protractor, Puzzles.
III. Procedure
Contents
Teacher and Students’
activities
1. Introducing terms used in the lesson
(C): circle
I: center of the circle
B, C, D: points on the circle
IB: radius of the circle (a line segment
joining the center of the circle to any point on
the circle itself; or the length of such a segment,
which is half a diameter).
R: the length of the radius
T: Deliver the 1st
worksheet to ask the
students to fill in the
name of objects in a
given picture.
S: Work in pairto finish
the task in 3 mins.
T: correct the answers.
(C)
R k
I
B
C
D
k: tangent line of the circle (a straight line
that touches the circle at a single point).
CD: chord (a line segment whose endpoints
lie on the circle).
CmD: minor arc (connectedpart of the
circle's circumference).
Line CD: Secant(an extended chord, a
straight
line cutting the circle at two points).
2. Reading the passage to getacquainted with the language
Fill in the gaps using the given words (a
word may be used one more time)
radius distance tangent points
centre Common
A circle is the set of all (1) in a
plane that are a given (2) from a
given point, the (3). The distance
between any of the points and the centre is
called the(4). A(5) of a
circle
touches the circle at one point and the
distance from the center of the circle to the
tangent is equal to the radius of the circle.
T: deliver the 2st
worksheet to the students.
S: Work in pair to finish
the task in 3 mins
T: correct the answer and
take note some sentences
that should learn by heart
“the distance from.to”
“to be equal to”
3. Introduce the lesson
Problem 1. In the coordinate plane, given a
point I(a; b) and a positive real number R. On
what conditions that the point M(x; y) is on the
circle C(I; R) ?
Solution: M is on the circle if and only if
the distance from the point M to the point I is
equal to R, that is IM R IM R= Û =
uuur
( ) ( )2 2x a y b RÛ - + - =
( ) ( )2 2 2x a y b RÛ - + - =
I. Standard form of the equation of a
circle
T: State problem
S: Find the solution to the
problem
Q: What is the formulae to
calculate the distance
between two points?
What is the coordinates of
the vector IM?
The circle with center I(a; b) and radius R is
the set of all points (x;y) satisfying the equation
( ) ( )2 2 2 (1)x a y b R- + - =
The equation (1) is called the Standard form
of the equation of a circle
Example 1. Determine the
coordinate of the center and the radius of a
circle in the following cases:
i) ( ) ( )2 22 1 3x y- + + =
ii) ( )22 1 4 0x y+ + - =
iii) 2 2 4 2 1 0x y x y+ - - + =
iv) 2 2 4 2 6 0x y x y+ - - + =
Answer:
i) ( )2; 1 ; 3I R- =
ii) ( )0; 1 ; 2I R- =
iii) Completing the square form of x and y,
we
obtain
( ) ( )2 22 1 4x y- + - = Then ( )2;1I and 2R =
iv) ( ) ( )2 22 1 1x y- + - = - then this is not the
equation of a circle
Problem 2. Given the equation
( )2 2 2 2 0 2x y ax by c+ - - + =
On what conditions does the equation (2) be
the equation of a circle. In this case, determine
the coordinates of the center and the formula to
calculate the radius.
Solution:Write the left-hand sideof the
equation (2) in completed square form
( ) ( )2 2 2 2 0x a y b a b c- + - - - + =
Moving on the constant to the right-hand
side
( ) ( )2 2 2 2x a y b a b c- + - = + -
The equation (3) is in the standard form of a
T. State the equation of a
circle officially.
T. Deliver the 3rd
Worksheet.
S. Finish the task
individually.
T. Correct the answer.
Q. Convert the given
equation to the standard
form to identify the
coordinates of the center
and the radius in each
cases.
Q. Completing the
squares and moving on
the constant to the right
S.Convert the given
equation to the standard form
of equation.
State the condition that the
equation exist. Then,Identify
the coordinates of the center
circle, the equation exits if and only if the right-
hand side is positive, that is 2 2 0a b c+ - > .
In this case, the coordinates of the center are
(a; b) and the radius is 2 2a b c+ -
II. General equation of a
Circle
The equation 2 2 2 2 0x y ax by c+ - - + =
When 2 2 0a b c+ - > , is the general equation
of a circle with center I(a; b) and the
radius of the circle is 2 2R a b c= + -
and the length of the radius in
this case
T. Correct the answer and
state the general equation
of a circle officially.
4. Examples
Example 1. Given two points A(1; 2) and
B( -1; 4).
Find the equation of the circle with the
diameter AB.
Answer:
The center of the circle is the midpoint I of
the
segment AB, then the coordinates of I(0; 3)
and the radius 2R IA= = . Hence, the equation
of the circle with the diameter AB is
( )22 3 3x y+ - =
Example 2.Given three points A(1;2);
B(2;5);
C(4;1). Find the equation of the circumcircle
of the triangle ABC.
Answer:
T. Deliver the 4rd
worksheet.
S.Finish the task in 5
mins.
T.Correct the answer.
Q. What is the center of
the circle? How to calculate
the coordinates of the
midpoint of a
segment?
( )1;3AB =
uuur
; ( )2; 4BC = -
uuur
; ( )3; 1AC = -
uuur
Notice that . 0AB AC =
uuur uuur
then the triangle ABC
is
right triangle at A. Hence, the center of the
circumcircle is the midpoint of the hypotenuse
BC. So, the center I(3; 3) and the radius 5R =
and the equation of the circle is
( ) ( )2 23 3 5x y- + - =
Example 3.Find the equation of a circle
which
touches the x-and y-axes and passes through
the point A( -1; 3)
Answer:
 Let I be the center of the circle with
coordinates
(a; b).The circle touches the x and y –axes
that means d(I;Ox) = d(I; Oy) then a b R= =
Case 1: If a = b then we obtain the equation
( ) ( )2 2 2x a y a a- + - =
The point A is on the circle then the
coordinates
satisfying the equation, then we have the
equation
( ) ( )2 2 2 21 3 4 10 0a a a a a- - + - = Û - + =
The equation has no roots in this case.
Case 2: If a = -b then the equation is
( ) ( )2 2 2x a y a a- + + =
The point A is on the circle then the
coordinates
satisfying the equation, we obtain the
equation
( ) ( )2 2 21 3a a a+ + + = 2 8 10 0a aÛ + + =
Solving this equation gives 4 6a = - ±
.Therefore the solution is
( ) ( ) ( )2 2 24 6 4 6 4 6x y+ + + - - = - -
Notice the character of
the triangle?
Locate the center of a
right triangle.
Under what condition that
a line touchesa circle?
( ) ( ) ( )2 2 24 6 4 6 4 6x y+ - + - + = - +
5. Summary the lesson
- Review the terms learned during the lesson through flashcards.
- Summary the knowledge focus
6. Homework
Exercise 1:Find the equation of the circle (C) centered at I(1; 2) and tangent
to the line (d) with the equation: 3x –4y + 15 = 0
Exercise 2: Find the equation of the circle (C) passing through three points
A( -2; 4); B(5; 5); C(6; -2).
Exercise 3: Given the circle (C): 2 2 2 2 2 0x y x y+ - - - = and the line
(d): x + 2y –1 = 0. Find the points of intersection of the line and the circle.
PHỤC LỤC 3 ( Sử dụng padlet hoặc azota )
Test
P1.(4 point) Given a line D with equation ( )2 3
4 5
x t
t R
y t
= +ì
Îí = - +î
 and a point
( )3;6M - .
a) Find the coordinates of a point on D and a direction vector of D .
b) Find the parametric equation of the line d passing through M and being
parallel to D .
P2. (2 point) Find the equation of the circle (C) centered at ( )5,1I and passing
through the point ( )2,5M .
P3. (4 point) Which of the following equations are the equations of the circles?
Find the centers and the radius if they are.
( )2 2 2 4 4 0 1x y x y+ - - - =
( )2 2 6 2 20 0 2x y x y+ - + + =
Answer
P1a) Let 0t = , we have the coordinates of a point ( )0 2, 4M - on D . (1 p)
A direction vector of D is: ( )3,5a =
r
(1 p)
b) The parametric equation of the line d passing through M and being parallel
to D is 3 3
6 5
x t
y t
= - +ì
í = +î
 (2 p)
P2. We have
( ) ( )2 22 5 5 1 9 16 25 5R IM= = - + - = + = = (1p)
The equation of the circle (C):
( ) ( )2 25 1 25x y- + - = (1 p)
P3a) ( )2 2 2 4 4 0 1x y x y+ - - - =
We have: 1, 2, 4a b c= = = -
2 2 1 4 4 9 0a b c+ - = + + = >
Hence, (1) is the equation of the circle with center ( )1,2I (1 p)
and the radius 3R = (1 p)
b) ( )2 2 6 2 20 0 2x y x y+ - + + =
We have: 3, 1, 20a b c= = - =
2 2 9 1 20 10 0a b c+ - = + - = - < (1 p)
Hence, (2) is not the equation of a circle. (1 p)