1 Content of the lessons
Page 2 Lesson 1 900 counter clockwise rotation ‒ CCW
Page 3 Lesson 2 Algebra through arithmetic, common sense and practice Page 4 Lesson 3 Cars moving along the S axis
Page 5 Lesson 4 Perpendicular lines
Page 6 Lesson 5 In the following graphs, the car travels at 40 mi/hr
Page 7 Lesson 6 Finding the address of point in space – rectangular coordinates Page 8 Lesson 7 Galileo’s Imaginary Experiment – Law of Inertia
Page 9 Lesson 8 Atmosphere Page 10 Lesson 9 Longitude and latitude Page 11 Lesson10 Area of a triangle and rectangle
Page 12 Lesson 11 Breeze at the seashore reverses direction at nighttime Page 13 Lesson 12 As air speeds up, its pressure decreases
Page 14 Lesson 13 Fractions Page 15 Lesson 14 Word problem Page 16 Lesson 15 Word problems
Page 17 Lesson 16 How lakes freeze in the winter
Page 18 Lesson 17 Fahrenheit and Celsius Temperature Scales
Page 19 Lesson 18 Horizontal circles measure latitude. Vertical semicircles measure longitude.
Page 20 Lesson 19 Having fun with math Page 21 Lesson 20 Vectors
Page 23 Lesson 21 Stretching the imagination – did you realize that water floats in water?
Page 24 Lesson 22 Critical thinking – the Big Bang
Page 25 Lesson 23 Solids push down, liquids push against wet surfaces, and gasses expand pushing in all directions Page 26 Lesson 24 Percentage
Page 27 Lesson 25 Putting a satellite in orbit
Page 28 Lesson 26 In 200 BC Eratosthenes calculated the circumference of the Earth Page 29 Lesson 27 Moment of a force
Page 30 Lesson 28 Linear functions
Page 31 Lesson 29 Rotation of tangents and secant in a circle
Page 32 Lesson 30 Reflection on a mirror ‒ Image of an object a distance d away from a mirror Page 33 Lesson 31 Angles in a triangle
Page 34 Lesson 32 An object is thrown downwards with initial velocity v0
Page 35 Lesson 33 Why cross-multiplying works
Page 36 Lesson 34 Solving constant rate problems using proportions Page 37 Lesson 35 Logic: Statements or propositions are either true or false Page 39 Lesson 36 Transition from Arithmetic to Algebra
Page 41 Lesson 37 Given two points (x1, y1) and (x2, y2), find their midpoint (x1 + x2)/2, (y1 +y2)/2 Page 43 Lesson 38 Unit circle ‒ radius one
Page 45 Lesson 39 A point (x, y) on a circle of radius 1 is ONE unit away from its center. x2 + y2 = 1 Page 47 Lesson 40 Vectors and dot product
Page 49 Lesson 41 Vecror equation of a line
Page 51 Lesson 42 Distance from a point to a line ‒ using vectors Page 53 Lesson 43 Unit vectors in 2D and 3D
Page 55 Lesson 44 Equation of a plane Page 57 Lesson 45 Definition of vector product Page 59 Lesson 46 Vector perpendicular to a plane Page 61 Lesson 47 System of equations
Page 64 Lesson 48 Cosine and Sine Curves ‒ passenger in a Ferris wheel Page 67 Lesson 49 Critical thinking – the Periodic Table
Page 68 Lesson 50 Three Ferris wheels of different size Page 71 Lesson 51 Rotational Motion
Page 73 Lesson 52 Parabola
Page 75 Lesson 53 Projectiles Page 77 Lesson 54 Completing the square in a quadratic equation
Page 79 Lesson 55 Properties of Parabolas
Page 81 Lesson 56 Equation of a Parabola: y – k = a(x – h)2 or Y = aX2 Page 84 Lesson 57 Logarithms
Page 86 Lesson 58 Applications of Logarithms
Page 88 Lesson 59 Laws of Logarithms or Laws of Exponents Page 90 Lesson 60 Exponential Functions
Page 92 Lesson 61 Calculus Page 95 Lesson 62 Congruent triangles
Page 98 Lesson 63 Medians of a triangle Page 100 Lesson 64 Application of Calculus
Page 103 Lesson 65 The incredible number of Blood Vessels in our Body Page 105 Lesson 66 Temperature Scales
Page 108 Lesson 67 Coriolis Effect
Page 111 Lesson 68 It all starts with the Big Bang!
Page 114 Lesson 69 Applications of Pressure
Page 116 Lesson 70 Graphs of Objects moving vertically in the Atmosphere Page 119 Lesson 71 Rotation of axis
Page 123 Lesson 72 A parallelogram consists of two congruent triangles Page 125 Lesson 73 Conics or Curves on the surface of a Cone
Page 128 Lesson 74 Physics Formulas
Page 131 Lesson 75 Different Coordinate Systems Page 134 Lesson 76 Newton’s 2nd Law “F = ma”, 3rd Law “Action and Reaction”
Page 137 Lesson 77 Reflection and Refraction of Light Page 141 Lesson 78 Differentials
Page 144 Lesson 79 Processes Involving Energy Page 147 Lesson 80 Quadrilaterals
Page 149 Lesson 81 Evolution and the Big Bang violate laws of nature.
Page 150 Lesson 82 Is God in control of the events in this world?
Lesson 1 900 counter clockwise rotation ‒ CCW
There were three professors whose class notes were excellent, Dr. Miller at Queens University, Dr. Sida at Carleton university both in Canada, and Dr. Young at Iowa State University. The written word is excellent for communicating ideas.
Q1 The product of the slopes of CA and BD is ‒ 1. T/F Q2 When ΔOCA is rotated 900 CCW it overlaps with triangle ______ . Q3 Rotating ➔OB 2700 CCW is the same as rotating ➔OB 900 ___
Q4 Find the run and rise from A to C and from D to B.
Q5 m = (y2 ‒ y1) / (x2 ‒ x1) = (y1 ‒ y2) / (x1 ‒ x2). T/F
Answer T, from C to A, run = 8, rise = 6, slope 3/4. From B to D, run = ‒ 6, rise = 8, slope = ‒ 4/3. The product of the slopes is ‒ 1.
Q2 ODB Q3 CW. Q4 ‒ 8, ‒ 6, slope 3/4, ‒ 6, 8, slope = ‒ 4/3. Q5 T, multiply numerator and denominator by ‒ 1.
On occasions, in the same lesson, different concepts may be discussed.
The wheel of a car ‒ do all parts have the same speed?
From O(0, 0) to A(5, 2) the change in x, 5, is called the run and the change in y, 2, is called the rise. The vector OA, ➔OA = [2, 5]. (2. 5) is a point and [2, 5] is a change in position. The change in position from C(‒ 3, ‒ 4) to A(5, 2) is run = 5 ‒ (‒ 3) = + 8 and the rise = 2 ‒ (‒ 4) = 6. ➔CA = [8, 6]
Here is how we do a 900 counterclockwise ‒ CCW ‒ rotation, of ➔AC about the origin O. A 900 rotation of a vector is very useful. When ➔OA = [5, 2] is rotated 900 CCW around O, its length remains the same and becomes ➔OB = [‒ 2, 5]. Please see the figure. The length and width of the grey and yellow rectangles are interchanged. The image of A(5, 2) is B(‒ 2. 5).
When ➔OC = [‒ 3, ‒ 4] is rotated 900 CCW, it lands in the 4th quadrant, The image of C(‒ 3, ‒ 4) is point D(4, ‒ 3). The orange rectangle is rotated into the blue rectangle, again their length and width are interchanged.
When AC is rotated 900 CCW it becomes BD; they are perpendicular.
900 counter clockwise rotation ‒ CCW
‒ rotation
A wheel has the forward velocity V of the car and a rotation around the axle at O.
Let me show you why the speed of rotation of the ream is also V.
At point T, where the rubber meets the road, there are two motions, the forward velocity of the car V and the motion due to the rotation. T is momentarily at rest.
Q1 Is point T going forward or backward?
Q2 Why is the top of the wheel moving twice as fast as the O?
Answer Q1 At T the bottom of the wheel begins to move forward Q2 Vcar + V® = 2V V
V® V
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Lesson 2 Algebra through arithmetic, common sense and practice
1. = 6 the number inside the box must satisfy the equality. It is obvious that the number is 6. In algebra, we have the equation x = 6.
2. + 2 = 8 to isolate the , ‒ 2 is added to both sides. + 2 ‒ 2 = 8 ‒ 2, hence = 6. In algebra we solve x + 2 = 8.
3. To solve the equation 3x + 2 = 8, we first isolate 3x. Now we have a new equation to solve, 3x = 6.
The final result is x = 2, which satisfies the original equation 3x + 2 = 8.
Suppose fractions appear in an equation
4. (1/2)x + 3 = 6. First, let us eliminate the fraction by multiplying by 2.
Simplifying the last equation, x + 6 = 12, x = 6, which satisfies (1/2)x + 3 = 6.
Note every term of (1/2)x + 3 = 6 is multiplied by 2.
5. A = (1/2)bh is the area of a triangle with base b and height h Suppose we want to solve for h.
To eliminate the fraction 1/2, multiply both sides of the equation by 2.
2A = 2(1/2)bh
2A = bh To solve for h, keep in mind that b and h are multiplying.
Hence h = 2A/b. Check the units; if b and h are in meters, A is in m2, and h in m2/m = m.
6. A car travels a distance s in a time t at a constant speed v = s/t Let’s solve for t.
Equals multiplied t = s/v
by equals. To travel 90 miles, it takes a car traveling at 45 mph,
One more step 2 hr. = 90 miles / 45 mph’
A ball rotating around O in a circle at a constant speed
If equals are added, the results are equal. Note to isolate x in x + 2 = 8; ‒ 2 is added to both sides.
3x + 2 = 8 ‒ 2 = ‒ 2 3x = 6 3x = 6 equals divided
3 = 3 by equals 3x/3 = 6/3
x + 2 = 8 ‒ 2 = ‒2 x = 6
(1/2)x + 3 = 6 2 = 2 2(1/2)x + 2(3) = 2(6)
2A = bh b = b 2A/b = bh/b
v = s/t t = t tv = t(s/t)
tv = s v = v tv/v = s/v
A ball is rotating around O in a circle. One observer is at O measuring the rotation of OT and the other observer at T measuring <STP, the rotation of the chord TP, which is following the ball.
Q1 The initial value of <STP is ___ .
Q2 The velocity of the ball, the arrow, is always tangent to the circle. T/F Q3 When <STP = 450, the ball, as seen from O, has rotated ____ degrees, Q4 The arc intercepted by <TRP = 450. T/F
Q5 When the ball is at Q its velocity has rotated _____ degrees.
Q6 When the ball is at Q, the chord TP has rotated ______ degrees.
Q7 As seen from R, the ball is rotating as fast as seen from O. T/F Answer Q1 00 Q2 T Q3 900 Q4 F, 900 Q5 1800 Q6 900 Q7 F, half as fast
Equals divided
Equals multiplied Equals divided
Equals added
A ball is rotating around O, it is at P. A person at P studies its motion, while a person at T observes how TP rotates.
Lesson 3 Cars moving along the S axis
Angles in a circle
Q1 The slope of a line never changes. T/F Q2 In which direction do these cars move?
Q3 Describe the motion of car 2.
Q4 The equation of this horizontal line is _____
Q5 How can we calculate the speed of car 3?
Q6 Using any two of these points, (1, 1), (2, 2), (4, 4) calculate the slope of the graph of car 3.
Q7 Can the points (0, 0) and (4, 3) be used to find the speed of car 3?
Q8 Which is the initial position of car 1?
Q9 Using the points (0, 4) and (4, 0), find the speed of car 1.
Q10 using the equation S ‒ S1 = m(t ‒ t1), show that for car 2, S = 2, car 3, S = t, car 1, S = 4 ‒ t.
Answer Q1 T. Q2 along S Q3 it is always 2 units away from (0, 0). Its speed or slope is zero. Q4 S = 2 or 0t + S = 2. Q5 by selecting two points on the graph of car 3. Q6 (2 ‒ 1) / (2 ‒ 1) = 1, (4 ‒ 1) / (4 ‒ 1) = 1.
Q7 No, (0, 0) is on the graph, but not (4, 3). Q8 S0 = 4 Q9 (0 ‒ 4) / (4 ‒ 0) = ‒ 1. Q10 car 2, S ‒ 2 = (2 ‒ 2) / (5 ‒ 2)(t ‒ 2), S = 2. Car 3, (S ‒ 0) = (2 ‒ 0) / (2 ‒ 0)(t ‒ 0), car 2, S ‒ 4 = (‒ 1)(t ‒ 0)
Angle of two intersecting chords equals (1/2)(sum of the angles of the two chords).
Q1 Chords 1‒6 and 3‒4 intersect at 9; angle = (1/2)(40 + __) Q2 Chords 1‒6 and 8‒4 intersect at 10; angle = (1/2)(90 + __) Q3 Chords 4‒3 and 4‒8 intersect at 4; angle = (1/2)(50 + __) Q4 The line tangent at 4 makes with the chord 4‒3 an acute angle of 1300. T/F
Q5 Find the obtuse angle formed by the tangent at 4 and chord 4‒3. __
Q6 The arcs contained by the sides 4‒11 and 4‒5 of the 1800 angle at 4 are ____ and ____.
Vertex 11 is exterior to the circle.
Q7 <11 = (1/2)(addition or subtraction of the arcs)?
Q8 When are the arcs added?
Answer Q1 900 Q2 900 Q3 00 Q4 F 1300/2 Q5 1150 Q6 3600, 00 Q7 subtraction Q8 when the chords intersect inside the circle
It is a good idea to imagine different situations to see when the formula applies or not. For example, as point 11 recedes from the circle <11 becomes smaller and the two arcs come close to each other. this makes sense.
500
11 S
As seen from B, as the chord AB rotates around E, point B sees chord BD rotate an angle (1/2)(arc BD).
As seen from A, as the chord AB rotates around E, point A sees chord AC rotate an angle (1/2)(arc AC).
Both rotations around E are clockwise equal to (1/2)(arc AC + arc BD).
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Lesson 4 Perpendicular Lines
Q1 In 2x + 3y = 6 both x and y can increase. T/F
Q2 Which of these vectors [2, 3], [‒ 3, 2], [ 3, ‒ 2] is parallel to the line 2x + 3y = 6?
Q3 The slopes of the vectors [‒ 3, 2] and [3, ‒ 2] are equal. T/F Q4 Show that (y2 ‒ y1) / (x2 ‒ x1) = (y1 ‒ y2) / (x1 ‒ x2)?
Q5 What is the difference between (3, 4) and [3, 4]?
Q6 Find a vector perpendicular to x + 0y = 0, the y axis, and a vector perpendicular to 0x + y = 0, the x axis.
Answer Q1 F, if both increase 2x + 3y > 6. Q2 [‒ 3, 2] and [ 3, ‒ 2]. Q3 T Q4 y2 ‒ y1) / (x2 ‒ x1) = ‒ (y1 ‒ y2) / ‒ (x1 ‒ x2) Q5 (3, 4) is a point; [3, 4] is a run and rise from (a, b) to (a + 3, b + 4), where (a, b) is any point. Q6 [1, 0], [0, 1].
Note [1, 0] is perpendicular to the X axis; its equation is 1x + 0y = 0; Likewise 0x + 1y = 0 is the equation of the Y axis.
Vector perpendicular to a line
[a, b] is perpendicular to ax + by = c; [0, 1] is perpendicular to 0x + y = 0; [1, 0] is perpendicular to x + 0y = 0.
Evolution
Recently I asked a family Dr. in Toronto, “Is the chemical composition of a monkey similar to that of a human?” She immediately said yes. In the internet, I discovered that a pig is the animal whose digestive system most resembles that of a human. Afterwards, I asked a Ph. D. chemistry professor, graduated from MIT, if the chemical compounds in the stomach of a monkey could evolve into those of a human, without violating a law of chemistry? Yes, he replied.
Pictures only show the external evolution of the monkey. Have you considered the internal evolution of the monkey?
In the sun, at a temperature of 15 million degrees two atoms of hydrogen combine to form an atom of helium. Many years ago, I asked our family Dr. in Miami, if the elements in nature appeared together or if they evolved from a single element?
He paused for a while and finally replied, good question. Has anyone explained how the sun and planets can be so different although they were all the result of the Big Bang? How did the sun become the source of energy and center of our solar system?
Segments in a circle ‒ similar triangles
1. 2x + 3y = 6 intersects the y axis, x = 0, at (0, 2) and the x axis, y = 0, at (3, 0).
From (0, 2) to (3, 0) run = 3, rise = ‒ 2. Slope = ‒ 2/3. The line perpendicular to 2x + 3y = 6 has slope = 3/2. Note, (2/3)(‒ 3/2) = ‒ 1.
Run and rise form vector [run, rise]. Slope ‒ 2/3 = 2/(‒ 3) can be associated with the vector [‒ 3, 2] and slope 2/3 with the vector [3, 2].
Since ‒ 2/3 is the slope of the line 2x + 3y = 12, [‒ 3, 2] is parallel to the line and [2, 3], slope 3/2, perpendicular to the line. Please, study the figure.
2. If 3x + 4y = 12, [3, 4] is a vector perpendicular to the line and [‒ 3, 4] or [3, ‒ 4] is a vector parallel to the line with slope = ‒ 3/4.
3. If 3x ‒ 2y = 6, [3, ‒ 2] is a vector perpendicular to the line and [2, 3] is a vector parallel to the line which has slope = 3/2.
[a, b] is perpendicular to ax + by = c and [‒ b, a] or [b, ‒ a] is parallel to it. Vectors make it very easy to find the slope of a line and its perpendicular.
Since ΔCAE is similar, angles are equal, to ΔBDE the sides are proportional.
a / d = c / b, ab = cd or a/c = d/b, ab = cd.
Q1 If E is at O, the center of the circle, ab = cd = r2. T/F Q2 If E is A, ab = cd = 0. T/F
Answer Q1 T Q2 T
Q3 Is a / c = d / b or is a / d = c / b?
Cross multiply the above.
Note a, b, c, d are the sides of a Δ.
Lesson 5 In the following graphs, the car travels at 40 mi/hr. during the 1st hour and at 60 mi/hr. the remaining hrs.
mi/hr. mi/hr. mi/hr.
40 40 40
0 1 2 hr. 0 1 2 3 hr. 0 1 2 3 4 hr.
Fig 1 Fig 2 Fig 3
With these graphs, I hope you will understand how to measure the average speed using areas.
Average speed = distance travelled / time = shaded are / time.
Q1 In Fig 1, calculate the average speed.
Q2 What is the average speed in Fig 2?
Q3 What is the average speed in Fig 3?
Answer Q1 (40 + 60) / 2 = 50 Q2 (40 + 120) / 3 = 53.3 Q3 220 / 4 = 55 The longer a car travels at 60 mi/hr. the closer to 60 mi/hr. is the average speed.
Translation of a triangle
Humor – Jaimito
This is an amusing story about Jaimito, who is Venezuela’s Dennis the Menace. In a 3rd-grade science class, the teacher asked him to tell the class about his new scientific discovery. Jaimito said, “I have discovered that the moon is older than the sun.” The teacher asked him to explain how he arrived at that conclusion. “The moon can come out at night, during the day and occasionally runs away for several days; but the sun is only allowed to come out during the day. Teacher, no one has ever seen the sun come out at night”.
Fractions
1. T/F 3 1/2 equals a) 3 + 1/2 b) 3(1/2) c) (6 + 1)/2 d) 3/6 e) 3 and 1/2.
2. a) How many half pies in 3 pies? b) How many quarters of a dollar in $3.00?
3. T/F a) 3 1/2 = 3 + 1/2 b) 3 1/2 = 6/2 + 1/2 c) 7/ 2 = (6 + 1)/2 d) 7/2 = (3 + 3 + 1)/2 e) (3 + 3 + 1)/2 = 3/2 + 3/2 + 1/2 4. T/F a) 10/3 = (3 + 3 + 3 + 1)/3 b) 10/3 = (3x3 + 1)/3 c) (3 + 3 + 3 + 1)/3 = 1 + 1 + 1/3 d) (3x3 + 1)/3 = 3 + 1 5. T/F a) (1/2) / 2= (1/2 / (2/2), both top and bottom of 1/2 are divided by 2 b) In (1/2)/2 = 1/(2)(2) c) (1/2)/2 =1/4 6. a) (1/2)(1/2) = _____ , one half of a half pie is _____ of a pie b) A third of a pie cut in three pieces equals _____ pie.
7. T/F a) 3(1/2) = 3x1/2 b) 3(1/2) = 3x1/3x2 c) 3(1/2) = 1/2 + 1/2 + 1/2 d) 3(1/2) = 3/2
8. a) (2/3)((1/2) = 2(1/3)(__) b) (2/3)((1/2) = 2(1/3)(1/2) = 2/6 T/F c) (2/3)((1/2) = 1/3 T/F d) (2/3)((1/2) =(1/3)(__) 9. a) (2/3)(3/5) = (2x3)(__ /3)(___/5) b) (2/3)(3/5) = (2/5)(___) c) (2/3)(3/5) = 6/___)
10. a) The number of 1/4 of a pie in 1/2 pie is (1/2)/(1/4) = ___
Answer 1. a) T b) F c) T d) F e) T 2. a) 6 b) 12 3. a) T b) T c) T d) F 4. a) T b) T c) F d) F 5. a) F b) T c) T 6. a) 1/4, quarter b) 1/9 7. a) T b) F c) T d) T 8. a) (1/2) b) T c) T d) 1 9. a) 1, 1 b) 1c) 15 10. 2
In the following examples, I’ll show how to apply algebra to fractions ‒ reciprocals will be used frequently 11. Multiply both sides of (1/2)/(1/4) = x by (1/4), obtain ____ = ___x. So x = ____ .
12. Multiply both sides (1/3)/(2/3) = x by (2/3), → _____ = (2/3)x. multiply both sides 1/3 = (2/3)x by 6 .So x = ___
60 60 60
Triangle ABC is translated to ΔPRQ, which is identical or congruent to ΔABC.
Q1 Find the run and rise of ➔AP, ➔BR, ➔CQ.
If the run = ‒ 2 and the rise 5,
Q2 which of the following expressions can be used to find the translation of (x, y)?
a) (x, y) ➔ (x + run, y + rise) b) (x, y) ➔ (x + 5, y ‒ 2) c) (x, y) ➔ (x ‒ 2, y + 5).
Q3 Using the expression (x, y) ➔ (x ‒ 2, y + 5), find the translation of P(3, ‒ 3), R(6, ‒ 3), Q(6, ‒ 1).
Q4 If ΔPRQ is the translation of ΔABC, complete the expression (x, y) ➔ (x ____ , y ____ ).
Answer Q1 [ 2, ‒ 5] Q2 a) c) Q3 A, B, C Q4 + 2, ‒ 5.
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Lesson 6 Finding the address of point in space – rectangular coordinates Z Back wall
E D F •P (2, 3, 4)
Left wall 4 O
2 C Y A B
3 Floor X
When did the laws of nature originate?
Did the laws of nature appear before, during or after the Bib Bang? Can an explosion occur if there are no laws to control it? The laws of nature and mathematics must have a common origin. Why are we so interested in the origin of life and the universe, but not at all interested in the origin of mathematics or the laws of nature? Every explosion that occurs in nature is controlled by its laws. When a tree falls it produces a crashing sound whether it is heard or not. Nature can not violate its laws.
Galileo and the Leaning Tower of Pisa
By simultaneously dropping a 1 kg and a 10 kg mass from the Leaning Tower of Pisa, in Italy, Galileo, 1564 – 1642, proved they hit the ground together. About 2000 years earlier, Aristotle stated the heavier object would hit the ground first.
Why did it take so long to show Aristotle was wrong?
Unit Circle
Corner O is a point common to the right wall, left wall and floor. The equation of the:
black wall is x + 0y + 0z = 0, which intersects the X axis at x = 0.
left wall is 0x + y + 0z = 0, which intersects the Y axis at y = 0.
floor is 0x + 0y + z = 0, which intersects the Z axis at z = 0.
Q1 The left wall includes all possible values of x and z. T/F Q2 Can you identify each of the planes?
a. x + 0y + 0z = 2 b. 0x + y + 0z = 3 c. 0x + 0y + z = 4 Point P(2, 3, 4) belongs to the planes x = 2, y = 3 and z = 4.
Q3 Find the coordinates of these points 1. A 2. B 3. P
Q4 To which wall are these vectors perpendicular: [1, 0, 0], [0, 1, 0], [0, 0, 1].
Answer Q1 T that is why x and z are multiplied by 0 Q2 a. ABF b. BCD c. EFD Q3 1. (2, 0, 0) 2. (2, 3, 0) 3. (2, 3, 4) Q4 Black wall, left wall, floor.
Galileo had an ingenious idea; Instead of dropping the object, he allowed the object to roll down an inclined plane.
Q1 What is the advantage of an inclined plane?
Answer Q1 it slows down the process
Q1 The circumference of a circle of radius r is C = _____.
Q2 In the figure, hat does π/2 measure?
Q3 Simplify
a) 30/360 = _____ b) 90/360 = _____ c) 150/360 = _____ d) 270/360 = _____
Q4 What do these expressions measure?
a) (30/360)(2π) = _____ b)( 90/360)(2π) = _____ c) (150/360)(2π) = _____
d) (270/360)(2π) = _____ e) (360/360)(2π) = _____
Q5 If the radius is r instead of 1, what do the following expressions measure?
a) (30/360)(2πr) = _____ b)( 90/360)(2πr) = _____ c) (150/360)(2πr) = _____
d) (270/360)(2πr) = _____ e) (360/360)(2πr) = _____
Q6 If the arc length is L, angle α and radius r, the formula L = (α/360)(2πr) can be used to _______.
Q7 In the expression L = (α/360)(2π), the variables are _____ .it is __ function.
Answer Q1 2πr Q2 arc Q3 a) 1/12 b) 1/4 c) 5/12 d) 3/4 Q4 a) 300 arc b) 900 arc Q5 a) 300 arc b) 900 arc Q6 measure L for different α Q7 L, α, a linear
Lesson 7 Galileo’s Imaginary Experiment – Law of Inertia Our imagination, properly used, can be a very powerful tool.
Galileo, 1564 – 1642, introduced the idea of doing experiments, which established the basis of the scientific method; A hypothesis must be verified through an observable experiment or event so it may become a theory. Yet, what he did here is not an observable phenomenon!
In the first two figures the ball can reach its original height provided the ball does not lose energy. In the second figure, the angle of the inclined plane is decreased so the ball travels a longer distance to reach the original height. Now here is where he showed his ingenuity – If the angle continues to decrease the ball travels a longer distance to reach the original height. In the last figure, the ball never stops rolling because it never reaches its original height. Cool!
You might wonder, how is this related to the Law of Inertia? When a ball rolls along the floor it comes to a stop because of the presence of friction. In the absence of friction or resistance, the ball rolls forever because there is no force to change its motion.
Inertia means resistance to change.
This experiment without the use of the imagination makes no sense. There are no frictionless surfaces. However, in space where only 4% is occupied by matter, there is no resistance to motion. Our planet does not encounter friction.
Law of Inertia
An object either at rest or moving in a straight line continues in that state unless another object exerts a force on it. This implies that on its own, an object cannot change its state of rest or motion. As the ball rolls downhill or uphill, it is interacting with the earth through gravity. On its own, the ball cannot reach a height greater than its original height H. Can an organism through evolution improve itself, on its own? Would not this contradict the law of entropy? It states that in any event in nature there is a loss of energy. An improvement implies a more complex structure, which requires additional energy.
Doesn’t the Big Bang violate the Law of Inertia and Entropy?
Suggestion
One should become familiar with Newton’s second and third law before discussing the first law.
Similar triangles
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Lesson 8 Atmosphere ‒ An important law of nature
Sealed straw Air
Water
Air pushing up
Barometer ‒ Under normal atmospheric conditions the column of mercury is 76 cm high
The orange tree that wanted to be uprooted
A farmer planted and orange tree which produced delicious oranges. On a certain occasion the orange tree told the farmer that it wanted to be uprooted so that it could be free to move around like him. The farmer tried to explain, without success, that the roots make the tree stable and nourished it. The farmer loved the tree and would not uproot it. One day a stranger, who had little regard for trees, came by; he was willing to uproot the tree. His lifestyle was – If you like it do it–
.Like the tree many of us want to uproot our lives, and like the stranger, who only lives for the pleasures of the moment, do things without considering the consequences.
Pressure inside a water hose
The force pushing the liquid inside a hose changes direction as the hose is bent. How can this force change direction?
Brain Teaser
When a straw, with water inside of it, is removed from a glass with water, the water remains inside the straw provided its top is sealed. How can the air under the straw overcome the force of gravity on the water?
Air always tries to expand; it hungers for space. When you lift a straw from a glass filled with water with its top sealed, perhaps by your thumb, some of the water remains inside the straw. The air above the straw pushes down on the thumb, not the water. Since air tries to expand, the air under the straw pushes the water up and keeps it inside! When the thumb is removed, the air on top of the straw pushes down on the water cancelling the upward push from the air below the straw. Now the weight of the water, the resultant force, pulls it out.
The small amount of air trapped under the thumb exerts an insignificant pressure.
.
A column of water 34 ft (10.4 m) high and area (1in)(1in) = 1in2 weighs 14.7 lb. (6.7 kg).
A column of air 20 miles (30 km) high and area (1in)(1in) = 1in2 weighs 14.7 lb.
A column of mercury 2.5 ft = 76 cm high and area (1in)(1in) = 1in2 weighs 14.7 lb.
Q1 Superman is standing on top of a 50-foot building with a 50-foot straw of area (1in)(1in) = 1in2 hoping to drink water from a pool on the ground floor. When he sucks out all the air, the water rises ___ ft.
Answer Q1 the downward force of the air in the atmosphere is 14.7 lb. on every square inch of the pool. This force is transmitted throughout the liquid pushing up the water inside the straw to a height of 34 ft, not 50 ft.
Sorry, Superman!
If the pool is full of mercury, the column of mercury in the straw reaches a height of 76 cm. Mercury is 13.6 times heavier than air, so the column of mercury is 13.6 times smaller:34/13.6 = 2.5 ft = 76 cm.
Q2 If the area of the straw is reduced to one half, would the liquid inside the straw reach the same height?
The force per unit area exerted by the atmosphere on the mercury spreads throughout it and sustains the column of mercury.
vacuum Mercury weight
Pressure or force on a unit area exerted on a liquid or gas is transmitted throughout the fluid.
Plungers, used to unplug toilets, rely on this principle.
Force diffused by mercury Weight of the atmosphere
Lesson 9 Longitude and latitude
Basic geometry ‒ parallel lines and exterior angle of a triangle
Two secants intersecting outside the circle
The Prime Meridian goes through Greenwich, a
charming and historic area of London, capital of England.
North and South America are located west of the Prime Meridian, whose longitude is 00.
To locate point B, whose longitude is 900 West and latitude 300 North, start at M and move along the equator to A. <MOA = 900. Both A and B have longitude 900. Now move north to B, <AOB = 300. B is in New Orleans.
Q1 The Prime Meridian is a circle. T/F
Q2 When a door swings 900 every point on it rotates 900. T/F
Imagine there are many people at equator who are going to walk north at the same pace.
Q3 How many degrees have they walked when they reach the North Pole?
Q4 If they walk 300, they form a ______
When walking west, time is set back one hour every 150. Q5 When walking half a circle, 1800, time is set back _____ hours.
Answer Q1 F, a semicircle Q2 T Q3 900 Q4 circle Q5 180/15 = 12 hours.
Q1 If the alternate interior angles are equal, which lines are //?
Q2 Which line segment is the transversal for the alternate interior angles?
Q3 Which lines have the same inclination?
Q4 Is AB a transversal?
Q5 1 + 2 + 3 = 1800, hence 1 + 2 = 1800 ‒ ____
Q6 4 + 5 = 180 ‒ ____
Q7 Conclusion __________.
Answer Q1 AC and BE Q2 CB Q3 AC and BE, they are // Q4 yes Q5 3
Q6 3 Q7 1 + 2 = <CBD Corresponding angles
Alternate interior
<CBO, exterior Angle = <A + <C
We want to prove that a1b2 = c1d2 or a1 / d2 = c1 / b2
Are ΔEAD and ΔECB similar?
<E is common, <CDA = <CBA = (1/2)(arc CA). If two angles are similar,
<EAD = <ECB.
Since the angles are equal, the sides are proportional, a1 / d2 = c1 / b2, better yet, ab = dc.
Q1 If ED is tangent at D, where AD = a1, and a12 = ____ .
Q2 If AD is tangent at D, and if EB is tangent to the circle at E, a12 = a12. T/F Answer Q1 c1d2 Q2 T
11
Lesson 10 Area of a triangle and rectangle
Area = bh, is the area of a parallelogram of base b and height h and the area of a rectangle with base b and height h.
Let me explain why the alternate interior angles are equal.
Walk from D to A and there turn towards C; at C you return to a direction parallel to AD. Since the final direction CB is parallel to DA, <DAC = <ACB. They are called alternate interiors angles. As you turn and return, you do and undo, To better appreciate why ΔADC ≅ ΔCBA, rotate ΔADC 1800 clockwise, A overlaps with C, D with B, and C with A.
Comparing the two triangles; AC = CA, AD = CB, <DAC = <ACB, which are the contained angles, SAS.
The two equal sides, S, are two slices of bread, and the angle, A, is the filling of the sandwich SAS.
Since ΔADC ≅ ΔCBA, the corresponding sides are equal. DC = AB. Note the symbol ≅ means congruent.
Q1 Prove that DC // AB.
Basic math
1. The following expressions are either true or false.
a. 1/2
1/2= 1/4
1/4 b. 1/2
1/2= 1 c. 1/2
1/2= (1
2)(2
1) d. 1/4
1/2= (14) (𝟐𝟏) 𝒆. 1/21/3 > 1 f. 1/2
1/4 < 1 2. a. 1/2
1/4= ___ b. 1/4
1/2= __ c. 1/3
1/2= (1
3) (21) = d. 1/23 = 1 e. 1
1/4= ____ f. 2/3
1/3= ___
3. in the fraction (1/2)/3, does 3 divide 1, 2 or both?
Answer 1. a. T. b. T c. T d. T e. T f. F 2. a. 2 b. 1
2 c. 2
3 d. 6 e. 4 f. 2 3. If 3 divides both, (1/2)/3 = (1/3)(2/3) = 1/2. If (1/2) pie is divided among 3, each one would get 1/6. (1/2) / 3 = 2(1/2) / (2)(3) = 1 / (2)(3).
In physics we frequently encounter (1m/sec) / (1 sec) = (1sec)(1m/sec) / (1sec)(1sec) = (1m) / (1sec)(1sec) = 1 m / 1sec2. The rectangle DEFC has base b and height h.
Q1 If b = 10 cm and h = 10 cm. how many cm2 cover the area of the rectangle DEFC?
Q2 Since ΔDAE ≅ ΔCBE, area of rectangle DEFC, bh, equals area of //gm DABC. T/F Answer Q1 100 cm2, there are 10 rows with 10 1 cm2 in each row. Q2 T
Given that in ABCD, AD = BC and AD // BC. Let’s prove that AB = CD.
The given is marked in the figure.
Proof
In order to prove that AB = DC, we need to find two congruent triangles. Drawing AC forms ΔABC and ΔCDA.
Since AD // BC, <DAC = <BCA.
ΔABC ≅ ΔCDA SAS, hence AD = CD. Note ≅ means congruent.
Q1 b is the base common to ΔABC and ΔABD. T/F Q2 Is line CD parallel to AB?
Q3 If the height is drawn at D, it is perpendicular to AB extended. T/F Answer Q1 T Q2 we don’t know Q3 T
Lesson 11 Breeze at seashore reverses direction at nighttime ‒ keep in mind, hot air rises cold air sinks Since water has a larger heat capacity than sand, it requires more heat to change its temperature. It heats and cools more slowly than land and sand because it has to absorb or release more heat to change its temperature. Air warms five times faster than water. Sand warms 5 times faster than water,
D C
AO BS
During the day land is warmer and heat flows towards the ocean. At night the ocean is warmer and heat flows towards the land. Regions near the coast will stay cooler in the summer during the day and warmer in winter during the night.
Convection cycle
The above circular movement of the air at the beach is known as a convection cycle. The word convection refers to the movement of masses of air due to changes in temperature or density. Heat is transferred through convection. Convection cycles are also present when water boils.
Q1 When water is boiling in a pot, why do bubbles only seem to come from the bottom?
Q2 Hot water is denser than cold water. T/F
Q3 As water boils, how does the cold water at the top get heated?
Answers Q1 the bottom of the pot is being heated Q2 F at the beach the water is cooler at the bottom Q3 As the hot water rises it leaves an empty space which is replaced by colder water from the top. Also, the rising hot water mixes with cold water.
Math riddle
a) How many four quarters in a dollar?
b) Suppose you spend $12.00 and purchase x lb. of bananas at &2/lb. and y lb. of apples at $3/lb, so 2x + 3y = 12. Can x and y can increase simultaneously?
c) A balloon filled with air expands and explodes when placed over a stove, but not too close to it, but if the balloon is filled wit water, nothing happens.
Repeat the experiment but place thermometers inside the balloons, which are transparent.
Can you explain why the balloon filled with air exploded?
Which one, air or water has a greater heat capacity?
Answer air heats up 1000 times faster than water. Water has a much greater heat capacity.
How did air and water acquire these distinct properties? As we try to understand the laws of nature, the more we will be convinced that our universe can not be the result of an accidental explosion.
Use your imagination. Point AO is on the surface of the ocean and point BS is on the surface of the sand at the beach.
Q1 During the day, since the sand warms faster than the water, hot air over the sand rises before it does over the ocean. T/F
Q2 As air over the sand rises, it draws in air from the ocean. T/F Q3 As air rises from B to C, its temperature _______.
Answers Q1 T, sand warms faster Q2 T, the rising air leaves behind an empty space drawing air from Ao Q3 decreases ‒ it gets cooler. During the day, heat rises from B to C, and eventually flows towards D.
13
Lesson 12As air speeds up, its pressure decreases
As a result of the collision of randomly vibrating air particles on surface A, a force F is exerted on both sides of it. The ratio F/A measures the pressure exerted on both sides of area A. As the speed of air increases, there are less randomly vibrating particles colliding with the surface, thus pressure decreases.
Sucking out the air in a straw in a glass of water reduces the pressure in the upper end of a straw, and water is pushed up by the higher pressure at the bottom of the straw. Perhaps this will help to understand why an airplane is sustained in the air by the difference in pressure on opposite sides of its wing. Let me explain. The shape of the wing forces the air on top of the wing to go faster and decreases its pressure. Since the speed of the air in the flat bottom of the wing is slower, the pressure of the air is greater. Consequently, the plane is being sucked up like the water in a straw!
Lift of an airplane wing
Similarity between the wing of an airplane and the sail of a sailboat
The shape of the wing of an airplane is like the shape of the sail of a sailboat. Since the boat sails into the wind, the air in front of the sail is moving faster than the air behind the sail. The lower pressure in the front sucks the sailboat forward.
That is cool!
Water flowing through a pipe
Volume of a pipe V = (area)(length). V1= (4 in2)(10 in) = 40 in3. V2 = (1 in2)(40 in) = 40 in3 Assume water flowing through section 1 fills volume V1 in 1 sec.
a. Water emerging through section 2 fills volume V2 in ___ sec.
b. In volume V1 the water entering travels ___ inches in 1 sec.
c. What is the speed of the water in the first pipe?
d. The speed of water in the second pipe is four times greater. T/F e. Is the amount of water between 1 and 2 always the same?
f. What do (4 in2)(10 in/1sec) and (1 in2)(40 in/1sec) measure?
g. Why is the pressure at 2 smaller than the pressure at 1?
Answer a. 1 b. 10 in c. 10 in/sec d. T, 40 in/sec e. Yes, amount entering = amount leaving; the flow is steady f. The volume of water entering and leaving the pipe per sec is (40 in3)/1sec. Area times speed is the same along the pipe and measures volume per sec entering or leaving. g. Speed increases.
The small diameter of a nozzle at the end of a water hose forces the water to emerge at a greater speed.
Riddle
a) The volume of water entering a hose per unit time equals the volume leaving the hose per unit time regardless of the size of the nozzle at the end of the hose. T/F
b) As a sailboat sails into the incoming wind, why isn’t the sail pushed back?
c) Why does the speed of air exiting a balloon increase when you squeeze it? Perhaps it has no choice!
Lift
Since air cannot be not stored, the volume of air entering through AB per second equals the volume leaving through CD per second. The flow is steady. This also happens when water flows between two rocks in a river, or when wind blows between two buildings.
Because of the curved upper portion of the wing and the stagnant air, the air encounters a narrow gap, forcing it to speed up. The air circulating under the wing does not encounter any obstacles and its speed and pressure remain the same.
The lower pressure at the top sucks the wing upwards.
Remember, as air speeds up there are fewer randomly vivrating particles, and pressure decreases.
The layer of air above CB, is stagnant. This layer together with the top of the wing squeezes the air flowing above it, thus forcing it to speed up.
C B Llift
D A
A
Area = 1 in2
D A
Ai Air flows less rapidly and evenly under the wing. r
Lift
Lesson 13 Fractions
If person A paints 1/4 of a circle in 1 day, it takes person A ___ days to paint the circle.
Note Though this problem is easy to solve, the procedure is worth remembering.
Solution 1
The symbol => stands for therefore.
Solution 2
Let x be the number of days required to paint the circle.
(x days to paint the circle)(1/4 of a circle per day) = the whole circle (4/4) => x(1/4) = 1, x/4 = 1 Q1 If 𝑥
4 = 1, x is either greater than 4, equal to 4 or smaller than 4?
x = 4(1) = 4
Answer x is equal to 4.
Q2 (1
4)(x) = 1: In words, one-quarter of what number is one?
Answer 4 Solution 3 The reciprocal of (
𝟏 𝟒)𝐜𝐢𝐫𝐜𝐥𝐞
𝟏𝐝𝐚𝐲 is 𝟏𝐝𝐚𝐲
(𝟏𝟒)𝐜𝐢𝐫𝐜𝐥𝐞 = 𝟒𝐝𝐚𝐲
𝟏𝐜𝐢𝐜𝐥𝐞 Note units help solve and understand problems.
Example 2
If John paints 2/3 of a circle in one day, how many days does it take him to paint the circle?
(2/3) of a circle / 1 day is the speed at which he paints.
The reciprocal of (2/3) circle / 1 day is 1 day / (2/3) circle = (3/2) day / 1 circle Interesting procedure.
It will take 3/2 = 1.5 days to paint the circle Exercise
1. (x days to complete the job)(2/3 of the job per day) = 1, the complete job. x = ___ . 2. In one day 2/3 of the job is done. The remaining 1/3 is done in ____ .
3. (1/3 of the circle needs to be painted) / (2/3 of the circle painted per day) = _____
Answer 1. x = 3/2 2. 1/2 day 3.1/2 day to finish the job
(Job to be done) ÷ (speed or job/day) = numbers of days to complete the job Reviewing fractions using angles in the unit circle
Q1 How many times is 3600 greater than 900? 900 is _____ of 3600. Simplify 90/360.
Q2 How many times is 1800 greater than 300? 300 is ____ of 1800. Simplify 30/180.
Q3 How many times is 900 greater than 300? 300 is ____ of 900. Simplify 30/90.
Q4 How many times is 1800 greater than 600? 600 is ___ of 1800. Simplify 60/180.
Q5 900 is one-half of 1800, which is one-half of 3600. 900 = (__)(1/2)3600. Q6 300 is one-third of 900, which is one-quarter of 3600. 300 = (1/3)(__)3600. Q7 300 is one-half of 600, which is one-sixth of 3600. Hence 300 = (1/2)(__)3600. Q8 600 is one-third of 1800, which is one-half of 3600. Hence 600 = (1/3)(__)3600.
In four days, the portion painted is 1
4+ 1
4+ 1
4+ 1
4 = 4
4 , the whole circle.
To add fractions, they must have the same denominator
4 4 = 4(1
4) = 1. Note since 4(1
4) = 1, 4 and 1
4 are reciprocals.
(x days to complete the job)(fraction of the job done per day, 1/4) = x(1/4) = x/4 = 1 = 4
4 (total job)
15
Lesson 14 Word problem The sum of numbers A and B is 10.
1. Which of the following equations is true?
a. A + B = 10 b. A – B = 10 c. A = 10 – B d. A = B – 10 e. B = 10 – A
2. If A + B = 10, as A increases by 1 a. B remains the same b. B increases by 1 c. B decreases by 1.
Answer 1. a. c. e. 2. c.
3. Ordered pair (A, B)
(0, 10) • (5, 5) • (10, 0) Equations 1st rule Isolating x
1. If x + 5 = 10, which term is greater x or 10? Does x = 10 – 5 or x = 10 + 5?
Note If 5 is subtracted from both sides, x + 5 – 5 = 10 – 5, => x = 5 2. If x – 5 = 10, which term is greater x or 5? Does x = 10 − 5 or x = 10 + 5?
Note If + 5 is added to both sides, x – 5 + 5 = 10 + 5, => x = 15
3. x + 5 = 15 may be interpreted as follows: after receiving 5 dollars you have 15 dollars. Solving the equation x + 5 = 15 allows you to find how much money you had.
4. x – 6 = 18 may be interpreted as follows: If after spending 6 dollars you still have 18. How much money did you have originally?
Answer 1. 10, x = 10 – 5 2. x, x = 10 + 5 3. x = 15 – 5 = 10 4. x = 18 + 6 = 24
2nd rule Transferring a factor to the other side of an equation.
5. 𝑥
4 = 5. a. Which is greater x or 5? b. How many times is x greater than 4?
6. 𝑥
4 = 5. Is x = 5
4 or x = 4(5)?
Answer 5. a. x, 5 b. 4 6. x = (4)(5)
Someone said: “Do not argue with an idiot. He will drag you down to his level and beat you with experience.”
Signs of maturity in a person
1. Get involved – say no to inertia – defend what is right.
2. Accept responsibility – take ownership – avoid excuses.
3. Pursue the vision – do not lose sight of the goal.
Winston Churchill, 1874 – 1965
British Prime Minister Winston Churchill provided an incredible leadership during the Second World War. I think his sharp tongue and sense of humor, helped sustain a spirit of hope during the war.
Which statement, 1 or 2, conveys more information?
1. If x = 3 then x2 = 9
2. If x2 ≠ 9 then x ≠ 3 and x ≠ ‒ 3 1. If it rains, then the grass will get wet.
2. If the grass did not get wet, then it did not rain, there was no dew, no one watered the grass, etc.
If 3x = 12, 12 is 3 times greater than x and x = 12
3. 3, which multiplies x on the left, divides 12 on the right.
If 𝑥
2 = 8, 8 is half the size of x, so x = 2(8).
The first number of the ordered pair (0, 10) is 0, and is represented by A, the second number 10, is represented by B.
a. From (0, 10) to (10, 0) A has increased by ___ and B has ____ by 10.
b. When number A increases by 5, B decreases by ___
(5, 5) is another pair of numbers whose sum 10.
c. From (0, 10) to (5, 5) the ratio (5 – 10) / (5 – 0) = ____
Answer 3. a. 10, decreased b. 5 c. – 1
In an equation, when a term changes side it changes sign: 5 + 3 = 8, 5 = 8 – 3
Lesson 15 Word problems A’s age + B’s age = 50
If A’s age is x then B’s age is (50 – x).
1. What does the number 50 represent?
Answer The sum of their ages 2. If A is 20 years old, how old is B?
Answer 50 – 20 = 30
If A’s age is x and B’s age is (50 – x), do we know who is older?
Answer No, there are many numbers whose sum is 50.
3. The sum of the ages x + (50 – x) = ___.
Answer 0x + 50 = 50
4. If their ages are equal, x = 50 – x, and x = ____.
Answer x = 50 – x => x + x = 50 => 2x = 50 => x = 25 A is 25 years old.
Note x + x = x(1 + 1) = 2x
5. If A is 10 years older than B, which statement is true?
a. A’s age – B’s age = 10 b. A’s age + 10 = B’s age c. A’s age = B’s age + 10 Answer a. c.
In the equations in problem 5, we replace A’s age by x and B’s age by (50 – x).
a. x – (50 – x) = 10 b. x + 10 = 50 – x c. x = 50 – x + 10
Solving these equations and determine which ones are correct a. x – (50 – x) = 10 Note – (50 – x) is the negative of 50 – x, which is – 50 + x.
x – (50 – x) = 10 => x – 50 – (– x) = 10 => x – 50 + x = 10 => 2x = 10 + 50 => 2x = 60 =>x = 30 b. x + 10 = 50 – x Note To solve, place terms in x in one side and numbers on the other
x + 10 = 50 – x => x + x + 10 = 50 => 2x + 10 = 50 => 2x = 50 – 10 => 2x = 40 => x = 20 Note Not true, A is 10 years older.
c. x = 50 – x + 10
x = 50 – x + 10 => x + x = 50 + 10 => 2x = 60 => x = 30 Note a. and c. are correct
Relationship between an angle at the circumference and an angle at the center of a circle
Q1 Why is 2x + w = 1800? For the same reason 2y + v = 1800. Q2 <AOD = 2(x + y) = 2<ACB T/F Q3 Arc AOC has 2x + 2y degrees. T/F
Q4 The degrees in <ACB is half of the degrees in arc ADB.
Answer Q1 adjacent angles or angles is a Δ Q2 T Q3 T Q4 T
When a car drives in a circle, it is always pointing along the tangent to the circle. At A <TAO = 900, where AT is tangent at A. As the car drives from A to B, the tangent AT rotates angle 2(X + Y), which is equal to <AOB. Since ΔADB is isosceles. <DAB = X + Y = <ACB.
When B is at A, <DAB = 00 = <ACB. When <BAD = 900, <ACB = (1/2)1800 = 900. As seen from C,
<TAB, measures the rotation of chord AB, which the rate of rotation of AB as seen from C. Both A and C are on the circumference.
17
Lesson 16 How lakes freeze during winter The following imaginary simple experiment helps to understand how lakes freeze in winter.
These three jars contain water at 100, 20 and 40C.
100C 100C 20C 40C 20C
40C
Assume that when poured into a large container, water at 100C, water at 20C, and water at 40C are separated, as shown here. Do you know why water at 40 is at the bottom and water at 100 is at the top?
Now suppose the temperature of the lake is 100C. We are going to drop two water balls into the lake, one at 150C and the other at 50C. These two special balls retain their temperatures and don’t mix with water at a different temperature. The ball at 50 sinks but the ball at 150 floats.
Now suppose the whole lake is at 20C. We are going to drop two water balls into the lake, one at 80C and the other at 40C.
The ball at 80 floats but the ball at 40 sinks.
Now suppose the whole lake is at 40C. We are going to drop two water balls into the lake, one at 80C and the other at 20C.
The outcome, both balls float.
This occurs because
When liquids of different densities are mixed, the densest or most compact one is at the bottom and the least dense at the top.
Suppose that while the lake is at 40C the temperature of the air around the lake decreases from 40C to 20C. The upper part of the lake is at 20C and the bottom remains at 40C.
If the temperature of the atmosphere continues to decrease, the upper part of the lake will freeze, but the water at the bottom of the lake remains at 40C.
The unfrozen bottom of the lake makes life possible, both for fish and plants.
Perhaps we need to be more cautious in science
1. For centuries, it was believed that the earth was flat and that it was the center of our solar system.
2. After nuclear energy was discovered, many scientists were convinced it would be so safe and cheap that it would drastically change our industrialized world. However, two events created doubts about the convenience and safety of nuclear energy.
a. The thermal pollution of rivers in extremely warm weather caused the shutdown of nuclear plants in France and expensive electricity was purchased from neighboring countries.
b. The tragic events at nuclear plants in Russia and Japan, 3. How can we protect nuclear plants from a terrorist attack?
Mankind has always made mistakes and will continue to make them; we are humans.
1cm3 of water at 40C is heavier than 1cm3 of water at any other temperature.
Lesson 17 Fahrenheit and Celsius Temperature scales
Q1
a. The freezing temperature of water is ___ 0C or ___ 0F.
b. The boiling temperature of water is ___ 0C or ___ 0F.
c. On the Fahrenheit scale, each small division corresponds to ___ degrees.
d. 500C corresponds to ___ 0F.
e. When the temperature rises from 00C to 500C, it rises ___0F.
f. What does 𝟗𝟎𝑭
𝟓𝟎𝑪 measure?
Q2
Fill in the blanks.
00 x 10000C 320 y 21200F
Q3 F = (9/5)C + 32. Find F a. when C = 00 b. C = 1000.
Q4 From the equation F = (9/5)C + 32, C = _________
Answer Q1. a. 0, 32 b. 100, 212 c. 2 d. 122 e. 90 f. increase in degrees F per unit increase in degrees C Q2. a. 180 b.
180 c. 100 d. T e. x = 100
180(𝑦 − 32) Q3. a. 320F, b. 2120F Q4 F – 32 = 9
5 C, 5(F – 32) = 9C. C = 5
9(C – 32) Proportions ‒ why cross multiplication works Change the following proportion 2
3 = 𝑥
5 into a new proportion where the denominators are equal.
Answer: 10
15 = 3𝑥
15 Note since the denominators are equal, 10 = 3x.
We can cross multiply and obtain the same result. Do you understand why cross multiplication works?
Source of energy for the atom
It is amazing how much energy is stored in the nucleus of atoms. Where is the source of this energy? That is, where did it come from? It could not have been from the Big Bang; it was an explosion. The Big Bang theory was conceived in 1927 by a Belgian priest named Georges Lemaitre, who was a physics professor. Since the neutron had not been discovered, the knowledge of the nucleus was incomplete. Do we know how nature assembles an atom?
Have you considered how the Big Bang stored so much energy? The crucial question is this, can a single source of energy produce nuclear energy, fossil fuel energy, wind energy, solar energy, …?
a. 𝒙−𝟎
𝟏𝟎𝟎−𝟎= 𝒚−𝟑𝟐 b. 𝒙−𝟎
𝒚−𝟑𝟐=𝟏𝟎𝟎−𝟎 c. 𝒚−𝟑𝟐
𝒙−𝟎 =𝟏𝟖𝟎 d. if 𝒙
𝟏𝟎𝟎= 𝒚−𝟑𝟐
𝟏𝟖𝟎 , x = (5
9)(y – 32) T/F. e. 𝒚−𝟑𝟐
𝟏𝟖𝟎= 𝒙
𝟏𝟎𝟎 , x = __
19
Lesson 18 Horizontal circles measure latitude. Vertical semicircles measure longitude
Q1 The circle 00 latitude corresponds to the _____.
Q2 900 latitude north corresponds to _______.
Q3 The semicircles 150, 450, .., 600, that measure longitude, are drawn between the ____.
Q4 The semicircles 1800W and 1800E coincide. T/F
Q5 What is the longitude of the IDL or International Date Line?
Q6 When crossing the IDL, does the date change?
Q7 If the earth is flat there is only one time zone. T/F
Q8 Two persons at the equator are 150 apart.As they walk north along their lines of longitude, their angular separation decreases. T/F
Answer Q1 equator Q2 NP Q3 poles Q4 T Q5 1800 W or 1800 E Q6 T one day Q7 T Q8 F, always 150
If a ship traveling west crosses the International Date Line, its calendar is moved behind one day, and if it crosses the IDL moving east its calendar is moved ahead one day.
Latitude and longitude determine the angular position of a point on the surface of the earth. The address of a point on the surface of the earth is the intersection of a circle with a semi-circle.
The point on a sphere is specified by the radius of the sphere, latitude and longitude.
Critical thinking does not come naturally ‒ we have more nonsense than common sense
If Mars orbits around the Earth and the earth is the center of our solar system, Mars, as seen from the earth, always moves in a forward direction. But both Earth and Mars orbit around the sun. The Earth’s orbit is smaller than the orbit of Mars and takes 365 days to make a complete rotation, while Mars takes 688 days. Suppose Mars is ahead of the Earth.
As the Earth catches up and passes it, planet Mars begins to move in the opposite direction. The retrograde motion of Mars finally convinced scientists that the sun, not the earth, is the center of our solar system. This sounds simple, but brilliant men debated this for years. Many astronomical observations had to be made before they were able to arrive at the right conclusion.
Rotation of tangent and secant in a circle
Rotation of a chord
Lines 12 and 34 are diameters.
Q1 Sine lines 15 and 37 are tangent to the circle, <51O = 900 and <62O
= 900. T/F
Q2 Since 15 is tangent at 1, it is ⊥ to all the lines passing through 1. T/F Q3 The tangents at 1 and 3 make an angle equal to <1O3. T/F Q4 The observer at 1 sees the secant 13 rotate <1O3 as a ball at 1 moves along the arc of the circle to 3. T/F
Q5 <1O3 = (1/2)(arc 13 + arc 24). T/F
Q6 As seen from O the tangents to the circle at 1 and 3 change at the same rate as secant 13.
Answer Q1 T Q2 F only one perpendicular Q3 T Q4 half the rate Q5 T Q6 half the rate
As seen from B, as the chord AB rotates around E to position CD, point B sees chord BD rotate angle (1/2)(arc BD).
As seen from A, as the chord AB rotates around E, point A sees chord AC rotate angle (1/2)(arc AC).
Both rotations around E are clockwise and are equal to (1/2)(arc AC + arc BD).
