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Slope Concept through Middle School to College• Slope is a ratio or a proportion – the ratio of the
rise to the run• Slope = Rate of change• Velocity = rate of change = distance/time• Slope corresponds to “instantaneous velocity”• How would we talk about the slope of a “curve”?• Derivatives
Slope Concept through Middle School to College• Differential equations• Physics – engineering – forensic science -
geology – biology – anything that needs to comprehend rates of change(If you tie a string to a rock and you swing it around your head and let it go, does it continue to travel in a circle?)
• Zooming in on the graph• Local linearity• Understanding lines and linear equations• Brings us back to slope
The concept of Area• Area is based on square units.• We base this on squares, rectangles and
triangles.• Area of a square: s2
• Area of a triangle: ½ bh• Area of a triangle (Heron’s Formula): triangle
has sides of length a, b, and c. Let s = (a + b + c)/2. Then ( )( )(Are )a s s a s b s c
2sin
2n
n
n Approx n Approx8 2.828427124 40 3.128689302
10 2.938926262 60 3.135853896
12 3.0 80 3.138363830
14 3.037186175 100 3.139525977
16 3.061467460 150 3.140674029
18 3.078181290 180 3.140954703
20 3.090169944 200 3.141075908
P is the point at which the tangent line to the curve is parallel to the secant QR.
Where does the line intersect the parabola?
2ax mx b 2
2 2
1 2
0
4 4,
2 2
ax mx b
m m ab m m abx x
a a
Points of IntersectionNow, we can find the points of intersection
of the line and the parabola, Q and R.
2 22
2 2 2 2
2 2 2
4 2, )
4
(2
2 24 2
,
m mQ x y ax
a
m
m ab aby
m ab m abm m aQ
ab
a
2 22
1 1 1 1
2 2 2
4 2, )
4
(2
2 24 2
,
m mR x y ax
a
m
m ab aby
m ab m abm m aR
ab
a
Slope of the Tangent Line
The slope of the tangent line at a point is twice the product of a and x.
2m ax
2x
ma
22
1 2, ) )( ( , ,2 4m m
P p x axa a
p
Area of Triangle
It does not look like we can find a usable angle here.
What are our options? (1)Drop a perpendicular from P to QR and then
use dot products to compute angles and areas.
(2)Drop a perpendicular from Q to PR and follow the above prescription.
(3)Drop a perpendicular from R to PQ and follow the above prescription.
(4)Use Heron’s Formula.
Use Heron’s Formula2 2
1 2 1 2( , ) ( ) ( )x xp d Q R y y 2 2( 4 )(1 )m ab
pa
m
2 21 1 1 2( , ) ( ) ( )x pq d P R y p
2 2 2( 4 )(4 4 5 4 4 )
4
m ab ab mq
m m ab
a
2 21 2 2 2( , ) ( ) ( )p xr d P Q p y
2 2 2( 4 )(4 4 5 4 4 )
4
m ab ab mr
m m ab
a
Now, the semiperimeter is:2
sp q r
2
2
2 2
2 2
44 1
8
4 4 5 4 4
4 4 5 4 4
m abm
a
ab m m m ab
ab m ab
s
m m
Uh – oh!!!!Are we in trouble? Heron’s Formula states
that the area is the following product:
( )( )( )s s p s q rK s
This does not look promising!!
2 22 2 2
4
2 2 2
2 2 2 2
2 2 2
2 2 2
2 2
( 4 )4 1 4 4 5 4 4
4096
4 4 5 4 4 4 1
4 4 5 4 4 4 4 5 4 4
4 1 4 4 5 4 4
4 4 5 4 4 4 1
4 4 5 4 4 4 4 5
m abK m ab m m m ab
a
ab m m m ab m
ab m m m ab ab m m m ab
m ab m m m ab
ab m m m ab m
ab m m m ab ab
2 21/2
4 4m m m ab
2 3
4
2 3/2
2
(6
(
)44
4 )8
mK
a
mK
ab
baa
and then a miracle occurs …
Note then that:
2 3/2
264 ) 4
3(m
A Kaab
MATH 6101 29
How did Archimedes do this?
10-Sept-2008
Claim: 8PQR PQS
What do we mean by “equals” here?
What did Archimedes mean by “equals”?
MATH 6101 30
What good does this do?
10-Sept-2008
What is the area of the quadrilateral □QSPR?
18
A K K
MATH 6101 31
A better approximation
10-Sept-2008
What is the area of the pentelateral □QSPTR?
T
SP
Q
R
MATH 6101 32
The better approximation
10-Sept-2008
1
1 1 18 8 4
A K K K K K
Note that the triangle ΔPTR is exactly the same as ΔQSP so we have that
MATH 6101 34
The next approximation
10-Sept-2008
1 2
3 4
1area( ) area( ) area( )
81 1 18 8 64
1area( ) area( ) area( )
81 1 18 8 64
QZ S SZ P QSP
K K
PZ T TZ R PTR
K K
Let’s go to the next level and add the four triangles given by secant lines QS, SP, PT, and TR.
MATH 6101 35
The next approximation
10-Sept-2008
2 1
4 1 164 4 16
A A K K K K
What is the area of this new polygon that is a much better approximation to the area of the sector of the parabola?
MATH 6101 36
The next approximation
10-Sept-2008
What is the area of each triangle in terms of the original stage?
3 2 1 3
1 1 1 1 1 18 8 8 8 8 8 8
KK K K K
What is the area of the new approximation?
3 2 3
1 1 18
4 16 64A A K K K K K
MATH 6101 37
The next approximation
10-Sept-2008
How many triangles to we add at the next stage?
Okay, we have a pattern to follow now.
What is the area of each triangle in terms of the previous stage?
8
3 2
18
K K
MATH 6101 38
The next approximation
10-Sept-2008
What is the area of the next stage?
We add twice as many triangles each of which has an eighth of the area of the previous triangle. Thus we see that in general, 1 1 1
4 16 4n nA K K K K
This, too, Archimedes had found without the aid of modern algebraic notation.
MATH 6101 39
The Final Analysis
10-Sept-2008
Now, Archimedes has to convince his readers that “by exhaustion” this “infinite series” converges to the area of the sector of the parabola.
Now, he had to sum up the series. He knew
14
1 1 1 1 41 ...
4 16 4 1 3n
MATH 6101 40
The Final Analysis
10-Sept-2008
Therefore, Archimedes arrives at the result
43
A K
Note that this is what we found by Calculus.Do you think that this means that Archimedes knew the “basics” of calculus?
1. What is the area of a sector of a circle whose central angle is θ radians?
2. Why must the angle be measured in radians?
3. What is a “radian”?