SLOPES AND AREAS:
YOU REALLY DO TEACH CALCULUS
•
•
•
•
•
•
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
Area
• Area of your hand
1 × 1 grid: Area =
½ × ½ grid: Area =
¼ × ¼ grid: Area =
Area
13 < A < 25
56/4 < A < 76/4
•
•
•
•
•
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
Area  s(s  a)(s  b)(s  c )
1 r
3r 2 3
Area  3    r 3 
2 2
4
T2
T3
r/2 T1
r
r 3
2
3r 2 3
2
 Area  3r 3
4
r
4r 2  C  8r
2r  Area  4r
2  4
2
2
r
5 2 5 5 2
A
r
8
3r
2
2
3
 Area  6r 2 3
2r
2
2  Area  16r
2

r
2 1

n
 2 
  sin 

2
n


n
8
Approx
2.828427124
n
40
Approx
3.128689302
10
12
14
2.938926262
3.0
3.037186175
60 3.135853896
80 3.138363830
100 3.139525977
16
18
3.061467460
3.078181290
150 3.140674029
180 3.140954703
20
3.090169944
200 3.141075908
Area of a Parabolic Sector
R
y  ax
2
Q
P
y  mx  b
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?
ax  mx  b
2
ax  mx  b  0
2
m  m  4ab
m  m  4ab
x1 
, x2 
2a
2a
2
2
Points of Intersection
Now, we can find the points of intersection
of the line and the parabola, Q and R.
m2  m m2  4ab  2ab
Q  ( x2 , y2 )
y2  ax 
2a
 m  m2  4ab m2  m m2  4ab  2ab 
Q
,



2a
2a


2
2
m2  m m2  4ab  2ab
R  ( x1 , y1 )
y1  ax 
2a
 m  m2  4ab m2  m m2  4ab  2ab 
R
,



2
a
2
a


2
1
Slope of the Tangent Line
The slope of the tangent line at a point is
twice the product of a and x.
m  2ax
m
x
2a
2


m
m
2
P  ( p1 , p2 )  ( x , ax )   ,

 2a 4a 
Area of the Parabolic Sector
R
Q
x2
x1
Calculus Answer
(m  4ab)
A
2
6a
2
3/2
Archimedes - Area of ΔPQR
R
p
q
Q
r
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 Formula
p  d(Q, R)  ( x1  x2 )2  ( y1  y2 )2
(m2  4ab)(1  m2 )
p
a
q  d( P, R)  ( x1  p1 )2  ( y1  p2 )2
(m2  4ab)(4ab  4  5m2  4m m2  4ab )
q
4a
r  d( P, Q)  ( p1  x2 )2  ( p2  y2 )2
(m2  4ab)(4ab  4  5m2  4m m2  4ab )
r
4a
pqr
Now, the semiperimeter is: s 
2
m2  4ab
s
4 1  m2
8a

 4ab  4  5m  4m m  4ab
2
2

 4ab  4  5m  4m m  4ab 

2
2
Uh – oh!!!!
Are we in trouble? Heron’s Formula states that
the area is the following product:
K  s(s  p)(s  q)(s  r)
This does not look promising!!
 (m2  4ab)2 
2
2
2
K 
4
1

m

4
ab

4

5
m

4
m
m
 4ab

4
 4096a 

 4ab  4  5m2  4m m2  4ab   4 1  m2

 4ab  4  5m2  4m m2  4ab  4ab  4  5m2  4m m2  4ab  

 4 1  m2  4ab  4  5m2  4m m2  4ab



 4ab  4  5m2  4m m2  4ab   4 1  m2


 4ab  4  5m2  4m m2  4ab  4ab  4  5m2  4m m2  4ab  

1/2
and then a miracle occurs …
(m  4ab)
K
64a4
2
(m  4ab)
K
8a2
2
3
3/2
Note then that:
(m2  4ab)3/2
4
 A K
2
6a
3
How did Archimedes know this?
R
Claim: PQR  8PQS
Q
10-Sept-2008
S
P
MATH 6101
28
How did Archimedes do this?
Claim: PQR  8PQS
What do we mean by “equals” here?
What did Archimedes mean by “equals”?
10-Sept-2008
MATH 6101
29
What good does this do?
What is the area of the quadrilateral □QSPR?
1
AK K
8
10-Sept-2008
MATH 6101
30
A better approximation
What is the area of the pentelateral □QSPTR?
R
T
Q
10-Sept-2008
S
P
MATH 6101
31
The better approximation
Note that the triangle ΔPTR is exactly the
same as ΔQSP so we have that
1
1
1
A1  K  K  K  K  K
8
8
4
10-Sept-2008
MATH 6101
32
An even better approximation
R
Z4
T
Q
Z1
10-Sept-2008
S
P
Z3
Z2
MATH 6101
33
The next approximation
Let’s go to the next level and add the four
triangles given by secant lines QS, SP, PT, and
TR.
1
area( QZ 1 S )  area( SZ 2 P )  area( QSP )
8
11  1
  K
K
8  8  64
1
area( PZ 3T )  area( TZ 4 R)  area( PTR)
8
11  1
  K
K
8  8  64
10-Sept-2008
MATH 6101
34
The next approximation
What is the area of this new polygon that is a
much better approximation to the area of the
sector of the parabola?
4
1
1
A2  A1 
KK K K
64
4
16
10-Sept-2008
MATH 6101
35
The next approximation
What is the area of each triangle in terms of
the original stage?
1
1  1  1  1  1  K
K 3  K2   K1     K    3
8
8  8  8  8  8  8
What is the area of the new approximation?
1
1
1
A3  A2  8 K 3  K  K  K 
K
4
16
64
10-Sept-2008
MATH 6101
36
The next approximation
Okay, we have a pattern to follow now.
How many triangles to we add at the next
stage?
8
What is the area of each triangle in terms of
the previous stage?
1
K3  K2
8
10-Sept-2008
MATH 6101
37
The next approximation
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
An  K  K  K 
4
16
1
 nK
4
This, too, Archimedes had found without the
aid of modern algebraic notation.
10-Sept-2008
MATH 6101
38
The Final Analysis
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
1 1
1  
4 16
10-Sept-2008
1
1
4
 n  ... 

1
4
1 4 3
MATH 6101
39
The Final Analysis
Therefore, Archimedes arrives at the result
4
A K
3
Note that this is what we found by Calculus.
Do you think that this means that
Archimedes knew the “basics” of calculus?
10-Sept-2008
MATH 6101
40
Surface area of a Cylinder
r
2πr
h
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”?
360° or 2π
180° or π
60° or π/3
2π - θ
90° or π/2
270° or 3π/2
θ
Surface area of a Cone
s
s  h r
2
h
r
2
2
Surface area of a Cone
s
θ
Area ?
Surface area of a Cone
s
θ
Area 

2
s 
2

r

2
2
h
2

Volume of a Cone
Do you believe that 3 of these fit into 1 of these?
h
h
r
1 2
V  r h
3
r
V   r 2h
Archimedes Again
Cylinder: radius R and height 2R
Cone: radius R and height 2R
Sphere with radius R
Volcone : Volsphere : Volcylinder = 1 : 2 : 3