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 hand1 × 1 grid: Area = ½ × ½ grid: Area = ¼ × ¼ grid: Area =

Area

56/4 < A < 76/413 < A < 25

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

r

T3

T1

T2

r/23

2r

21 3 32

3 32 4

rr a rrA e

223 3 3 3

4r Area r

r

2 2

42 Area

84

42

2 C rrr r

r

25 2 5 58

rA

223 3 6 3

2r Area r

r

2 22 2 16 2 1r Area r

2sin2n

n

n Approx n Approx8 2.828427124 40 3.12868930210 2.938926262 60 3.13585389612 3.0 80 3.13836383014 3.037186175 100 3.13952597716 3.061467460 150 3.14067402918 3.078181290 180 3.14095470320 3.090169944 200 3.141075908

Area of a Parabolic Sector

P

Q

R

2y ax

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?2ax mx b

2

2 2

1 2

0

4 4,2 2

ax mx b

m m ab m m abx xa a

Points of IntersectionNow, we can find the points of intersection

of the line and the parabola, Q and R.2 2

22 2 2 2

2 2 2

4 2, )

4

(2

2 24 2,

m mQ x y axa

m

m ab aby

m ab m abm m aQa

ba

2 22

1 1 1 1

2 2 2

4 2, )

4

(2

2 24 2,

m mR x y axa

m

m ab aby

m ab m abm m aRa

ba

Slope of the Tangent LineThe slope of the tangent line at a

point is twice the product of a and x.

2m ax

2x m

a

22

1 2, ) )( ( , ,2 4m mP p x axa a

p

Area of the Parabolic Sector

Q

R

x2 x1

Calculus Answer2 3/2

2(

64 )abmAa

Archimedes - Area of ΔPQR

p

r

q

P

Q

R

Area of TriangleIt 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 abp

am

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 aba

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 aba

Now, the semiperimeter is: 2s p q r

2

2

2 2

2 2

4 4 18

4 4 5 4 4

4 4 5 4 4

m ab ma

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 44096

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 aba

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

mKa

mK

ab

baa

and then a miracle occurs …

Note then that:

2 3/2

264 ) 4

3(m A K

aab

MATH 6101 28

How did Archimedes know this?

10-Sept-2008S P

Q

RClaim: 8PQR PQS

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

S P

Q

R

MATH 6101 32

The better approximation

10-Sept-2008

11 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 33

An even better approximation

10-Sept-2008

Z4

Z3

Z2

Z1

T

S P

Q

R

MATH 6101 34

The next approximation

10-Sept-2008

1 2

3 4

1area( ) area( ) area( )8

1 1 18 8 64

1area( ) area( ) area( )8

1 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 14 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 31 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 31 1 184 16 64

A 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

3 218

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 4

3A K

Note that this is what we found by Calculus.Do you think that this means that Archimedes knew the “basics” of calculus?

Surface area of a Cylinder

r

h

2πr

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 2π180° or π 60° or

π/3

270° or 3π/2

90° or π/2

2π - θθ

Surface area of a Cone

r

sh

2 2 2s h r

Surface area of a Cone

s

θArea ?

Surface area of a Cone

s

θ

2 2 2

2 2A a s hre r

Volume of a Cone

r

h

r

h

2V r h213

V r h

Do you believe that 3 of these fit into 1 of these?

Archimedes Again

Cylinder: radius R and height 2RCone: radius R and height 2RSphere with radius R

Volcone : Volsphere : Volcylinder = 1 : 2 : 3