lesson 22: areas and distances (handout)

Section 5.1Areas and Distances

V63.0121.006/016, Calculus I

New York University

April 13, 2010

Announcements

I Quiz April 16 on §§4.1–4.4

I Final Exam: Monday, May 10, 12:00noon

Announcements

I Quiz April 16 on §§4.1–4.4

I Final Exam: Monday, May10, 12:00noon

V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 2 / 30

Objectives

I Compute the area of aregion by approximating itwith rectangles and lettingthe size of the rectanglestend to zero.

I Compute the total distancetraveled by a particle byapproximating it as distance= (rate)(time) and lettingthe time intervals over whichone approximates tend tozero.


Notes

Notes

Notes

1

Section 5.1 : Areas and DistancesV63.0121.006/016, Calculus I April 13, 2010

Outline

Area through the CenturiesEuclidArchimedesCavalieri

Generalizing Cavalieri’s methodAnalogies

DistancesOther applications


Easy Areas: Rectangle

Definition

The area of a rectangle with dimensions ` and w is the product A = `w .

`

w

It may seem strange that this is a definition and not a theorem but wehave to start somewhere.


Easy Areas: Parallelogram

By cutting and pasting, a parallelogram can be made into a rectangle.

b b

h

So

Fact

The area of a parallelogram of base width b and height h is

A = bh


Notes

Notes

Notes

2


Easy Areas: Triangle

By copying and pasting, a triangle can be made into a parallelogram.

b

h

So

Fact

The area of a triangle of base width b and height h is

A =1

2bh


Easy Areas: Other Polygons

Any polygon can be triangulated, so its area can be found by summing theareas of the triangles:


Hard Areas: Curved Regions

???


Notes

Notes

Notes

3


Meet the mathematician: Archimedes

I Greek (Syracuse), 287 BC –212 BC (after Euclid)

I Geometer

I Weapons engineer


118

18

164

164

164

164

Archimedes found areas of a sequence of triangles inscribed in a parabola.

A = 1 + 2 · 1

8+ 4 · 1

64+ · · ·

= 1 +1

4+

1

16+ · · ·+ 1

4n+ · · ·


We would then need to know the value of the series

1 +1

4+

1

16+ · · ·+ 1

4n+ · · ·

But for any number r and any positive integer n,

(1− r)(1 + r + · · ·+ rn) = 1− rn+1

So

1 + r + · · ·+ rn =1− rn+1

1− r

Therefore

1 +1

4+

1

16+ · · ·+ 1

4n=

1− (1/4)n+1

1− 1/4→ 1

3/4=

4

3

as n→∞.


Notes

Notes

Notes

4


Cavalieri

I Italian,1598–1647

I Revisited thearea problemwith adifferentperspective


Cavalieri’s method

y = x2

0 1

1

2

Divide up the interval into piecesand measure the area of theinscribed rectangles:

L2 =1

8

L3 =1

27+

4

27=

5

27

L4 =1

64+

4

64+

9

64=

14

64

L5 =1

125+

4

125+

9

125+

16

125=

30

125Ln =?


What is Ln?

Divide the interval [0, 1] into n pieces. Then each has width1

n. The

rectangle over the ith interval and under the parabola has area

1

n·(

i − 1

n

)2

=(i − 1)2

n3.

So

Ln =1

n3+

22

n3+ · · ·+ (n − 1)2

n3=

1 + 22 + 32 + · · ·+ (n − 1)2

n3

The Arabs knew that

1 + 22 + 32 + · · ·+ (n − 1)2 =n(n − 1)(2n − 1)

6

So

Ln =n(n − 1)(2n − 1)

6n3→ 1

3as n→∞.V63.0121.006/016, Calculus I (NYU) Section 5.1 Areas and Distances April 13, 2010 15 / 30

Notes

Notes

Notes

5


Cavalieri’s method for different functions

Try the same trick with f (x) = x3. We have

Ln =1

n· f(

1

n

)+

1

n· f(

2

n

)+ · · ·+ 1

n· f(

n − 1

n

)=

1

n· 1

n3+

1

n· 23

n3+ · · ·+ 1

n· (n − 1)3

n3

=1 + 23 + 33 + · · ·+ (n − 1)3

n4

The formula out of the hat is

1 + 23 + 33 + · · ·+ (n − 1)3 =[12n(n − 1)

]2So

Ln =n2(n − 1)2

4n4→ 1

4as n→∞.


Cavalieri’s method with different heights

Rn =1

n· 13

n3+

1

n· 23

n3+ · · ·+ 1

n· n3

n3

=13 + 23 + 33 + · · ·+ n3

n4

=1

n4

[12n(n + 1)

]2=

n2(n + 1)2

4n4→ 1

4

as n→∞.So even though the rectangles overlap, we still get the same answer.


Outline





Notes

Notes

Notes

6


Cavalieri’s method in general

Let f be a positive function defined on the interval [a, b]. We want to find the areabetween x = a, x = b, y = 0, and y = f (x).

For each positive integer n, divide up the interval into n pieces. Then ∆x =b − a

n.

For each i between 1 and n, let xi be the nth step between a and b. So

a bx0 x1 x2 . . . xixn−1xn

x0 = a

x1 = x0 + ∆x = a +b − a

n

x2 = x1 + ∆x = a + 2 · b − a

n· · · · · ·

xi = a + i · b − a

n· · · · · ·

xn = a + n · b − a

n= b


Forming Riemann sums

We have many choices of how to approximate the area:

Ln = f (x0)∆x + f (x1)∆x + · · ·+ f (xn−1)∆x

Rn = f (x1)∆x + f (x2)∆x + · · ·+ f (xn)∆x

Mn = f

(x0 + x1

2

)∆x + f

(x1 + x2

2

)∆x + · · ·+ f

(xn−1 + xn

2

)∆x

In general, choose ci to be a point in the ith interval [xi−1, xi ]. Form theRiemann sum

Sn = f (c1)∆x + f (c2)∆x + · · ·+ f (cn)∆x

=n∑

i=1

f (ci )∆x


Theorem of the Day

Theorem

If f is a continuous function orhas finitely many jumpdiscontinuities on [a, b], then

limn→∞

Sn = limn→∞

n∑i=1

f (ci )∆x

exists and is the same value nomatter what choice of ci wemade.

x1x1 x2x1 x2 x3x1 x2 x3 x4x1 x2 x3 x4 x5x1 x2 x3 x4 x5 x6x1 x2 x3 x4 x5 x6 x7x1x2x3x4x5x6x7x8x1x2x3x4x5x6x7x8x9x1x2x3x4x5x6x7x8x9x10x1x2x3x4x5x6x7x8x9x10x11x1x2x3x4x5x6x7x8x9x10x11x12x1x2x3x4x5x6x7x8x9x10x11x12x13x1x2x3x4x5x6x7x8x9x10x11x12x13x14x1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x16x1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x16x17x1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x16x17x18x1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x16x17x18x19x1x2x3x4x5x6x7x8x9x10x11x12x13x14x15x16x17x18x19x20a b


Notes

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Notes

7


Analogies

The Tangent Problem(Ch. 2–4)

I Want the slope of a curve

I Only know the slope of lines

I Approximate curve with aline

I Take limit over better andbetter approximations

The Area Problem (Ch. 5)

I Want the area of a curvedregion

I Only know the area ofpolygons

I Approximate region withpolygons

I Take limit over better andbetter approximations


Outline





Distances

Just like area = length× width, we have

distance = rate× time.

So here is another use for Riemann sums.


Notes

Notes

Notes

8


Application: Dead Reckoning


Example

A sailing ship is cruising back and forth along a channel (in a straightline). At noon the ship’s position and velocity are recorded, but shortlythereafter a storm blows in and position is impossible to measure. Thevelocity continues to be recorded at thirty-minute intervals.

Time 12:00 12:30 1:00 1:30 2:00

Speed (knots) 4 8 12 6 4

Direction E E E E W

Time 2:30 3:00 3:30 4:00

Speed 3 3 5 9

Direction W E E E

Estimate the ship’s position at 4:00pm.


Solution

We estimate that the speed of 4 knots (nautical miles per hour) ismaintained from 12:00 until 12:30. So over this time interval the shiptravels (

4 nmi

hr

)(1

2hr

)= 2nmi

We can continue for each additional half hour and get

distance = 4× 1/2 + 8× 1/2 + 12× 1/2

+ 6× 1/2− 4× 1/2− 3× 1/2 + 3× 1/2 + 5× 1/2

= 15.5

So the ship is 15.5 nmi east of its original position.


Notes

Notes

Notes

9


Analysis

I This method of measuring position by recording velocity was necessaryuntil global-positioning satellite technology became widespread

I If we had velocity estimates at finer intervals, we’d get betterestimates.

I If we had velocity at every instant, a limit would tell us our exactposition relative to the last time we measured it.


Other uses of Riemann sums

Anything with a product!

I Area, volume

I Anything with a density: Population, mass

I Anything with a “speed:” distance, throughput, power

I Consumer surplus

I Expected value of a random variable


Summary

I We can compute the area of a curved region with a limit of Riemannsums

I We can compute the distance traveled from the velocity with a limitof Riemann sums

I Many other important uses of this process.


Notes

Notes

Notes

10


lesson 22: areas and distances (handout)

Education

n n3 n n3 n n

n n2 n

distances v63

easy areas

f nn nn n n

n pieces

hard areas

6n33as n