Welcome to Math 103:207Integral Calculus with Applications to the Life

Sciences

Cole Zmurchokhttp://www.math.ubc.ca/~zmurchok

January 6, 2015

January 6, 2015

Introductions & Information

Integral Calculus & the Life Sciences

Areas and sums

Introductions & Information

http://www.math.ubc.ca/~zmurchok

Course Information:

I Schedule & Important Dates

I Notes

I WebWork & Old-School Homework (OSH)

I Pizza

I Resources

Office Hours: Wednesday 9:30 to 10:30 and Thursday 9:00 to10:00 in LSK 300BExpectations:

I Phones & laptops

I Homework professionally completed

I Lecture preparation: read & work

Integral Calculus & cats

Cats tend to land on their feetwhen falling (0.3 seconds to flipover).

Question

What is the minimum heightfrom which a cat can fall toensure it lands on its feet?

Annals of Improbable ResearchDiamond, J. (1988) Why cats have nine lives, Nature 332.Thanks to Joseph M. Mahaffy for the falling cat idea.

Boardwork: Falling Cat

Newton’s law of motion provides a model:

ma = −mg

I m is the mass of the cat

I a is the acceleration of cat

I −mg is the force of gravity (g = 980.7 cm/sec2)

Let h(t) be the height of the cat at time t. Initially the cat is atrest at height h0:

I h(0) = h0

I v(0) = h′(0) = 0 (initial velocity)

h′′ = −g.

Boardwork: Falling Cat

Main theme of this course is “integral calculus”. The integral isthe inverse operator of the derivative:∫

f(x)dxddx−−→ f(x)

ddx−−→ f ′(x)

Integrating h′′ = −g, we find

v(t) = h′(t) = −gt,

which satisfies v(0) = 0. Integrating h′ = −gt, we find theheight of the cat at time t:

h(t) =−gt2

2+ h0,

which satisfies h(0) = h0.

Boardwork: Falling Cat

The height of the cat at time 0.3 seconds is h(0.3):

h(0.3) =−g(0.3)2

2+ h0 = −44.1315 + h0,

so the cat must be above 44.1315 cm to have sufficient time toflip over before hitting the ground.

Note

This is consistent with the Annals of Improbable Research data,which suggests that a cat dropped upside down from a height of2 to 6 feet (30.84 cm to 182.88 cm) will always land on its feet.

Boardwork: Areas and sums

Self-study

Review how to find the area, volume, perimeters of basic shapes(Section 1.1 to 1.3)

Integral calculus originated as a method to calculate the areaand volume of everyday objects (size of plot of land, volume ofbarrel, etc.).The fundamental idea is to cut up the geometric shape intosmaller pieces, and approximate those smaller pieces by regularshapes that can be easily quantified.

Areas and sums

Example Area under y = x2 from 0 to 1.

Dissect the area under the curve into N pieces, each with width1N . The kth piece can be approximated by a rectangle with

dimensions 1N by

(kN

)2.

Areas and sums

Thus, the area, A, is approximately

A ≈ 1

N

(1

N

)2

+1

N

(2

N

)2

+ · · ·+ 1

N

(N − 1

N

)2

+1

N

(N

N

)2

=1

N3

(12 + 22 + 32 + · · ·+ (N − 1)2 + N2

)=

1

N3

N(N + 1)(2N + 1)

6

=1

6

(1 +

1

N

)(2 +

1

N

).

As N →∞, A→ 13 . This is the key idea behind integration,

which we will make precise next week.

Sums

Summation notation: a.k.a. sigma notation

ak + ak+1 + · · ·+ an =

n∑j=k

aj

Here, the sum is indexed by the auxiliary variable j, which wecould choose to be anything, e.g. ♣ or ∆. k signifies the termwhich starts off the series, and n signifies when the series ends.

Example Write the sum 3 + 6 + 11 + 18 in sigma notation.

3 + 6 + 11 + 18 =4∑

i=1

(i2 + 2)

works. Can you find another expression that works?

SumsExample

I Find the value of 3 + 6 + 11 + 18 =∑4

i=1(i2 + 2).

3 + 6 + 11 + 18 = 38

or

4∑i=1

(i2 + 2) =

4∑i=1

i2 +

4∑i=1

2 =4(4 + 1)(2 · 4 + 1)

6+ 4 · 2 = 38.

Important Formula

n∑i=1

i =n(n + 1)

2,

n∑i=1

i2 =n(n + 1)(2n + 1)

6,

n∑i=1

i3 =

(n(n + 1)

2

)2

Gauss’ formula

Theorem (Gauss’s formula)

n∑i=1

i =n(n + 1)

2

Proof.

2

n∑i=1

i = 1 + 2 + 3 + · · ·n− 2 + n− 1 + n

+ n + n− 1 + n− 2 + · · ·+ 3 + 2 + 1

= n + 1 + n + 1 + n + 1 + · · ·+ n + 1 + n + 1 + n + 1

= n(n + 1)

SumsExample

n∑i=1

i3 + i2 + i + 1 =

n∑i=1

i3 +

n∑i=1

i2 +

n∑i=1

i +

n∑i=1

1

=

(n(n + 1)

2

)2

+n(n + 1)(2n + 1)

6

+n(n + 1)

2+ n

Example

5∑k=2

(k − 1)3

Use a substitution (change summation index), i.e., let j = k− 1:

5∑k=2

(k − 1)3 =4∑

j=1

j3 =

(4(4 + 1)

2

)2

= 100

Sums

Example

n∑i=1

(i− 1)2 =

n∑i=1

i2 − 2i + 1 =

n∑i=1

i2 −n∑

i=1

2i +

n∑i=1

1

=n(n + 1)(2n + 1)

6− 2

n∑i=1

i + n

=n(n + 1)(2n + 1)

6− n(n + 1) + n

Alternately:

n∑i=1

(i− 1)2 =

n−1∑j=0

j2 = 0 +

n−1∑j=1

j2

=(n− 1)(n− 1 + 1)(2(n− 1) + 1)

6=

n(n− 1)(2n− 1)

6which is the same!

Sums

Example Sums can have infinitely many terms, yet can still bere-indexed

1− 2 + 3− 4 + 5− 6 + · · · =∞∑j=1

(−1)j+1j =

∞∑j=0

(−1)j(j + 1)

Example Telescoping sum

n∑k=1

1

k(k + 1)=

n∑k=1

(1

k− 1

k + 1

)= 1− 1

2+

1

2− 1

3+

1

3− 1

4+

1

4− 1

5+ · · ·+ 1

n− 1

n + 1

= 1− 1

n + 1

Sums

Example If∑n

k=1 ak = A,∑n

k=1 bk = B, and c, d ∈ R, find∑nk=1 cak + dbk.

n∑k=1

cak + dbk =

n∑k=1

cak +

n∑k=1

dbk = c

n∑k=1

ak + d

n∑k=1

bk = cA+ dB

This property is called linearity.

Lecture 1 Ending

Recap:

I Areas and sums

I Prelude to integration

I Gauss’ formula

I Working with summation notation

Questions?

For next class

Try to do the following problems, we will start Thursday’s classwith them.

I Find the sum of 1 + 12 + 1

4 + 18 + · · ·+ (12)n.

I Find the sum of 1 + 2 + 22 + 23 + · · ·+ 2n.

I Review formulas for areas & volumes.

I Read Chapter 1.