Calculus for the Biological Sciences

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Ahmed Kaffel,

([email protected])

Department of Mathematics and Statistics Marquette University Milwaukee, WI 53233

Spring 2021

L e c t u r e N o t e s – L i n e a r M o d e l s

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Outline

1 Linear Models

Chirping Crickets and Temperature 2 Equation of Lines

Slope-Intercept Point-Slope Two Points - Slope

Parallel and Perpendicular Lines Intersection of Lines

3 Inverse Linear Function 4 Other Linear Models

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The line creates a mathematical model

The t e m p e r a t u r e , T as a function of the rate snowy tree crickets chirp, C h i r p R a t e , N

There are several Biological and Mathematical questions about this Lin ear C rick et M o d e l

There is a complex relationship between the biology of the problem and the mathematical model

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C h i r p i n g C r i c k e t s a n d T e m p e r a t u r e

Biological Questions – Cricket Model 2

When can this model be applied from a practical perspective?

Biological thermometer has limiteduse

Snowy tree crickets only chirp for a couple months of the year and mostly at night

Temperature needs to be above 50^◦F

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O t h e r L i n e a r M o d e l s

Mathematical Questions – Cricket Model 1

Over what range of temperatures is this modelvalid?

Biologically, observations are mostly between 50^◦F and 85^◦F

Thus, limited r a n g e of temperatures, so limited r a n g e on the Lin ear M o d e l

R a n g e of Lin ear fu n ctions affects itsD o m a i n From the graph, D o m a i n is approximately 50–200 C h i r p s / m i n

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L i n e a r M o d e l s E q u a t i o n o f L i n e s M e t r i c S y s t e m C o n v e r s i o n I n v e r s e L i n e a r F u n c t i o n J u v e n i l e H e i g h t O t h e r L i n e a r M o d e l s

C h i r p i n g C r i c k e t s a n d T e m p e r a t u r e

Mathematical Questions – Cricket Model 2

How accurate is the model and how might the accuracy be improved?

Data closely surrounds Bessey M o d e l –No more than about 3^◦F away fom line

D o l b e a r M o d e l is fairly close though not as accurate – Sufficient for rapid temperature estimate

Observe that the temperature data trends lower at higher chirp rates – compared against linear model

Better fit with Q u a d r a t i c fu nctio n –Is this really significant?

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Equation of Line – Slope-Intercept Form

The S lo pe -In te rc e p t form of theLine y = m x + b

The variable x is the i n d e p e n d e n t variable The variable y is the d e p e n d e n t variable The slope is m

The y-intercept isb

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S lo p e - I n t e r c e p t P o i n t - S l o p e T w o P o i n t s - S l o p e

P a r a l l e l a n d P e r p e n d i c u l a r L i n e s I n t e r s e c t i o n o f L i n e s

Equation of Line – Cricket-Thermometer

The folk/Dolbear model for the cricketthermometer N

T = 4 + 40

The independent variable is N , chirps/min The dependent variable is T , the temperature

Thus, the temperature can be estimated from counting the number of chirps/min

Equivalently, the temperature (measurement) depends on how rapidly the cricket is chirping

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Equation of Line – Point-Slope Form

The P o in t-S lo p e form of the Line is often the most useful form y − y0 = m(x − x0)

or

y = m(x − x0) +y0

The slope is m

The given p o i n t is (x0, y0)

Again the independent variable is x, and the dependent variable is y

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Equation of Line – Two Points

Given two points (x0, y0) and (x1, y1), the slope is given by

m = y1 − y0

x1 − x0

Use the previous p o in t-slope form of the line satisfies y = m(x − x0) +y0

where the slope is calculated above and either point can be used.

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Example – Slope and Point

Find the equation of a line with a slope of 2, passing through the point (3, −2).

What is they-intercept?

E x a m p l e

The point-slope form of the equation gives:

y − (−2) = 2(x − 3) y + 2 = 2x − 6

y = 2x − 8

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Example – Two Points

Find the equation of a line passing through the points (−2, 1) and (3, −2)

E x a m p l e

The slope satisfies

m = 1 − (−2)

−2 − 3 = − 3 5

From the point-slope form of the line equation, using the first point

y − 1 = − (x + 2)3

y = − x − 5

3 1

5 5

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Parallel and Perpendicular Lines

Consider two lines given by theequations:

y = m1x + b1 and y = m2x + b2

The two lines are parallel if the slopes are equal,so m1 = m2

and the y-intercepts are different.

If b1 = b2, then the lines are the same.

The two lines are p e r p e n d i c u l a r if the slopes are negative reciprocals of each other, that is

m1m2 = − 1

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Example – Perpendicular Lines 1

Find the equation of the line perpendicular to the line 5x + 3y = 6

passing through the point (−2, 1)

E x a m p l e

Solution: The line can be written

3y = −5 x + 6 y = − x + 25

3

The slope of the perpendicular line (m2) is the negative reciprocal

m2 = 53

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Example – Perpendicular Lines 2

The point slope equation of the perpendicular line is

y − 1 = 5³(x + 2) y = ³₅x + ¹¹₅

−30 −2 −1 1 2 3

5

4

3

2

0 x

y

1 L : y = −(5/3) x + 21

L : y = (3/5) x + (11/5)2

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Example – Intersection of Lines 1

Find the intersection of the line parallel to the line y = 2x passing through (1, −3) and the line given by the formula

3x + 2y = 5

S k ip E x a m p l e

Solution: The line parallel to y = 2x has slope m = 2, so satisfies

y + 3 = 2(x − 1) y = 2x − 5

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Example – Intersection of Lines 2

C o n t i n ue d: Substitute y into the formula for the secondline

3x + 2(2x − 5) = 5

7x = 15 or x = 15 7 Substituting the x value into the first line equation gives

y = 2 15

7 − 5 = − 5 7 The point of intersection is

(x, y) = 15,− 5

7 7

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Example – Temperature 1

C o n v e r t T e m p e r a t u r e Fa h r e n h e i t t o Celsius The freezing point of water is 32^◦F and 0^◦C, so take

(f0, c0) = (32,0)

The boiling point of water is 212^◦F and 100^◦C (at sea level), so take

(f1, c1) = (212, 100)

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Example – Temperature 2

C o n v e r t T e m p e r a t u r e Fa h r e n h e i t t o Celsius Solution: The slope satisfies

m = c1 − c0 = 100 − 0

= 5 f1− f0 212 − 32 9 The point-slope form of the line gives

c − 0 = 5(f − 32) 9

c = 5(f − 32) 9

The temperature f in Fahrenheit is the independent variable The equation of the line gives the dependent variable c in Celsius

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Inverse Linear Function

Linear functions always have a n Inverse (Provided m /= 0)

Consider the line

y = mx + b Solving for x

mx = y − b x = y − b

m The In verse Line satisfies

x = 1

m y − b m

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Example - Inverse Line

The equation for converting ^◦F to ^◦C is 5

c = 9(f − 32)

So, 9

f − 32 = 5c The equation for converting ^◦C to ^◦F is

9 f = 5c + 32

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Scuba Diving Model 1

T h e p r e s s u r e of air d elivered by t h e r e g u l a t o r t o a S c u b a diver varies linearly w i t h t h e d e p t h of t h e w a t e r

The regulator delivers air to a Scuba diver as follows: Air Pressure at 29.4 psi when at 33 ft

Air Pressure at 44.1 psi when at 66 ft

Find the pressure of air delivered at the surface (0 ft.), at 50 ft., and at 130 ft. (the maximum depth for recreational diving).

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Scuba Diving Model – Solution 2

Solution: The linear model

p = md + p0

where p is the pressure (psi) and d is the depth in feet The data give two points

(d0, p0) = (33,29.4) and (d1, p1) = (66, 44.1) The slope is

m = 44.1 − 29.4 = 14.7

66 − 33 33 = 0.445 psi/ft

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Scuba Diving Model – Solution 3

S o lu tio n (co n t): The linear model satisfies

p− 29.4 = 0.445(d − 33) p = 0.445 d + 14.7

At the surface,d = 0 and the air pressure is 14.7 psi At a depth ofd= 50 ft, the air pressure is 36.95 psi At a depth of d = 130 ft, the air pressure is 72.55 psi

Note these assume we are at sea level and diving in sea water

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Below is a graph of the data and the Linear P r e s s u r e M o d e l

0 20 40 80 100 120

10 80 70 60 50 40 30 20

60 Depth (ft)

Pressure(psi)