Topic 6

Rates of Change I

Topic 6: New Q Maths Chapter 6.1 - 6.4, 6.7

Rates of Change I Chapter 8.2

concept of the rate of change calculation of average rates of change in both practical and

purely mathematical situations interpretation of the average rate of change as the gradient of

the secant intuitive understanding of a limit (N.B. – Calculations using limit

theorems are not required) definition of the derivative of a function at a point derivative of simple algebraic functions from first principles

Model : A cyclist travels 315 km in 9 hours. Express this in m/sec

315 km in 9 hours = 35 km in 1 hour

= 9.72 m/sec (2dp)

Read e.g. 3 Page 187

s

m

Hr

Km

1

35

EXAMPLE 3: page 187

Volume (L) 5 12 15 20 25

Mass (Kg) 4.1 9.3 12 15.7 19.75

Kg/L

EXAMPLE 3: page 187

Volume (L) 5 12 15 20 25

Mass (Kg) 4.1 9.3 12 15.7 19.75

Kg/L 0.82 0.78 0.80 0.79 0.79

Within experimental error, these variables are related by a fixed rate (≈

0.79 Kg/L)

Calculator Steps for Linear Regression

TI – 83 (Enter data via Stat – Edit)

2nd Stat Plot

Turn plot 1 on

Choose scatter plot

X list: L1

Y list: L2

Set window

Graph

TI – 89 (Enter data via APPS – option 6 – option 1). You may need to set up a variable if you’ve never used this function before.

F2 (plot setup)

F1 (define)

Plot type → scatter

x: C1 y: C2

Frequency: no

Enter to save

You’ll return to this screen (ESC)

Set window

Add a Regression Line

TI – 83

Turn on DiagnosticOn (via catalog)

Stat – Calc

4: LinReg

LinReg L1, L2, Y1

Enter (examine stats)

Graph

TI – 89

F5: calc

Calc type → 5: LinReg

x: C1 y: C2

Store regEQ → y1

Freq → no

Enter to save

Graph

Exercise

NewQ P 188

Exercise 6.1

Rates of Change

The rate of change of a second quantity w.r.t. a firstquantity is the quotient of their differences:

Read e.g. 4 Page 190 (Do on GC)N.B. If the rate of change is constant, the

graph will be a straight line.

12

12

1quantity in change

2quantity in change change of Rate

xx

yyx

y

Consider the following situation:

A car travels from Bundaberg to Miriamvale (100 km) at 50 km/h.How fast must he travel coming home to average 100 km/h for the entire trip?

N.B. Average speed = total distance

total time

Consider the following situation:

A car travels from Bundaberg to Miriamvale (100 km) at 80 km/h.How fast must he travel coming home to average 100 km/h for the entire trip?

Exercise

NewQ P 193

Exercise 6.2

Use CBR to emulate motion graphs

Use your GC to find the rate of change ofy = 2 + 4x – 0.25x2 from x = 3 to x = 5

x

y

-2 -1 0 1 2 3 4 5 6 70

5

10

15

Rate of change = 4/2 = 2

(3 , 11.75)

(5 , 15.75)

Find VALUESDraw graph

Exercise

NewQ P 198

Exercise 6.3No. 1, 2, 4, 6(a&b), 7

Exercise

NewQ P 204, 212

Exercise 6.4 no. 2, 5, 6 & 7

6.6 no. 1-3, 6, 9

y = x2 + 2

y = x3 –x2 -4x + 4

Finding Tangents

An algebraic approach Differentiation by First Principles

Differentiation 1: (11B)

-Tangent applet

Let P[ x, f(x) ] be a point on the curve y = f(x)

P[x, f(x)]

and let Q be a neighbouring point a distance of h further along the x-axis from point P.

x+h - x

f(x+h) – f(x)

Q [ x+h,

f(x+h) ]

Q [ x+h, f(x+h) ]

h

P(x,f(x))

Q[x+h,f(x+h)]

f(x+h) – f(x)

x+h - x

Gradient of tangent = lim f(x+h) – f(x) h0 h

xhx

xfhxf

xx

yy

x

ympq

)()(12

12

Let P[ x, f(x) ] and let Q[ x+h, f(x+h) ]

First Principl

es

0

2

0

2

0

2

0

0

0

( ) ( )Gradient PQ lim

(2 ) 4lim

4 4 4lim

4lim

(4 )lim

lim 4

4

h

h

h

h

h

h

f x h f x

h

h

h

h h

h

h h

hh h

hh

x

y

-4 -3 -2 -1 0 1 2 3 40

5

10

15

Model Find the gradient of the tangent to f(x)= x2 at the point where x = 2

Let P be the point (2, 4) and let Q be the point [(2+h), f(2+h)]

P

Q

Models

Use first principles to find the gradient of the curve

(a) y = 2x2 – 13x + 15 at x=5 (b) y = x2 + 3x - 8 at any point

Differential Graphing Tool

Scootle: First Principles

Exercise

NewQ P 262

Exercise 8.2 2-5

Q: Differential Functions (11B)

Q: Differentiate Polynomials (11B)

Differentiation 1: (11B)

-Tangent applet- 3 derivative puzzles