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
35 Km
m

1Hr
s
= 9.72 m/sec (2dp)
Read e.g. 3 Page 187
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 first
quantity is the quotient of their differences:
changein quantity2
Rate of change
changein quantity1
y

x
y2  y1

x2  x1
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.
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 of
y = 2 + 4x – 0.25x2 from x = 3 to x = 5
Draw graph
Find VALUES
Rate of change = 4/2 = 2
y
(5 , 15.75)
15
(3 , 11.75)
10
5
x
0
-2
-1
0
1
2
3
4
5
6
7
Exercise
NewQ P 198
Exercise 6.3
No. 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
Differentiation 1: (11B)
 An algebraic approach
 Differentiation by First Principles
-Tangent applet
Let P[ x, f(x) ] be a point on the curve y = f(x)
and let Q be a neighbouring point a distance
of h further along the x-axis from point P.
Q [ x+h, f(x+h) ]
Q [ x+h, f(x+h) ]
f(x+h) – f(x)
P[x, f(x)]
h
x+h - x
Let P[ x, f(x) ]
and
let Q[ x+h, f(x+h) ]
y y2  y1
m pq 

x x2  x1
Gradient of tangent = lim
h0
f(x+h)
– f(x)
f (x 
h) 

h
f ( x)
xh  x
Q[x+h,f(x+h)]
f(x+h) – f(x)
P(x,f(x))
x+h - x
First
Principles
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)]
f ( x  h)  f ( x )
Gradient PQ  lim
h 0
h
(2  h) 2  4
 lim
h 0
h
4  4h  h 2  4
 lim
h 0
h
4h  h 2
 lim
h 0
h
h(4  h)
 lim
h 0
h
 lim 4  h
h 0
4
y
15
Q
10
5
P
x
0
-4
-3
-2
-1
0
1
2
3
4
Scootle: First Principles
 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
Exercise
NewQ P 262
Exercise 8.2
Differentiation 1: (11B)
2-5
-Tangent applet
- 3 derivative puzzles
Q: Differential
Functions (11B)
Q: Differentiate
Polynomials (11B)