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Topic 6
Rates of Change I
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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
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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
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EXAMPLE 3: page 187
Volume (L) 5 12 15 20 25
Mass (Kg) 4.1 9.3 12 15.7 19.75
Kg/L
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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)
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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
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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
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Exercise
NewQ P 188
Exercise 6.1
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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
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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
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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?
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Exercise
NewQ P 193
Exercise 6.2
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Use CBR to emulate motion graphs
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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
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Exercise
NewQ P 198
Exercise 6.3No. 1, 2, 4, 6(a&b), 7
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Exercise
NewQ P 204, 212
Exercise 6.4 no. 2, 5, 6 & 7
6.6 no. 1-3, 6, 9
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y = x2 + 2
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y = x3 –x2 -4x + 4
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Finding Tangents
An algebraic approach Differentiation by First Principles
Differentiation 1: (11B)
-Tangent applet
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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
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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) ]
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First Principl
es
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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
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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
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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