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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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Rates Model : Mike earns $120 in an 8 hour shift. (a) What is his rate of pay? (b) How much would he earn in 7 hours? (a) Rate of pay = 120/8 $/hr = $15/hr (b) In 7 hours, he earns 7 x 15 = $105

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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

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EXAMPLE 3: page 187 Volume (L)512152025 Mass (Kg)4.19.31215.719.75 Kg/L

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EXAMPLE 3: page 187 Volume (L)512152025 Mass (Kg)4.19.31215.719.75 Kg/L0.820.780.800.79 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) 2 nd 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: C1y: 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, Y 1 Enter (examine stats) Graph TI – 89 F5: calc Calc type → 5: LinReg x: C1y: C2 Store regEQ → y 1 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 first quantity 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.

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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 of y = 2 + 4x – 0.25x 2 from x = 3 to x = 5 Rate of change = 4/2 = 2 (3, 11.75) (5, 15.75) Find VALUESDraw graph

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Exercise NewQ P 198 Exercise 6.3 No. 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 = x 2 + 2

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y = x 3 –x 2 -4x + 4

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Finding Tangents An algebraic approach Differentiation by First Principles Differentiation 1: (11B) -Tangent appletTangent 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) h 0 h Let P [ x, f(x) ] and let Q [ x+h, f(x+h) ]

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First Principles

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Model Find the gradient of the tangent to f(x)= x 2 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 = 2x 2 – 13x + 15 at x=5 (b) y = x 2 + 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 appletTangent applet - 3 derivative puzzles 3 derivative puzzles

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