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The Difference Quotient

The Difference Quotient . Thursday, Jan 30th. Goal: to develop a general equation fo r rate of change. y. rise. run. Goal: to develop a general equation fo r rate of change (aka. slope of a secant). x. y. rise. Slope =. r ise r un. run.

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The Difference Quotient

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  1. The Difference Quotient Thursday, Jan 30th

  2. Goal: to develop a general equation for rate of change

  3. y rise run Goal: to develop a general equation for rate of change (aka. slope of a secant) x

  4. y rise Slope = rise run run Goal: to develop a general equation for rate of change (aka. slope of a secant) x

  5. y rise Slope = rise run run Goal: to develop a general equation for rate of change (aka. slope of a secant) = f(b) – f(a) b – a a b x

  6. y Slope = rise rise run run = Goal: to develop a general equation for rate of change (aka. slope of a secant) f(b) – f(a) b – a Notice: b = a + h h a b x

  7. y Slope = rise run rise = f(b) – f(a) b – a run Goal: to develop a general equation for rate of change (aka. slope of a secant) Slope = Notice: b = a + h f(a + h) – f(a) a + h – a h a b x

  8. y Slope = rise run rise = f(b) – f(a) b – a run Goal: to develop a general equation for rate of change (aka. slope of a secant) Slope = Notice: b = a + h f(a + h) – f(a) a + h – a = h f(a + h) – f(a) h a b x

  9. y rise Slope = f(a + h) – f(a) h run We’ve derived the Difference Quotient! h a b x

  10. y f(a + h) – f(a) h Slope = rise run How could we use this equation for the slope of a secant to determine the slope of a tangent? Let h  0 h a b x

  11. y f(a + h) – f(a) h Slope = How could we use this equation for the slope of a secant to determine the slope of a tangent? Let h  0 h a b x

  12. Example from Physics! Miss Timan – in a fit of marking madness – threw the Advanced Functions off her balcony (not a true story). The height of the exams above the ground (in metres) can be modelled by: f(t) = 15 – 4.9t2. Develop an expression for the instantaneous rate of change (velocity) of the falling exams.

  13. Example from Physics! Miss Timan – in a fit of marking madness – threw the Advanced Functions off her balcony (not a true story). The height of the exams above the ground (in metres) can be modelled by: f(t) = 15 – 4.9t2. Develop an expression for the instantaneous rate of change (velocity) of the falling exams. f(a + h) – f(a) h Slope = f(a) = 15 – 4.9a2 f(a + h) = 15 – 4.9(a + h)2

  14. Example from Physics! Miss Timan – in a fit of marking madness – threw the Advanced Functions off her balcony (not a true story). The height of the exams above the ground (in metres) can be modelled by: f(t) = 15 – 4.9t2. We just discovered that v(t) = – 9.8t. What is the acceleration of the exams (aka. the rate of change of velocity)?

  15. Example from Physics! Miss Timan – in a fit of marking madness – threw the Advanced Functions off her balcony (not a true story). The velocity of the exams above the ground (in metres) can be modelled by: v(t) = – 9.8t Develop an expression for the instantaneous rate of change of the velocity of the falling exams. v(a + h) – v(a) h Slope = v(a) = – 9.8a v(a + h) = – 9.8(a + h)

  16. Interview-style assessments every Friday Assignment #1: Rates of Change Weekly Assignments

  17. Page 20 #10, 11, 16, 17 Homework

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