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# The Difference Quotient - PowerPoint PPT Presentation

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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Thursday, Jan 30th

Goal: to develop a general equation for rate of change

rise

run

Goal: to develop a general equation for rate of change (aka. slope of a secant)

x

rise

Slope =

rise

run

run

Goal: to develop a general equation for rate of change (aka. slope of a secant)

x

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

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

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

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

rise

Slope =

f(a + h) – f(a)

h

run

We’ve derived the Difference Quotient!

h

a

b

x

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

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

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.

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

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)?

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)

Assignment #1: Rates of Change

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