Recall from last day:Average rate of change is represented on a graph by the slope of the secant.

A secant is a line joining two points on a curve.

What is the difference between SLOPE and ARC?Slope IS an average rate of change – on a graph (just

based on numbers.) When we talk specifically about an ARC, it is usually

about quantities ... so it involves UNITS !!! (m/s, g/ml, etc)

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ARoC

We did this example yesterday...A dragster races down a 400m strip. It’s distance in

metres from the starting line after t seconds is given by d(t)=3t2+10t.

What is the average velocity in the last half of the time?

That will be the interval [5,10]

Average Rate of Change = =

= 55The average velocity is 55 m/s.

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dd

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ARoC

A dragster races down a 400m strip. It’s distance in metres from the starting line after t seconds is given by d(t)=3t2+10t.

So for the interval [5,10] , ARC[5,10] = 55 m/s.For the interval [0,5], ARC[0,5] = 25 m/s.What do you think was the actual speed when the clock said 0:05.0 ?How could we get it more accurately if we wanted to be absolutely sure?For the interval [5,6],

ARC[5,6] = 43 m/s.For the interval [4,5], ARC[4,5] = 37 m/s. For the interval [5,5.1], ARC[5,5.1] = 40.3 m/s.For the interval [4.9,5], ARC[4.9,5] = 39.7 m/s.For the interval [5,5.001], ARC[5,5.001] = 40.003 m/s.For the interval [4.999,5], ARC[4.999,5] = 39.997 m/s.Are you confident yet that the INSTANTANEOUS RATE OF CHANGE is 40 m/s?

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ARoC

How does this fit on the graph?

Here is d(t) = 3t2 + 10t

zooming in ...

What is happening?

We are approaching a TANGENT.

Using Intervals to Find Instantaneous Rates of ChangeWe found we could approximate the IRC by looking at the ARC of a small interval before (called the preceding interval) and after (called the following interval).

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IRC

h

h

)()(lim

)()()(

lim

0

0

If we use h to represent that tiny interval, then we are looking at the secant between points

( x , f(x) ) and

( x+h , f(x+h) ) for those intervals.

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xhxxfhxf

IRC

h

h

)()(lim

)()()(

lim

0

0

Another way to deal with this is to use a centred interval.

In this case, we are looking at the secant between points

( x-h , f(x-h) ) and

( x+h , f(x+h) ) for those intervals.

This formula LOOKS like it is only taking care of the following interval, but it actually handles the preceding interval, too. HOW?

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hxhxhxfhxf

IRC

h

h

2)()(

lim

)()()()(

lim

0

0

ex. Find the instantaneous rate of change of f(x) = 3x when x=4.

Use the preceding and following intervals, and then try the centred interval.

It is usually best if you use h = 0.01 or 0.001 at this point to APPROACH a gap with zero distance!

For the preceding interval, I will use points (4,81)

and (3.999, 80.91106...). ARC = 80.91106...81 3.999-4 = 88.9387...

For the following interval, I will use points (4,81) and (4.001,

81.089036...). ARC = 81.089036...81 4.001-4 = 89.0364...

ex. Find the instantaneous rate of change of f(x) = 3x when x=4.

For the preceding interval,

I will use points (4,81) and (3.999,

80.91106...). ARC = 80.91106...-81 3.999-4 = 88.9387...

For the following interval, I will use points (4,81) and (4.001,

81.089036...). ARC = 81.089036...-81 4.001-4 = 89.0364...

For the centred interval, I will use points (4.001,

81.089036...) and (3.999, 80.91106...). ARC = 81.089036..-80.91106... .002 = 88.98761...

The Instantaneous Rate of Change comes out to around 89.

It is NOT actually a whole number answer – don’t make the assumption that it will be!!!

(Answer to 7 decimal places is 88.9875953...)