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Recall from last day:

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.)

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Recall from last day:

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  1. 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)

  2. 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 = = = 55 The average velocity is 55 m/s.

  3. 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?

  4. How does this fit on the graph? Here is d(t) = 3t2 + 10t zooming in ... What is happening? We are approaching a TANGENT.

  5. Using Intervals to Find Instantaneous Rates of Change We 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). 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.

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

  7. 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...

  8. ex. Find the instantaneous rate of change of f(x) = 3x when x=4. 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 preceding interval, I will use points (4,81) and (3.999, 80.91106...). ARC = 80.91106...-81 3.999-4 = 88.9387... 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...

  9. 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...)

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