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BCC.01.3 - Using Limits to Find Instantaneous Rates of Change MCB4U - Santowski.

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Presentation on theme: "BCC.01.3 - Using Limits to Find Instantaneous Rates of Change MCB4U - Santowski."— Presentation transcript:

1 BCC Using Limits to Find Instantaneous Rates of Change MCB4U - Santowski

2 (A) Review We have determined a way to estimate an instantaneous rate of change (or the slope of a tangent line) which is done by means of a series of secant lines such that the secant slope is very close to the tangent slope. We accomplish this "closeness" by simply moving our secant point closer and closer to our tangency point, such that the secant line almost sits on top of the tangent line because the secant point is almost on top of our tangency point. Q? Is there an algebraic method that we can use to simplify the tedious approach of calculating secant slopes and get right to the tangent slope??

3 (B) Notations and Definitions We have several notations for tangent slopes We will add one more  where  x or h represent the difference between our secant point and our tangent point as seen in our diagram on the next slide

4 (B) Notations and Definitions- Graph

5 (C) Notations - Derivatives This special limit of is the keystone of differential calculus, so we assign it a special name and a notation. We will call this fundamental limit the derivative of a function f(x) at a point x = a The notation is f `(a) and is read as f prime of a We can further generalize the formula wherein we do not specify a value for x as:

6 (D) Limits and Differentiability One comment needs to be made about limits and derivatives  recall that the idea of derivatives come from limiting values of secant line slopes These secant lines can be drawn on either side (left or right) With the secant line slopes  the limit “from the right” must be the same as “limits from the left”  i.e. the secant slopes that we generate “from the right” must come to the same limiting value as the secant slopes that we generate “from the left” This is nicely illustrated and animated in the following applet: One-sided derivative from IES One-sided derivative Conclusion: If the limit fails to exist at a point in the function, then we can’t take a derivative at that point

7 (C) Notations - Derivatives Alternative notations for the derivative are : f `(x) y` y` dy/dx which means the derivative at a specific point, x = a KEY POINT: In all our work with the derivative at a point, please remember its two interpretations : (1) the slope of the tangent line drawn at a specified x value, and (2) the instantaneous rate of change at a specified point.

8 (D) Using the Derivative to Determine the Slope of a Tangent Line ex 1. Determine the equation of the tangent line to the curve f(x) = -x 2 + 3x - 5 at the point (-4,-33) Then the equation of the line becomes y = 11x + b and we find b to equal (-29) = 11(-4) + b  so b is 15 so y = 11x + 15

9 (D) Using the Derivative to Determine the Slope of a Tangent Line ex 2. A football is kicked into the air and its height is modelled by the equation h(t) = - 4.9t² + 16t + 1, where h is height measured in meters and t is time in seconds. Determine the instantaneous rate of change of height at 1 s, 2 s, 3 s. So the slope of the tangent line (or the instantaneous rate of change of height) is 6.2  so, in context, the rate of change of a distance is called a speed (or velocity), which in this case would be 6.2 m/s at t = 1 sec.  now simply repeat, but use t = 2,3 rather than 1

10 (D) Using the Derivative to Determine the Slope of a Tangent Line ex 3. A business estimates its profit function by the formula P(x) = x 3 - 2x + 2 where x is millions of units produced and P(x) is in billions of dollars. Determine the value of the derivative at x = ½ and at x = 1½. How would you interpret these derivative values?

11 (E) Homework Nelson text, page 213, Q2,5,7bdf,10, are concept questions and 12,13,14,17 are application questions


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