Presentation on theme: "Calculus I Chapter two1 To a Roman a “calculus” was a pebble used in counting and in gambling. Centuries later “calculare” meant” to compute,” “to figure."— Presentation transcript:
Calculus I Chapter two1 To a Roman a “calculus” was a pebble used in counting and in gambling. Centuries later “calculare” meant” to compute,” “to figure out.” Today in mathematics and sciences calculus is elementary mathematics enhanced by the limit process.
Calculus I Chapter two 8 In the tables below we’ll find the average rate of change of two functions in three intervals.
Calculus I Chapter two 9 The average rate of change DOES NOT change! In fact for a line the average rate is the same as instantaneous rate, and that is the slope! For this function, the average rate of change DOES change!
Calculus I Chapter two 10 If we are asked to find the instantaneous rate at x=2, then we can say that: since the rate does not change the rate at x=2 is 2! But, for this function the above question is a little tricky! Is the rate at x=2 3 or 5 or none? This is the question that was, finally, answered by Calculus!
Calculus I Chapter two 11 What if we had to answer the question without using calculus? Well, we can approximate the answer. But, how? We can say the rate is approximately 3 or 5, but can we do better than that? Yes! We use x values very close to 2, for example 2.1, 2.01, 2.001, or 2.0001! The closer the value to 2 the better the approximation!
Calculus I Chapter two13 Average velocities are approaching 0. So, we say the instantaneous velocity is 0 at π/2.
14Calculus I Chapter two Geogebra file: TangentLine2 In order to compute the slope of the tangent line to the graph of y = f (x) at (a, f (a)), we compute the slope of the secant line over smaller and smaller time intervals of the form [a, x].
15Calculus I Chapter two Thus we consider f (x)−f (a)/(x−a) and let x → a. If this quantity approaches a limit, then that limit is the slope of the tangent line to the curve y = f (x) at x = a. Geogebra file: TangentLine2
Calculus I Chapter two16 Slope of the tangent line is the number the averages of the slopes of the secant lines approach. In this case it is 2.
Calculus I Chapter two21 Area of Irregular Shapes Problem
Calculus I Chapter five22 The area of the rectangle(s) overestimates the area under the curve.
Calculus I Chapter five23 If we continue with this process of dividing the interval from zero to one to more and more partitions (more rectangles), then the sum of the areas of the rectangles becomes closer to the exact area for every rectangle we add. If we increase the number of rectangles, hypothetically, to infinity, then the sum of the rectangles would give the exact area! Of course we cannot literally do so! But, we can do so in our Imagination using the concept of limit at infinity! Below, the number of rectangles is 10, 20,50, and 100, and the exact answer we are approaching is 1/3! Geogebra File
Calculus I Chapter two24 Limit of a Function and One-Sided Limits
Calculus I Chapter two25 Suppose the function f is deﬁned for all x near a except possibly at a. If f (x) is arbitrarily close to a number L whenever x is suﬃciently close to (but not equal to) a, then we write lim f (x) = L. x→ a
Calculus I Chapter two26 Suppose the function f is deﬁned for all x near a but greater than a. If f (x) is arbitrarily close to L for x suﬃciently close to (but strictly greater than) a, then lim f (x) = L. x→a+ Suppose the function f is deﬁned for all x near a but less than a. If f (x) is arbitrarily close to L for x suﬃciently close to (but strictly less than) a, then lim f (x) = L. x→a−
Calculus I Chapter two27 It must be true that L = M.
Calculus I Chapter two46 The statement we are trying to prove can be stated in cases as follows: For x> 0, −x ≤ x sin(1/x) ≤ x, and For x< 0, x ≤ x sin(1/x) ≤ −x. Now for all x ≠ 0, note that −1 ≤ sin(1/x) ≤ 1 (since the range of the sine function is [−1, 1]). For x> 0 we have −x ≤ x sin(1/x) ≤ x For x< 0 we have −x ≥ x sin(1/x) ≥ x, which are exactly the statements we are trying to prove.
Calculus I Chapter two47 Since lim −|x| = lim |x| = 0, and since −|x| ≤ x sin(1/x) ≤ |x|, x→0 x→0 the squeeze theorem assures us that: lim x sin(1/x) = 0 as well. x→0