Announcements. Topics: Homework: - sections 2.2, 2.3, 4.1, and 4.2 * Read these sections and study solved examples in your textbook!

Transcription

1 Announcements Topics: - sections 2.2, 2.3, 4.1, and 4.2 * Read these sections and study solved examples in your textbook! Homework: - review lecture notes thoroughly - work on practice problems from the textbook and assignments from the coursepack as assigned on the course web page (under the SCHEDULE + HOMEWORK link)

2 Trigonometric Functions Trigonometric functions are used to model quantities that oscillate.

3 Trigonometric Models Example: Seasonal Growth A population of river sharks in New Zealand changes periodically with a period of 12 months. In January, the population reaches a maximum of 14, 000, and in July, it reaches a minimum of 6, 000. Using a trigonometric function, find a formula which describes how the population of river sharks changes with time.

4 Example: Trigonometric Models

5 Inverse Trigonometric Functions Since the 3 main trigonometric functions are not one-to-one on their natural domains we must first restrict their domains in order to define inverses.

6 Inverse of Sine Restrict the domain of f (x) = sin x to [ π, π ]. 2 2 Now the function is one-to-one on this interval so we can define an inverse.

7 Inverse of Sine The inverse of the restricted sine function is denoted by f 1 (x) = sin 1 x or f 1 (x) = arcsin x. Cancellation equations: arcsin(sin x) = x sin(arcsin x) = x x [ π, π ] 2 2 x [ 1,1] (domain of sin x) (domain of arcsin x) Calculate: arcsin( 1) 2 sin(arcsin( 5 )) 7 arcsin(sin(π))

8 Graphs of Sine and Arcsine y = sin x y = arcsin x domain: x [ π 2, π 2 ] domain: x [ 1,1] range: y [ 1,1] range: y [ π 2, π 2 ]

9 Inverse of Tangent Restrict the domain of f (x) = tan x to ( π, π ). 2 2 This portion of tangent passes the HLT so tangent is one-to-one here

10 Inverse of Tangent The inverse of the restricted tangent function is denoted by f 1 (x) = tan 1 x or f 1 (x) = arctan x. Cancellation equations: arctan(tan x) = x tan(arctan x) = x x ( π 2, π 2 ) x (, ) (restricted domain of tan x) (domain of arctan x) Calculate: arctan(1) tan(arctan10) arctan( 3)

11 Graphs of Tangent and Arctangent y = tan x y = cos x y = arctan x y = arccos x domain: range: x ( π 2, π 2 ) y (, ) domain: x (, ) range: y ( π 2, π 2 )

12 Real-life Use of Arctangent Example: Model for World Population One of the many models used to analyze human population growth is given by $ P(t) = & π % 2 arctan 2007 t 42 where t represents a calendar year and P(t) is the population in billions. ' ) (

13 Rates of Change The rate of change of a function tells us how the dependent variable changes when there is a change in the independent variable. Geometrically, the rate of change of a function corresponds to the slope of its graph.

14 Average Rate of Change = Slope of Secant Line The average rate of change of f(t) from t=t 1 to t=t 2 corresponds to the slope of the secant line PQ. m PQ = Δf Δt = f (t 2) f (t 1 ) t 2 t 1 t 2 f( ) t 1 f( ) y Q P t 1 t 2 secant y = f(t) t

15 Average Rate of Change = Slope of Secant Line Alternative Notation: The average rate of change of f(t) from the base point t=t 0 to t=t 0 +Δt is m PQ = Δf Δt = f (t + Δt) f (t ) 0 0 Δt t +Δt f( ) 0 t 0 f( ) y Q P t0 t 0+Δt secant y = f(t) t Δt

16 Average Rate of Change = Slope of Secant Line Example: Find the average rate of change of the function s(t) = lnt starting from time t=1 and lasting 1, 0.1, and 0.01 units of time. t 0 Δt t 0 +Δt f(t 0 +Δt) - f(t 0 ) Δt

17 Estimating the Slope of the Tangent Steps: 1. Approximate the tangent at P using a secant line intersecting P and a nearby point Q. y 2. Obtain a better approximation to the tangent at P by moving Q closer to P, but Q P. P Q 3 Q 2 Q 1 y = f(t) 3. Define the slope of the tangent at P to be the limit of the slopes of secants PQ as Q approaches P. t

18 Instantaneous Rate of Change = Slope of Tangent Line The instantaneous rate of change of f(t) at t=t 0 corresponds to the slope of the tangent line at t=t 0. Δf f '(t 0 ) = lim Δt 0 Δt = lim Δt 0 f (t 0 + Δt) f (t 0 ) Δt Note: The slope of the curve y=f(t) at P is the slope of its tangent line at P. t +Δt f( ) 0 t 0 f( ) y P tangent Q t0 t 0+Δt y = f(t) t

19 Instantaneous Rate of Change = Slope of Tangent Line Example: s(t) = lnt t 0 Δt t 0 +Δt f(t 0 +Δt) - f(t 0 ) Δt (a) Guess the limit of the slopes of the secants as Δt 0. (b) Use this to find the equation of the tangent line to s(t) when t=1.

20 The Limit of a Function y y=f(x) Notations: f (2) = 5 5 means that the y-value of the function AT x=2 EQUALS $ x 2 4 & if x 2 f (x) = % x 2 ' & 5 if x = 2 x lim f (x) = 4 x 2 means that the y-values of the function APPROACH 4 as x APPROACHES 2

21 The Limit of a Function Definition: lim x a f (x) = L the limit of f(x), as x approaches a, equals L means that the values of f(x) can be made as close as we d like to L by taking x sufficiently close to a, but not equal to a.

22 Limit of a Function Some examples: y y y x=3 y=g(x) y=h(x) y=f(x) x x Note: f may or may not be defined at x=a. Limits are only asking how f is defined NEAR a.

23 Left-Hand and Right-Hand Limits lim x a means f (x) L as x a from the left (x < a). lim x a + means f (x) L as x a from the right (x > a). ** The full limit exists if and only if the left and right limits both exist (equal a real number) and are the same value.

24 Left-Hand and Right-Hand Limits Example: Consider the function f (x) = 1 x, x < 1 x 2, -1 x 1 x 1, x >1 Determine the following limits, if they exist. (a) lim f (x) (b) lim f (x) x 1 x 1

25 Evaluating Limits We can evaluate the limit of a function in 3 ways: 1. Graphically 2. Numerically 3. Algebraically

26 Evaluating a Limit Numerically Example: Use a table of values to estimate the value of x 2 16 lim x 4 x 4 x f(x) undefined

27 BASIC LIMITS Limit of a Constant Function Limit of the Identity Function lim x a c = c, where c R lim x a x = a Example: Example: y lim 2 = 2 x 3 lim x = 3 x 3 y y=2 y=x x x

28 LIMIT LAWS Suppose that c is a constant and the limits exist. Then lim f (x) and limg(x) x a x a lim [ f (x) + g(x)] = lim f (x) + limg(x) x a x a x a lim [ f (x) g(x)] = lim f (x) limg(x) x a x a x a lim x a c f (x) [ ] = c lim x a f (x)

29 LIMIT LAWS Continued 4. lim [ f (x) g(x)] = lim f (x) limg(x) x a x a x a 5. lim x a [ f (x) g(x)] = lim x a f (x) limg(x), if lim g(x) 0 x a x a

30 Evaluating Limits Algebraically Example: Evaluate the limit and justify each step by indicating the appropriate Limit Laws. lim x 1 (x 2 5x + 6) = lim x 1 = lim x 2 lim x 1 5x + lim x 1 x lim x 5lim x + lim x 1 x 1 x 1 = (1)(1) 5(1) + 6 = 2 6 x 1 6

31 Direct Substitution Property From the previous slide, we have lim x 1 (x 2 5x + 6) = 2 f (x) f (1) Notice that we could have simply found the value of the limit by plugging in x=1 into the function.

32 Direct Substitution Property Direct Substitution Property: If f(x) is an algebraic, exponential, logarithmic, trigonometric, or inverse trigonometric function, and a is in the domain of f(x), then lim x a f (x) = f (a)

33 Evaluating Limits Algebraically Evaluate each limit or state that it does not exist. (a) lim x 4 x x + 4 (b) lim x 1 1 x x 1 (c) lim x 1 x + 2 x +1