Unit B Overview Derivatives ROC and Tangent Lines
Unit B Overview Derivatives
ROC and Tangent Lines
Average Rate of Change Average rate of change of a quantity over a period of time is the amount of change divided by the time it takes. In general, the average rate of change of a function over an interval is the amount of change divided by the length of that interval.
Example 1
Secant Line A secant line of a curve is a line that (locally) intersects two points on the curve. A secant line is a straight line joining two points on a function. It is also equivalent to the average rate of change, or simply the slope between two points. The average rate of change of a function between two points and the slope between two points are the same thing. Secant line = Average Rate of Change = Slope
Experimental Biology Example w/ Secant Line Experimental biologists often want to know the rates at which populations grow under controlled laboratory conditions.
Example 2 graphically Find the average rate of change from time t = 0 to time t = 8. The average rate of change is essentially the slope. So we will draw a secant line from the point at t = 0 to the point at t = 8. Then we find the slope of the secant line.
Secant Slope Formula
Instantaneous rate of change A moving objects instantaneous rate of change is the rate of change of the moving object at a given instant of time. The major issue is how to compute this. Since the elapsed time would be zero.
Tangent Line The tangent line (or simply the tangent) to a curve at a given point is the straight line that "just touches" the curve at that point. As it passes through the point where the tangent line and the curve meet, or the point of tangency, the tangent line is "going in the same direction" as the curve, and in this sense it is the best straightline approximation to the curve at that point.
Instantaneous rate of change graphically… Find the rate of change at the moment t = 0. Normally we find the slope of the secant line. But we don’t have two points to create a line. So lets estimate what the slope of a line at that point would be. This would be the slope of the tangent line. Using the slope of the tangent line we could guess about what the rate of change is.
Instantaneous rate of change graphically… Find the rate of change at the moment t = 0. Lets get a little more technical than guessing. Draw secant lines with points getting closer to point P. What happens as the points get closer and closer to the point we are interested in?
HW Pg. 104 # 5, 7, 9, 25 (part a), 27 (part a)
Derivatives
Definition: Slope of a Curve at a Point Definition of tangent line with slope m.
Example: The slope of the graph of a linear function
Example: Tangent Lines to the Graph of a Nonlinear Function
Example: Finding the equation of the tangent line. Point: (2, 5) Slope: 4 y – 5 = 4(x – 2) y = 4 x - 3
The derivative of a function
Use for derivative What if I asked find the instantaneous rate of change (slope of tangent line) for f(x) = x² at the points when x = 0, 1, 2, etc…. Would we want to go through the limit definition for every single one of those? This is when we can find use the derivative to make our lives easier.
Example 1
Derivative of x² is 2 x. Meaning 2 x is the slope of my tangent at any value of x. What if I asked find the rate of change (slope of tangent line) for f(x) = x² at the points x = 0, 1, 2… etc Now that we know the derivative we can just plug the x values in… 2(0) = 0 so the slope of the tangent line at 0 is 0 2(1) = 2 so the slope of the tangent line at 1 is 2 2(2) = 0 so the slope of the tangent line at 2 is 4
Example 2 Differentiate (that is, find the derivative of) f (x) = 2 x 2.
Derivative If f’(x) > 0 in (a, b), then: The graph of f(x) has a positive slope in (a, b) If f’(x) < 0 in (a, b), then: The graph of f(x) has a negative slope in (a, b) If f’(x) = 0, there will be a minimum, maximum or inflection point on f(x) an inflection point, point of inflection, or inflection (inflexion) is a point on a curve at which the curvature or concavity changes sign. The curve changes from being concave upwards (positive curvature) to concave downwards (negative curvature)
Derivative notation f’(x) y‘ “y prime” dy/dx “dy dx” or “the derivative of y with respect to x”
Phrases for derivative The following phrases all mean the same thing: the slope of f at a the derivative of f at a f ' (a) the slope of the tangent line to f at a the instantaneous rate of change of f at a
Right and Left Hand Derivatives
Derivative of a Constant Function
Derivative of x Find y’, if y = x y’ = 1
Power Rule for Positive Integer Powers of x
Negative Integer Powers of x
Constant Multiple
The Sum and Difference Rule
The Product Rule
The Quotient Rule
Derivatives of Trigonometric Functions
Example 1 Differentiating with Sine and Cosine Find the derivative. Remember that cos 2 x + sin 2 x = 1 So sin² x = 1 – cos 2 x
Example 4
Find
Find
Find y = ln (4 x)
Find
Official definition
Chain rule
Identifying outside and inside functions
EXAMPLE: (3 x²+1)² Derivative of the outside would be 2(3 x²+1)= 6 x²+2 Derivative of the inside would be 6 x Multiply them you get (6 x²+2) 6 x = 36 x³ + 12 x
x(t) = cos (t² + 1)
Power chain rule
HW Pg. 104 # 13, 14, 17 , 18 Pg. 115 # 3, 5, 7, 9, 13, 19, 23, 39, 43, 45 Pg. 126 # 1, 7, 39 Pg. 128 # 93, 97 Pg. 330 #45, 48 Pg. 357 # 35, 37 Pg. 366 # 37 Pg. 137 # 7, 9, 13, 41, 67, 73
Differentiability
Where might a derivative not exist? At a corner Where the one-sided derivatives differ Example: y = |x|; there is a corner at x = 0 At a cusp Where the slopes of the secant lines approach ∞ from one side and -∞ from the other Example: y = x 2/3 , there is a cusp at x = 0
Where might a derivative not exist? At a vertical tangent Where the slopes of the secant lines approach either ±∞ from both sides Example: f(x) = x 1/3 At a discontinuity Which will cause one or both of the one-sided derivatives be non-existent Example: a step function
Compare the right-hand left-hand derivatives to show that the function is NOT differentiable.
Thm 2. 1 If f is differentiable at x = c, then f is continuous at x = c.
Example Describe the x values at which f is differentiable (-∞, 0) U(0, ∞) (- ∞, -4) U (-4, 0) U (0, ∞) (-4, ∞) (-∞, 2) U(2, ∞)
HW Pg. 106 #81 -86
Curve Sketching
Graphing f and f’ f(x) = 2 x 2 Negative slope f’(x) = 4 x f’(x) > 0 Positive slope f’(x) < 0
How can you draw a derivative, given a function? You can find the slope at point x on f(x) and graph the point (xi , mi ) on the derivative graph.
Decide which f’ graph matches f
Matching F with f’
ROC, velocity, acceleration
The derivative gives us the slope, so find the derivative. y’ = 3 x² - 3 Then plug in x = 2. So the slope of the tangent line to y at x = 2 is 9.
Finding the tangent line
Finding Horizontal Tangents Does the curve y = x 4 – 2 x 2 + 2 have any horizontal tangents? If so, where? Calculate dy/dx 4 x 3 – 4 x Solve the equation dy/dx = 0 for x. 4 x 3 – 4 x = 0 4 x (x 2 – 1) = 0 4 x (x – 1) (x + 1) = 0 4 x = 0; (x – 1) = 0; (x + 1) = 0 x = 0, 1, -1
Second and High Order Derivatives dy/dx: first derivative f’(x) = y’ dy’/dx = second derivative f’’(x) = y’’ = d 2 y/dx 2 dy’’/dx = third derivative f’’’(x) = y’’’ = d 3 y/dx 3 etc….
Displacement and Average Velocity
Example A particle moves along a line so that its position at any time t≥ 0 is given by the function s(t) = t²-4 t +3, where s is measured in meters and t in seconds. Find the displacement of the particle in the first 2 seconds. Find the average velocity during the first 4 seconds.
s(t) = t²-4 t +3 Displacement: s(t) = t²-4 t + 3 s(2) – s(0) = (-1) – 3 = -4 This value means that the particle is 4 units left of where it started.
s(t) = t²-4 t +3 Average Velocity
Motion along a line Suppose that an object is moving a long a coordinate line so that we know its position on that is line is a function of time t: s(t)
Definitions Instantaneous Velocity The instantaneous velocity is the derivative of the position function s(t) with respect to time. At time t, the velocity is: Speed The absolute value velocity
acceleration The derivative of velocity with respect to time. If v(t) = ds/st, then the acceleration at time t is: a(t) = dv/dt = d 2 s/dt 2
Overall
Example: Modeling Vertical Motion A dynamite blast rock propels a heavy rock straight up with a launch velocity of 192 ft/sec (about 130. 91 mph). It reaches a height of s = 192 t – 16 t 2 after t seconds. a) How high does the rock go? The maximum height of the rock is 576 ft.
Example 2 Modeling Vertical Motion b) What is the velocity and speed of the rock when it is 512 ft above the ground on the way up? Way down? (t-4)(t-8) = 0
Example 2 Modeling Vertical Motion c) What is the acceleration of the rock at any time t during its flight (after the blast)?
Example 2 Modeling Vertical Motion d) When does the rock hit the ground? The rock will hit the ground 12 seconds after is blasts off.
Particle Motion When velocity is negative, the particle is moving left When velocity is positive, the particle is moving right When acceleration is negative the particle is decelerating. When acceleration is positive the particle is accelerating.
Particle Motion A particle moves along a line so that its position at any time t≥ 0 is given by the function s(t) = t²-4 t + 3, where s is measured in meters and t in seconds. (a) Find the instantaneous velocity when t = 4 (b) Find the acceleration when t = 4
(a) Find the instantaneous velocity when t = 4 s(t) = t²-4 t + 3 v(t) = ds/dt = 2 t – 4 So v(4) = 4 m /sec
(d) Find the acceleration when t = 4 s(t) = t²-4 t + 3 v(t) ds/dt = 2 t - 4 a(t) = dv/dt = 2 m/sec² So a(4) = 2
Particle Motion The accompanying figure shows the position s(t) of a particle moving on a coordinate line. 1 2 3 4 5 6 7 8 9 t= seconds (a) When is P moving to the left? Moving to the right? Standing still? �P is moving left from t =0 to t = 6. � P is moving right from t = 7 to t = 9 � P is standing still from t = 6 to t = 7
Particle Motion The accompanying figure shows the velocity v = f(t) of a particle moving on a coordinate line. 1 2 3 4 5 6 7 8 9 t= seconds (a) When does the particle move forward (when v(t) > 0)? backward? (v (t) <0)? When does the particle move forward? [0, 1) ( 5, 7) Backward? (1, 5)
Particle Motion The accompanying figure shows the velocity v = f(t) of a particle moving on a coordinate line. 1 2 3 4 5 6 7 8 9 t= seconds (b)When is the particle’s acceleration positive? Negative? Zero? When is the particle’s acceleration positive? (3, 6) Negative? [0, 2) (6, 7) Zero? (2, 3) ( 7, 9]
Particle Motion The accompanying figure shows the velocity v = f(t) of a particle moving on a coordinate line. 1 2 3 4 5 6 7 8 9 t= seconds (c)When does the particle stand still for more than an instant? (7, 9]
Jerk is the derivative of acceleration. If a body’s position at time t is s(t), the body’s jerk at time t is
Example A Couple of Jerks Two bodies moving in simple harmonic motion have the following position functions: s 1(t) = 3 cos t s 2(t) = 2 sin t – cos t Find the jerks of the bodies at time t. s 2(t) = 2 sin t – cos t velocity acceleration = -3 cost jerk velocity = 2 cost + sint acceleration = -2 sint + cost jerk = -2 cost - sint
HW Worksheet
Implicit Differentiation
Implicit Differentiation Process Differentiate both sides with respect to x Collect terms with dy/dx on one side Factor out dy/dx Solve for dy/dx
Find
Find dy/dx if y 2 = x
More Examples
Finding a Second Derivative Implicitly
Finding a Second Derivative Implicitly
Finding a Second Derivative Implicitly
HW Pg. 146 #1, 5, 33
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