Aim: What do these derivatives do for us anyway? Do Now: Find the average rate of change of f(x) = x2 on the interval [-1, 2] Average rate of change is the slope of the secant line connecting the two endpoints of the interval

Find avg. rate of change f(x) = x2 for [-1, 2] Average Rate of Change The average rate of change of a function f(x) on an interval [a, b] is found by computing the slope of the line joining the end points of the function on that interval. 2 – (-1) f(2) – f(-1) -1, f(-1) 2, f(2) Find avg. rate of change f(x) = x2 for [-1, 2]

Displacement, Velocity & Acceleration Average velocity = Velocity – speed and direction Position Function g = -32 ft/sec2 Freely Falling Object Speed – Absolute Value of velocity Velocity Function Instantaneous Rate of Change of distance with respect to time. 1st Derivative Acceleration Function 1st Derivative of Velocity Instantaneous Rate of Change in Velocity with respect to time. 2nd Derivative of Distance or Displacement

Displacement vs. Distance You drive 100 miles and then return 70 of those miles. What is your total distance traveled? What is your displacement? 100 + 70 = 170 miles total distance 100 miles: + displacement start return – 70 miles: – displacement finish 30 miles: net displacement

Aim: What do these derivatives do for us anyway? Do Now: Academic rigor is teaching, learning, and assessment which promotes student growth in knowledge of the discipline and the ability to any analyze, synthesize and critically evaluate the content under study. What does it look like? “Rigor” in the context of intellectual work refers to thoroughness, carefulness, and right understanding of the material learned. Rigor is to academic work what careful practice and nuanced performance is to musical performance, and what intense and committed play is to athletic performance. When we talk about a ‘rigorous course’ in something, it’s a course that examines details, insists on diligent and scrupulous study and performance, and doesn’t settle for a mild or informal contact with the key ideas. Robert Talbert

Aim: What do these derivatives do for us anyway? Do Now: Find dy/dx given sin x + 2cos(2y) = 1

Motion of Freely Falling Object A ball is dropped from a window 20 feet above the ground. Find the height of ball after 1 sec. Find the velocity of ball after 1 sec. When will the ball hit the ground? What is its speed at that moment? Position Function g = -32 ft/sec2 Freely Falling Object

Motion of Freely Falling Object A ball is dropped from a window 20 feet above the ground. Find the height of ball after 1 sec. Find the velocity of ball after 1 sec. When will the ball hit the ground? What is its speed at that moment?

Model Problem A football is punted into the air. Its displacement (directed distance) from the ground is a function of time t in seconds and is described by the equation y = -16t2 + 37t + 3. Find the velocity at t = 1 at t = 2 Find the acceleration at t = 1 at t = 2

Model Problem: y = -16t2 + 37t + 3. Find the velocity at t = 1 at t = 2 @ 1 sec @ 2 sec speed is faster a 2 sec. than at 1 sec. but velocity is less velocity is directed speed

Model Problem: y = -16t2 + 37t + 3. b. Find the acceleration at t = 1 at t = 2 @ 1 sec @ 2 sec negative acceleration means velocity is decreasing

Speeding Up or Slowing Down? If velocity and acceleration have the same sign, the object is speeding up If velocity and acceleration have the opposite signs, the object is slowing down. positive velocity positive acceleration negative velocity negative acceleration Same signs positive velocity negative acceleration negative velocity positive acceleration Opposite signs

Rectilinear Motion Rectilinear motion is motion along a straight line. Particle Velocity v(t) = 0 & a(t) isn’t = 0 v(t) > 0 v(t) < 0 Sign of v(t) changes particle at rest particle moves right or up particle moves left or down particle changes directions

Aim: What do these derivatives do for us anyway? Do Now: Find the equation of the normal to 3x2 – 4y2 + y = 9 when x = 2 in QI.

Model Problem A particle moves along a line with its position at time t given by s(t) = t3 – 3t + 2, t in seconds, s in feet. Find velocity function. Find v(0) and v(2). When is velocity zero? Where is the particle at that time? Draw a number line indicating the position and velocity of particle at t = 0, t = 1 and t = 2. Refer to the number line to write a description of the motion of the particle from t = 0 to t = 2. How far does it travel in from t = 0 to t = 4?

Model Problem A particle moves along a line with its position at time t given by s(t) = t3 – 3t + 2, t in seconds, s in feet. Find velocity function. Find v(0) and v(2). When is velocity zero? Where is the particle at that time? v(t) = s’(t) = 3t2 – 3 v(0) = -3 ft/sec and v(2) = 9 ft/sec 3t2 – 3 = 0: t = 1; s(1) = 0

Model Problem A particle moves along a line with its position at time t given by s(t) = t3 – 3t + 2, t in seconds, s in feet. d. Draw a number line indicating the position and velocity of particle at t = 0, t = 1 and t = 2. 2 4 s(t) Particle Motion Time t Position Velocity 1 2 s(t) = t3 – 3t + 2 2 4 v(t) = s’(t) = 3t2 – 3 -3 ft/s 0 ft/s 9 ft/s

Model Problem A particle moves along a line with its position at time t given by s(t) = t3 - 3t2 + 2, t in seconds, s in feet. e. Refer to the number line to write a description of the motion of the particle from t = 0 to t = 2. At t = 0, the particle is at s = 2 and is moving to the left at 3 ft/s. One second later at t = 1, the particle is at s = 0 and is at rest. The particle then turns and at t = 2, the particle is a s = 4 and moving to the right at 9 ft/s. 2 4 s(t) Particle Motion Time t Position Velocity 1 2 s(t) = t3 – 3t + 2 2 4 v(t) = s’(t) = 3t2 – 3 -3 ft/s 0 ft/s 9 ft/s

Model Problem e. How far does it travel from t = 0 to t = 4? We must divide the time interval into the time when the velocity is negative and positive and add the absolute values of each distance. 2 4 s(t) Particle Motion Time t Position Velocity 1 2 s(t) = t3 – 3t + 2 2 4 v(t) = s’(t) = 3t2 – 3 -3 ft/s 0 ft/s 9 ft/s

e. How far does it travel from t = 0 to t = 4? Model Problem e. How far does it travel from t = 0 to t = 4? We must divide the time interval into the time when the velocity is negative and positive and add the absolute values of each distance during those intervals. s(t) = t3 – 3t + 2 Interval [0, 1]; velocity negative Parametric graphing: Mode: t-range [0, 10]; increments of .1; x-range [0, 50]; y-range [0, 2]; x = t3 – 3t + 2; y = 1 compare with Function graph: x-min: -1; max: 5; y min -10, y max 60 Interval [1, 4]; velocity positive 56

Model Problem A particle moves in the x-direction (miles) in such a way that its displacement from the y-axis is x = 3t3 - 30t2 + 64t + 57, for t > 0 in hours. Find equations for its velocity and accel. Find velocity and accel. at t = 2, t = 4, and t = 6. At each time, state Whether x is increasing or decreasing and at what rate Whether the object is speeding up or slowing down c. At what times in the interval [0, 8] is x at a maximum? Is x ever negative in [0, 8]?

Model Problem x = 3t3 - 30t2 + 64t + 57, for t > 0 Find equations for its velocity and accel. Find velocity and accel. at t = 2, t = 4, and t = 6. t v(t) a(t) x is … Speeding/Slowing 2 4 6 -20 -24 decreasing at 20mph Speeding up -32 12 decreasing at 32mph Slowing down 28 48 increasing at 28mph Speeding up

x t Model Problem x = 3t3 - 30t2 + 64t + 57, for t > 0 c. At what times in the interval [0, 8] is x at a maximum? Is x ever negative in [0, 8]? x at t = - relative maximum; velocity changes from positive to negative at t = 8 - absolute maximum; x is never negative in interval t

Aim: What do these derivatives do for us anyway? Do Now: Given the position function x(t) = t4 – 8t2 find the distance that the particle travels from t = 0 to t = 4.

Model Problem Use implicit differentiation to find the derivative of

Model Problem Use implicit differentiation to find

Model Problem If the position function of a particle is find when the particle is changing direction.

Model Problem If the position function of a particle is find the distance that the particle travels from t = 2 to t = 5.

Model Problem A particle moves along a line with its position at time t given by s(t) = tsint, t in seconds, s in feet. Find velocity function and acceleration function b. When is velocity zero? Where is the particle at that time? Draw the path of the particle on a number line and show the direction of increasing t. Is the distance traveled by the particle greater than 10?