3.4 Solutions.notebook March 24, Horizontal Tangents

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1 Note Fix From 3.3 Horizontal Tangents Just for fun, sketch y = sin x and then sketch its derivative! What do you notice? More on this later 3.4 Velocity and Other Rates of Change A typical graph of the movement of an object is as follows: s s(t) s = position or displacement (in km's) t = time (in hours) If the slope of a function is the change in position over the change in time, this rate of change will be the average velocity in kms/hour. Please note that the speedometer in a car does not measure average velocity, but instantaneous velocity (the velocity at one moment in time). The formula below, based on the definition of the derivative, would give the instantaneous velocity from the above function at any time, t. Velocity is the first derivative of position. For this version of the slope formula, delta t is the same as h in the regular formula. t

2 (km/hour) If the slope of a function is the change in velocity over the change in time, this rate of change will be the average acceleration in km/hour 2. Acceleration is the first derivative of velocity, and the second derivative of position. If s(t) is the position/displacement of an object: v(t) = s'(t) a(t) = v'(t) = s''(t) {velocity is the first derivative of position} {acceleration is the first derivative of velocity and the second derivative of position.} j(t) = a'(t) = v''(t) = s'''(t) { jerk is the first derivative of acceleration, the second derivative of velocity, and the third derivative of position.} In mathematical terms and for the purpose of this course, velocity can be either positive or negative. Velocity to the right or up is considered positive. Velocity to the left or down is considered to be negative. Speed is the absolute value of velocity.

3 Terminology There are various ways to indicate velocity, acceleration and jerk. If we choose to use the "Leibniz terminology", these quantities can be expressed this way: or or Example: The position of an object can be expressed by the function s(t) = 3t 2 9t + 5, where t is in seconds and s is in meters. a) Find the position of the object at t = 5s. b) Find v(t), and then find the velocity at t = 7s. c) Find the speed at t = 1s. d) Find an expression for the object's acceleration, a(t). e) Find the acceleration at t = 10s.

4 If velocity and acceleration are both positive or negative, then a body is speeding up. V + a + V a Speeding Up If velocity and acceleration are opposite in sign, then a body is slowing down. V + a V a + Slowing Down

5 v velocity (in m/s) t time (in seconds) 1) When is the object (a) moving forward, (b) moving backward? 2) When is acceleration positive? Negative? Zero? 3) When is the object (a) speeding up, and (b) slowing down? 4) When does the particle move at its greatest speed? Example: A softball is launched from the Earth. The position, s(t), of the ball is given by the curve: where t represents the time in seconds since the launch. a) Determine the height of the ball after 8 seconds b) Determine the height of the ball after 17 seconds c) Determine the velocity of the ball after 8 seconds d) Determine the velocity of the ball after 17 seconds e) Determine the acceleration of the ball after 4 seconds

6 Example : Find velocity and acceleration when t = 3 seconds for the given function. S in meters, t in seconds. S = t 3 + 6t 2 1 Example: A baseball is thrown in the air, reaching a height of h = 20t 5t 2, with h in meters and t in seconds. a) Find the velocity function. b) When is the velocity 0? c) When is the ball at its greatest height? d) How long does it take to reach a velocity of 15 m/s? e) When, and with what velocity does the ball hit the ground?

7 Review: Solve the following quadratic inequality. Example: A particle is moving back and forth on a horizontal line. It's movement is defined by the position function S(t) = t 3 12t t 30, where t 0. a) Find the velocity and acceleration functions. b) When is the velocity zero? c) During what time interval is the velocity positive? d) When is the velocity negative?

8 Problem: A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of S = 160t 16t 2 feet after t seconds. a) How high does the rock go? b) What is the velocity of the rock when it is 256 ft above the ground on (i) the way up, and (ii) the way down? Problem: A dynamite blast propels a heavy rock straight up with a launch velocity of 160 ft/sec. It reaches a height of S = 160t 16t 2 feet after t seconds. c) What is the acceleration of the rock at time t? d) When does the rock hit the ground? e) With what velocity does the rock hit the ground?

9 Problem: The area of a circle is a function of its radius. a) Find the instantaneous rate of change in the area, A, with respect to the radius, r. b) Find the instantaneous rate of change in the area, A, when the radius is 4. Economists use rate of change to describe "marginal cost". If the cost of production, C(x) is the cost as it relates to the number of units produced, x, then marginal cost: is the rate of change of cost. Economists regard this as the cost to product 1 more unit when the production is at x items. Example: The cost of producing x washing machines is: Determine the marginal cost when x = 100 washing machines.

10 Example: The cost of producing a certain type of tire is: C(x) = x x 2 a) Determine the marginal cost at a production level of 500 tires b) Determine the actual cost of producing the 501st tire Homefun Page 136 #13,15,18, 9,10,11,19,20,21 Homefun Pages 135: # 1, 3, 8, 25, 27, 28