AP Calculus AB 2.2 Average and Instantaneous Velocity Objectives: Use derivatives to find rates of change

Distance Graphs Slope of Distance Graphs Slope represents velocity “Rate of Change” Distance Time

Average vs. Instantaneous Velocity At one point in time Average Distance Distance Slope of Secant = Average Velocity Slope of Tangent = Instantaneous Velocity Time Time

Directions Find the average rate of change of the function over the indicated interval. Compare this average rate of change with the instantaneous change at the endpoints of the interval.

1) 𝑓 𝑡 =2𝑡+7 𝑜𝑛 [1, 2] Average Rate of Change = 𝑓 2 −𝑓(1) 2−1 = 11−9 2−1 = 2 1 =2 Instant. Rate of Change 𝑓 ′ 𝑡 =2 𝑓 ′ 1 =2 𝑓 ′ 2 =2 Linear

2) 𝑔 𝑡 =2 𝑡 3 −1 𝑜𝑛 [0, 1] Average Rate of Change = 𝑔 1 −𝑔(0) 1−0 = 1−(−1) 1−0 =2 Instant. Rate of Change 𝑔 ′ 𝑡 =6 𝑡 2 𝑔 ′ 0 =6 (0) 2 =0 𝑔 ′ 1 =6 (1) 2 =6

Position Functions Position function in feet and time in seconds 𝑠 𝑡 =−16 𝑡 2 + 𝑣 𝑜 𝑡+ 𝑠 𝑜 s(t) = position as a function of time 𝑣 𝑜 = initial velocity Start from rest, 𝑣 𝑜 = 0 𝑠 0 = initial position On the ground, 𝑠 𝑜 = 0 Rate of change of position is velocity 𝑣(𝑡)= 𝑑𝑠 𝑑𝑡 = 𝑠 ′ 𝑡 =𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑠 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡𝑖𝑚𝑒

Example 1 A silver dollar is dropped 970 feet from the top floor of a hotel at time t = 0. When an object is dropped, the initial velocity is zero. Write the position as a function of time for this object. 𝑉 𝑜 =0 𝐷𝑟𝑜𝑝𝑝𝑒𝑑 𝑆 𝑜 =970 Use: 𝑆 𝑡 =−16 𝑡 2 + 𝑉 𝑜 𝑡+ 𝑆 𝑜 𝑆 𝑡 =−16 𝑡 2 +0𝑡+970 𝑆 𝑡 =−16 𝑡 2 +970

Average Velocity = 𝑆 2 −𝑆(1) 2−1 Example 1 A silver dollar is dropped 970 feet from the top floor of a hotel at time t = 0. When an object is dropped, the initial velocity is zero. b) Find the average velocity for 1≤𝑡≤2. Average Velocity = 𝑆 2 −𝑆(1) 2−1 = (−16 2 ) 2 +970 −(−16(1 ) 2 +970) 2−1 = 906−954 2−1 =−48 𝑓𝑡 𝑠

Must find the derivative of position first Example 1 A silver dollar is dropped 970 feet from the top floor of a hotel at time t = 0. When an object is dropped, the initial velocity is zero. c) Find the instantaneous velocity at t = 1. Must find the derivative of position first 𝑆 ′ 𝑡 =−32𝑡 𝑆 ′ 1 =−32 1 =−32 𝑓𝑡 𝑠

Example 1 A silver dollar is dropped 970 feet from the top floor of a hotel at time t = 0. When an object is dropped, the initial velocity is zero. d) How long will it take for the dollar to hit the ground? (This happens when s(t) = ? ) S(t)=0 𝑆 𝑡 =−16 𝑡 2 +970=0 16 𝑡 2 =970 𝑡 2 =60.625 𝑡=7.786 𝑠

Why is the instantaneous velocity negative? Example 1 A silver dollar is dropped 970 feet from the top floor of a hotel at time t = 0. When an object is dropped, the initial velocity is zero. e) Find the velocity of the dollar just before it hits the ground. This happens when 𝑡=7.786 𝑠 𝑆 ′ 7.786 =−32 7.786 =−249.158 𝑓𝑡 𝑠 Why is the instantaneous velocity negative? Moving Down

Position Functions Position function in meters and time in seconds 𝑠 𝑡 =−4.9 𝑡 2 + 𝑣 𝑜 𝑡+ 𝑠 𝑜 s(t) = position as a function of time 𝑣 𝑜 = initial velocity Start from rest, 𝑣 𝑜 = 0 𝑠 0 = initial position On the ground, 𝑠 𝑜 = 0 Rate of change of position is velocity 𝑣(𝑡)= 𝑑𝑠 𝑑𝑡 = 𝑠 ′ 𝑡 =𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑎𝑠 𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡𝑖𝑚𝑒

Example 2 A rock is thrown straight down from a 400 meter high bridge into a river with an initial velocity of −30 𝑚 𝑠 . Write the position as a function of time. 𝑉 0 =−30, 𝑆 𝑜 =400 𝑆 𝑡 =−4.9 𝑡 2 + 𝑉 𝑜 𝑡+ 𝑆 𝑜 𝑆 𝑡 =−4.9 𝑡 2 + −30 𝑡+400 𝑆 𝑡 =−4.9 𝑡 2 −30𝑡+400

Example 2 A rock is thrown straight down from a 400 meter high bridge into a river with an initial velocity of -30 m/s. What is the velocity after 4 seconds? 𝐼𝑛𝑠𝑡𝑎𝑛𝑡𝑎𝑛𝑒𝑜𝑢𝑠 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑆 ′ 𝑡 =−9.8𝑡−30 𝑆 ′ 4 =−9.8 4 −30=−69.2 𝑚 𝑠

Example 2 A rock is thrown straight down from a 400 meter high bridge into a river with an initial velocity of -30 m/s. What is the velocity after falling 300 feet? 𝑆 𝑡 =400−300=100 𝑚 𝑁𝑒𝑒𝑑 𝑡𝑖𝑚𝑒 𝑡𝑜 𝑔𝑒𝑡 𝑡𝑜 100 𝑓𝑡 𝑎𝑏𝑜𝑣𝑒 𝑔𝑟𝑜𝑢𝑛𝑑 100=−9.8 𝑡 2 −30𝑡+400 0=−9.8 𝑡 2 −30𝑡+300 𝑈𝑠𝑒 𝑡ℎ𝑒 𝑐𝑎𝑙𝑐𝑢𝑙𝑎𝑡𝑜𝑟 𝑆𝑜𝑙𝑣𝑒𝑟 𝑜𝑟 𝐺𝑟𝑎𝑝ℎ 𝑡=4.210 𝑠

Example 2 A rock is thrown straight down from a 400 meter high bridge into a river with an initial velocity of -30 m/s. What is the velocity after falling 300 feet? 𝑡=4.210 𝑠 𝑀𝑢𝑠𝑡 𝑝𝑙𝑢𝑔 𝑡𝑖𝑚𝑒 𝑖𝑛𝑡𝑜 𝑡ℎ𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑇ℎ𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑖𝑠 𝑡ℎ𝑒 𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒 𝑜𝑓 𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑆 ′ 4.210 =−9.8 4.210 −30=−71.258 𝑚 𝑠

Formative Assessment Day 3 Pg. 116-117 (67, 68, 89-94)

Use the graph to make a sketch of the corresponding velocity function Velocity is represented by the slope on a distance vs. time graph 𝐴𝑣𝑔. 𝑉𝑒𝑙 (4−0)= 10−0 4−0 = 5 2 𝐴𝑣𝑔. 𝑉𝑒𝑙 4−8 = 10−10 8−4 =0 𝐴𝑣𝑔. 𝑉𝑒𝑙 8−9 = 20−10 9−8 =10

Formative Assessment Day 4 Pg. 117-118 (95-100, 102, 104, 113, 114) Exit Question Join Code 13 Pg. 117 #90 What is the average rate of change? Type numeric answer into clicker Round to the closest decimal place