1. Today’s Learning Goals …
Make the connection between the average rate of change and the slope of a secant to a graph between two points and use this to calculate the average rate of change in practical applications.
Make the connection between the instantaneous rate of change and the slope of a tangent to a graph at a certain point and use this to calculate the instantaneous rate of change in practical applications.
NOTE: Special emphasis will be placed on calculating velocity.

2. 1.3 Rates of Change Average Rate of Change
The difference quotient Δ𝑦 Δ𝑥 = 𝑓 𝑏 −𝑓(𝑎) 𝑏−𝑎 is called the average rate of change in 𝑦 with respect to 𝑥 over the interval from 𝑥=𝑎 to 𝑥=𝑏.
Instantaneous Rate of Change
The instantaneous rate of change in 𝑦=𝑓(𝑥) when 𝑥=𝑎 is lim ℎ→0 𝑓 𝑎+ℎ −𝑓(𝑎) ℎ , provided that the limit exists.

3. Example #1: Velocity
A pebble is dropped from a cliff, 80m high. After 𝑡 seconds, the pebble is 𝑠 metres above the ground, where 𝑠 𝑡 =80−5 𝑡 2 , 0≤𝑡≤4.
a. Calculate the average velocity of the pebble between the times 𝑡=1 s and 𝑡=3 s
Average velocity = 𝑠 3 −𝑠(1) 3−1
𝑠 1 =80− =75
Average velocity = change in position change in time = ∆𝑠 ∆𝑡 = 𝑠 𝑏 −𝑠(𝑎) 𝑏−𝑎
𝑠 3 =80− =35
= 35−75 3−1
= −40 2
∴ the average velocity of the pebble between 𝑡=1 s and 𝑡=3 s is −20 𝑚/𝑠.
=−20 𝑚/𝑠

4. Example #1: Velocity
A pebble is dropped from a cliff, 80m high. After 𝑡 seconds, the pebble is 𝑠 metres above the ground, where 𝑠 𝑡 =80−5 𝑡 2 , 0≤𝑡≤4.
b. Calculate the instantaneous velocity of the pebble at 𝑡=1 s.
= lim ℎ→0 −10ℎ−5 ℎ 2 ℎ
𝑉 𝑡 = lim ℎ→0 𝑠 1+ℎ −𝑠(1) ℎ
= lim ℎ→0 80−5 (1+ℎ) 2 −75 ℎ
= lim ℎ→0 ℎ(−10−5ℎ) ℎ
= lim ℎ→0 80−5(1+2ℎ+ ℎ 2 )−75 ℎ
= lim ℎ→0 −10−5ℎ
=−10 𝑚/𝑠
= lim ℎ→0 80−5−10ℎ−5 ℎ 2 −75 ℎ
∴ the velocity of the pebble at 𝑡=1 s is −10 𝑚/𝑠.

5. Practice
A toy rocket is launched straight up so that it’s height 𝑠, in metres, at time 𝑡, in seconds, is given by 𝑠 𝑡 =−5 𝑡 2 +30𝑡+2, 0≤𝑡≤4. What is the velocity of the rocket at 𝑡=4?
-10 m/s

6. In summary … QUESTIONS: p.29-31 #1, 7bc, 9, 11, 12, 13, 15, 16
The average rate of change equivalent to the slope of the secant.
Δ𝑦 Δ𝑥 = 𝑓 𝑏 −𝑓(𝑎) 𝑏−𝑎
The instantaneous of change is equivalent to the slope of the tangent.
lim ℎ→0 𝑓 𝑎+ℎ −𝑓(𝑎) ℎ
Velocity is an example of a rate of change between distance and time.
QUESTIONS: p #1, 7bc, 9, 11, 12, 13, 15, 16