Section 2.2 Day 2 Basic Differentiation Rules & Rates of Change AP Calculus AB
Example 1: Tangent Line Write the equation of the tangent line for 𝑓 𝑥 = 𝑥 3 −2𝑥+10 at the point 𝑥=−2 1. Need a point: 𝑓 −2 =6 2. Need a slope: 𝑓 ′ 𝑥 =3 𝑥 2 −2 𝑓 ′ −2 =12 3. Put into point-slope form: 𝑦−6=12(𝑥+2)
Example 2: Tangent Line Write the equation of the tangent line for 𝑓 𝑥 =5 cos 𝑥 + 𝑥 2 at the point 𝑥=0 1. Need a point: 𝑓 0 =5 2. Need a slope: 𝑓 ′ 𝑥 =−5 sin 𝑥 +2𝑥 𝑓 ′ 0 =0 3. Put in point-slope form: 𝑦 −5=0(𝑥−0)
Example 3: Horizontal Tangents Does the curve 𝑦= 𝑥 4 −2 𝑥 2 +2 have any horizontal tangents? If so, where? 1. Horizontal tangents means the slope of the horizontal is 0. 2. Need slope function: 𝑦 ′ =4 𝑥 3 −4𝑥 3. Set it equal to 0 to find x values: 4 𝑥 3 −4𝑥=0 4𝑥 𝑥 2 −1 =0 4. Thus, it occurs at 𝑥=0, 𝑥=1 𝑎𝑛𝑑 𝑥=−1
Relationship between Position & Velocity Let’s take a look at a function that represents the position of an object. What are the units of the slope of the function? Thus, the derivative of the position function is the velocity function.
Example 4: Position & Velocity pg113 Ex 9 If a billiard ball is dropped from a height of 100ft, its height, 𝑠 at time 𝑡 is given by the position function 𝑠=−16 𝑡 2 +100 where 𝑠 is measured in feet and 𝑡 in seconds. A) Find the average velocity over the interval [1, 2] B) What is the velocity of the billiard ball at 4 seconds?
Example 4: Position & Velocity pg113 Ex 9 If a billiard ball is dropped from a height of 100ft, its height, 𝑠 at time 𝑡 is given by the position function 𝑠=−16 𝑡 2 +100 where 𝑠 is measured in feet and 𝑡 in seconds. A) Find the average velocity over the interval [1, 2] This is AROC: 𝑓 2 −𝑓 1 2−1 = 36−84 1 =−48 𝑓𝑒𝑒𝑡/𝑠𝑒𝑐𝑜𝑛𝑑
Example 4: Position & Velocity pg113 Ex 9 If a billiard ball is dropped from a height of 100ft, its height, 𝑠 at time 𝑡 is given by the position function 𝑠=−16 𝑡 2 +100 where 𝑠 is measured in feet and 𝑡 in seconds. B) What is the velocity of the billiard ball at 4 seconds? This is IROC: 𝑓 ′ 𝑥 =−32𝑡 𝑓 ′ 4 =−128 𝑓𝑒𝑒𝑡/𝑠𝑒𝑐𝑜𝑛𝑑
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