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Rates of Change & Tangent Lines Day 1

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1 Rates of Change & Tangent Lines Day 1
AP Calculus AB

2 Learning Targets Define and determine the average rate of change
Find the slope of a secant line Create the equation of the secant line Find the slope of a tangent line Explain the relationship of the slope of a secant line to the slope of a tangent line Create the equation of the tangent line Define and determine the normal line Define and determine the instantaneous rate of change Solve motion problems using average/instantaneous rates of change

3 Average Rate of Change: Definition
The average rate of change of a function over an interval is the amount of change divided by the length of the interval. In this class, we will often reference average rate of change as β€œAROC”. This is not necessarily a shortcut that is adapted across the board.

4 Example Find the average rate of change of 𝑓 π‘₯ = π‘₯ 3 βˆ’π‘₯ over the interval [1,3] 1. 1, 𝑓 1 & (3, 𝑓 3 ) 2. 𝑓 3 βˆ’π‘“ 1 3βˆ’1 = 24 2 12

5 Secant Line & Slope of the Secant Line Definition
A line that joins two points on a function The slope would be π‘‘βˆ’π‘ π‘βˆ’π‘Ž from the points (a,c) and (b,d)

6 Secant Line Connection to AROC & Example
Notice that the slope of the secant line is the average rate of change! We could say that f(x) is the function which describes the growth of flies. A is at 23 days, b is at 45 days, c is 150, and d is 340. We can see that the AROC is simple the slope of the secant line between the points (23, 150) and (45, 340)

7 How can we find the Slope of Tangent Line?
Let’s take a look at a general secant line What would be the slope of this secant line? 𝑓 π‘Ž+β„Ž βˆ’π‘“ π‘Ž π‘Ž+β„Ž βˆ’π‘Ž = 𝑓 π‘Ž+β„Ž βˆ’π‘“ π‘Ž β„Ž How could we use this to help us find the slope exactly at π‘₯=π‘Ž? Show applet

8 How can we find the slope of the Tangent Line?
Let’s make the distance, β„Ž, get smaller and smaller. Then, the slope would become more and more accurate. The slope of the tangent line would be lim β„Žβ†’0 𝑓 π‘Ž+β„Ž βˆ’π‘“ π‘Ž π‘Ž+β„Ž βˆ’π‘Ž = lim β„Žβ†’0 𝑓 π‘Ž+β„Ž βˆ’π‘“ π‘Ž β„Ž

9 Tangent Line & Slope of the tangent line definition
A straight line that touches the function at one point. The slope of the tangent line would be when π‘₯=π‘Ž lim β„Žβ†’0 𝑓 π‘Ž+β„Ž βˆ’π‘“ π‘Ž β„Ž

10 Slope of Tangent Example
Find the slope of the tangent line of 𝑓 π‘₯ =2π‘₯βˆ’1 at the instant π‘₯=1 1. Definition lim β„Žβ†’0 𝑓 1+β„Ž βˆ’π‘“ 1 β„Ž since a = 1 2. lim β„Žβ†’ β„Ž βˆ’1 βˆ’1 β„Ž = lim β„Žβ†’ β„Žβˆ’1βˆ’1 β„Ž = lim β„Žβ†’0 2β„Ž β„Ž = lim β„Žβ†’0 2 =2 3. Thus, the slope of the tangent line at π‘₯=1 is 2.

11 Instantaneous Rate of Change Definition
The rate of change at a particular instant/moment In other words, it is the slope of the tangent line to the curve. In this class, we will reference this as β€œIROC”

12 Instantaneous Rate of Change Example
Let 𝑓 π‘₯ =2π‘₯βˆ’1 represent the daily growth of a plant in inches. What does the IROC represent at the instant π‘₯=1. 1. Before, we saw that the slope at this point was 2. 2. Therefore, the IROC is that the plant is growing at a rate of 2 inches per day on the first day.

13 Practice 1 Find the average rate of change of the function 𝑓 π‘₯ = π‘₯ 3 +1 over the interval [βˆ’1,1]. Then, write the equation of the secant line that describes that average rate of change. Slope: 𝑓 1 βˆ’π‘“ βˆ’1 1βˆ’ βˆ’1 = =1 (can use β€œy-vars” to put in calculator) Use point slope form for equation: yβˆ’ 𝑦 1 =π‘š π‘₯βˆ’ π‘₯ 1 m = 1, π‘₯ 1 =βˆ’1 π‘œπ‘Ÿ 1, 𝑦 1 =0 π‘œπ‘Ÿ 2 yβˆ’0=1(π‘₯+1) or π‘¦βˆ’2=1(π‘₯βˆ’1)

14 Practice 2 Write the equation of the tangent for the function 𝑓 π‘₯ =3 π‘₯ 2 βˆ’1 at the point π‘₯=2 1. lim β„Žβ†’0 𝑓 2+β„Ž βˆ’π‘“ 2 β„Ž = lim β„Žβ†’0 3 2+β„Ž 2 βˆ’1 βˆ’11 β„Ž 2. lim β„Žβ†’ β„Ž+3 β„Ž 2 βˆ’1βˆ’11 β„Ž = lim β„Žβ†’0 12β„Ž+3 β„Ž 2 β„Ž = lim β„Žβ†’0 β„Ž 12+3β„Ž β„Ž 3. lim β„Žβ†’0 12+3β„Ž =12. Thus, m = The point is (2, 11) 5. Using point-slope form we get: π‘¦βˆ’11=12(π‘₯βˆ’2)

15 Practice 3 Determine whether the curve has a tangent line at the indicated point. If it does, give its slope. If not, explain why not. 𝑓 π‘₯ = 2βˆ’2π‘₯βˆ’ π‘₯ 2 , π‘₯<0 2π‘₯+2, π‘₯β‰₯0 at the point x = 0 1. Is the function continuous? Yes (check all 3 conditions) 2. Check slope from both sides using one-sided limits 3. lim β„Žβ†’ 0 βˆ’ 𝑓 0+β„Ž βˆ’π‘“ 0 β„Ž = lim β„Žβ†’ 0 βˆ’ 2βˆ’2β„Žβˆ’ β„Ž 2 βˆ’2 β„Ž =βˆ’2 4. lim β„Žβ†’ 0 + 𝑓 0+β„Ž βˆ’π‘“ 0 β„Ž = lim β„Žβ†’ 0 + 2β„Ž+2 βˆ’2 β„Ž =2 5. Limits do not match. Therefore, the slope does not exist at this point

16 Practice 4 Write the equation for the normal line to the curve 𝑓 π‘₯ =4βˆ’ π‘₯ 2 at π‘₯=1 (Normal line to a curve is the line perpendicular to the tangent at that point) 1. Normal is opp. recip. of tangent slope 2. lim β„Žβ†’0 𝑓 1+β„Ž βˆ’π‘“ 1 β„Ž =βˆ’2. Thus, π‘š= Use point slope: π‘¦βˆ’3= 1 2 (π‘₯βˆ’1)

17 Exit Ticket for Feedback
1. Find the average rate of change of the function 𝑓 π‘₯ = 4π‘₯+1 over the interval [10, 12] 2. Write the equation of the normal line to the curve 𝑓 π‘₯ =9βˆ’ π‘₯ 2 at π‘₯=2


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