Presentation is loading. Please wait.

Presentation is loading. Please wait.

1 10 X 8/30/10 8/31 2 10 XX X 3 Warm up p.45 #1, 3, 50 p.45 #1, 3, 50.

Similar presentations


Presentation on theme: "1 10 X 8/30/10 8/31 2 10 XX X 3 Warm up p.45 #1, 3, 50 p.45 #1, 3, 50."— Presentation transcript:

1

2 1 10 X 8/30/10 8/31 2 10 XX X 3

3 Warm up p.45 #1, 3, 50 p.45 #1, 3, 50

4 Section 2.1 The Tangent and Velocity Problem SWBAT SWBAT –Find the equation of a tangent line –Calculate Average velocity

5 The Tangent Problem What would it mean in general for a line to be tangent to a curve?

6 Example Find an equation of the tangent line to the parabola y = x 2 at the point P(1, 1). Find an equation of the tangent line to the parabola y = x 2 at the point P(1, 1). How do we write the equation of a line? How do we write the equation of a line? What is the slope m of the tangent line? What is the slope m of the tangent line?

7 Problem!!! We need 2 points to find slope Use a secant line. Use a secant line. Use another point Q, to find the slope. Use another point Q, to find the slope.

8 Example (cont’d) The tables show that m PQ approaches 2 ! Now we allow x to approach the value 1 without ever being 1, and calculate m PQ Now we allow x to approach the value 1 without ever being 1, and calculate m PQ

9

10 Example (cont’d) Now, plug in the point and slope: Now, plug in the point and slope: y – 1 = 2(x – 1) or y = 2x – 1

11 The Velocity Problem So, velocity is the slope of the position function So, velocity is the slope of the position function

12 Velocity Problem (cont’d) Suppose a ball is dropped from the top of a tower 450 m tall. Find the velocity after 5 seconds. Suppose a ball is dropped from the top of a tower 450 m tall. Find the velocity after 5 seconds. Position function is s(t) = 4.9t 2. Position function is s(t) = 4.9t 2.

13 Velocity Problem (cont’d) Since there is no time interval, Since there is no time interval, Compute the average velocity from t = 5 to t=5.1. Compute the average velocity from t = 5 to t=5.1.

14 Velocity Problem (cont’d) It appears that as we shorten the time period, the average velocity is becoming closer to 49 m/s. Therefore we define the instantaneous velocity at t = 5 to be v = 49 m/s.

15 Problems Compared (cont’d)

16 Review Two fundamental problems: Two fundamental problems: –secant line vs. Tangent line –Average Velocity vs. instantaneous velocity

17 Assignment 4 p. 97 1-7 odd p. 97 1-7 odd P.84 #5 P.84 #5

18


Download ppt "1 10 X 8/30/10 8/31 2 10 XX X 3 Warm up p.45 #1, 3, 50 p.45 #1, 3, 50."

Similar presentations


Ads by Google