3. Section 2.6 Slopes of tangents  SWBAT:  Find slopes of tangent lines  Calculate velocities

4. Review of Secant lines  Remember this!?! the slope of the secant line PQ :the slope of the secant line PQ :

5. Concept: take the limit x approaches a.

6. Slope of a Tangent to f(x) at point ( a, f(a) ) is:

7. Example  Start with y = x 2 at the point P(2, 4).  Use our definition of the slope of the tangent at a point.  Need a hint:

8. Example (cont’d)

9.  Using the point-slope form of the equation of a line y – 4 = 4(x – 2) So,y = 4x – 4

10. Another Expression (cont’d)

11. Another Expression  There is another way to define slope of a tangent: We let h = x – aWe let h = x – a Then x = a + hThen x = a + h Thus the slope of secant line PQ isThus the slope of secant line PQ is

12. A second definition is:

13. Example 2:  Use to calculate the slope of f(x) at a.

14. Example 3. Find the equation of the tangent line to the hyperbola y = 3/x at the point (3, 1)

15. Example (cont’d)

16.  Solution Let f(x) = 3/x. Then the slope of the tangent at (3, 1) is – ⅓.  Therefore an equation of the tangent line is y – 1 = – ⅓ (x – 3),  or y= ⅓ x+2.

17. Example (cont’d)

18. Velocity The average velocity over a time interval h equals the slope of the secant line PQ.

19. Velocity (cont’d)

20. Taking it to the limit we get... Instantaneous Velocity  This means that the velocity at time t = a is equal to thevelocity at time t = a is equal to the slope of the tangent line at P.slope of the tangent line at P.

21. Example  (Calculator active)  Suppose a ball is dropped from the top of a tower 450m. Tall. Find the velocity at 5 seconds.  Position function f(t) = 4.9t 2

22. Example (cont’d) a) The velocity after 5 s is v(t)=9.8t v(5) = (9.8)(5) = 49 m/s.

23. Part deux:  How fast is the ball going when it hits the ground?

24. Example (cont’d)  It hits the ground when it traveled 450m  The ball will hit the ground when, 4.9t 2 = 450.  Solving for t gives t ≈ 9.6 s.  The velocity of the ball as it hits the ground is therefore v(t) = 9.8t ≈ 94 m/s

25. Assignment 11  P. 145 1-13 odd

26. Other Rates of Change  In general, we can have other average rates of change:  These can lead to instantaneous rates of change.

27. Other Rates (cont’d)

28. Example  Temperature readings T (in °C) were recorded every hour starting at midnight on a day in Whitefish, Montana.  The time x is measured in hours from midnight.  The data are given in the table on the next slide:

29. Example (cont’d)

30. a) Find the average rate of change of temperature with respect to time i.from noon to 3 P.M. ii.from noon to 2 P.M. iii.from noon to 1 P.M. b) Estimate the instantaneous rate of change at noon.

31. Solution to (a) i. From noon to 3 P.M. the temperature changes from 14.3 °C to 18.2 °C, so ∆T = T(15) – T(12) = 18.2 – 14.3 = 3.9 °C while the change in time is ∆x = 3 h. Therefore, the average rate of change of temperature with respect to time is

32. Solution to (a) (cont’d) ii. From noon to 2 P.M. the average rate of change is iii. From noon to 1 P.M. the average rate of change is

33. Solution to (b)  We plot the given data on the next slide and sketch a smooth curve of the temperature function.  Then we draw the tangent at the point P where x = 12.  By measuring the sides of triangle ABC, we estimate the slope of the tangent line to be 10.3/5.5 ≈ 1.9.

34. Solution to (b) (cont’d)

35.  Therefore the instantaneous rate of change of temperature with respect to time at noon is about 1.9 °C.

36. Review  Use of limits to define Tangent linesTangent lines Instantaneous velocitiesInstantaneous velocities Other rates of changeOther rates of change