1. 1. Find the instantaneous velocity for the following 1. Find the instantaneous velocity for the following time—distance function at time = 5 seconds. time—distance function at time = 5 seconds. Quiz 10-3 4. Approximate : (the area under the function for the interval x = [1,4] for the interval x = [1,4] Use the RRAM with 3 rectangles. 41 6 2. How can you tell if you can use the “direct substitution” 2. How can you tell if you can use the “direct substitution” method to solve a limit problem? method to solve a limit problem? 3. Find the limit: 3. Find the limit: 5.Is the approximation of problem #4 larger or smaller problem #4 larger or smaller than the true area? than the true area?

2. HOMEWORK Section 10-4 Section 10-4 (page 832) (evens) 2-24 (12 problems)

3. 10-4 Numerical Derivatives and integrals

4. What you’ll learn Derivatives on a Calculator Definite Integrals on a Calculator Computing a Derivative from Data Computing a Definite Integral from Data … and why The numerical capabilities of a graphing calculator make it easy to perform many calculations that would have been exceedingly difficult in the past.

5. 6543210 0 5 10 15 20 25 30 35 Instantanious Velocity: the actual velocity at an instant in time. How do you find the slope at a point of a function with continously varying slope? Instantaneous slope at a point is the slope of the tangent line at that point. at that point.

6. f(a + h)6543 f(a)0 0 10 15 20 30 35 0 a 10 15 20 a+h 30 35 Instantanious Velocity at “a”: slope of the tangent line at “a” s(t)feet Time (seconds) Approximation: pick two points on the graph and find the slope: “a” and “a + h”: (a+h) – a = h f(a+h) – f(a) This approximation is not a very accurate not a very accurate measure of the slope measure of the slope of the curve at point “a”. of the curve at point “a”.

7. f(a + h)6543 f(a)0 0 10 15 20 25 30 35 0 a 10 15 a+h 20 25 30 35 Instantanious Velocity at “a”: slope of the tangent line at “a” s(t)feet Time (seconds) Better approximation: pick a point that is closer to “a”  make “h” smaller: closer to “a”  make “h” smaller: (a+h) – a = h f(a+h) – f(a) This approximation is better, but still does better, but still does not have the same not have the same slope as the tangent slope as the tangent line at point “a”. line at point “a”.

8. f(a + h)6543 f(a)0 0 10 20 25 30 35 0 a 10 a+h 15 20 25 30 35 Instantanious Velocity at “a”: slope of the tangent line at “a” s(t)feet Time (seconds) Better approximation: pick a point that is closer to “a”  make “h” smaller: closer to “a”  make “h” smaller: h f(a+h) – f(a) This approximation is even better, but still does even better, but still does not have the same not have the same slope as the tangent slope as the tangent line at point “a”. line at point “a”.

9. f(a + h)6543 f(a)0 0 10 20 25 30 35 0 a 10 a+h 15 20 25 30 35 Instantanious Velocity at “a”: slope of the tangent line at “a” s(t)feet Time (seconds) Better approximation: pick a point that is closer to “a”  make “h” smaller: closer to “a”  make “h” smaller: h f(a+h) – f(a) If we make “h” infinitesmially close to “a” then we would have an accurate slope of the tangent line at “a”. of the tangent line at “a”. We can’t make “h” equal to zero since the slope would zero since the slope would be undefined (division by 0) be undefined (division by 0)

10. Example: The distance on object travels is given by the relation: by the relation: For the interval: (2 sec --2.1 sec): For the interval: (2 sec --2.01 sec): For the interval: (2 sec --2.001 sec): For the interval: (2 sec --2.0001 sec): What is happening to the velocity as to the velocity as our time interval our time interval approaches zero? approaches zero? We can make “h” smaller and smaller smaller and smaller and the slope will and the slope will become more become more accurate. accurate.

11. Numerical Derivative f(a + h) a h f(a+h) – f(a-h) h a-ha+h f(a) f(a–h) A method used to calculate slope. run = 2h If we make “h” small enough it will be reasonably accurate.  h = 0.001 “symetric difference quotient”

12. Computing a Numerical Derivative Let: Calculate NDERf(3) by calculating the symetric difference quotient with h = 0.001 Translation: Use a numerical method to find the slope of f(x) at x=3 (using a symetric difference quotient) with h = 0.001

13. Your Turn: 1. f(3 + 0.001) = ? At x = 3 using h = 0.001 3.Compute a Numerical Derivative (slope) of the function at point “a” ( x = 3) using h = 0.001 point “a” ( x = 3) using h = 0.001 2. (3 – 0.001) = ?

14. TI-84 Derivative Method Access the derivative function by pushing: Scroll down to “nDeriv(“ then hit “enter” The format is: nDeriv( expression, variable, x-value of point where you want to find the slope, the incriment value (0.001) “math” (gets you into a list of functions) Example: Find the derivative of at x = 2 with h = 0.001 nDeriv( x³, x, 2, 0.001)

15. Your Turn: 2. Compute a Numerical Derivative for : At x = π using h = 0.001 Careful, what mode should your calculator be in?

16. TI-84 Integration 0 1 2 3 4 5 Time (sec) VELOCITY 50403020100 On the interval [1, 4] I’ll show you how to use the TI-84 to integrate.

17. TI-84 Integration (there are 2 methods) On the interval [1, 4] 0 1 4 Time (sec) VELOCITY 1.Enter your function into “y=“ 2.Access the integration function by: “shift” then hit “calculate” “shift” then hit “calculate” 3. Scroll down to the 7 th option and hit “enter” 4. It asks for the lower limit of the interval: hit “1” then “enter” 5. It asks for the upper limit of the interval: hit “4” then “enter”

18. Your turn: 3. Compute the integral for: On the interval [1, 4]

19. Numerical Integral Access the integration function by: Scroll down to “fnInt(“ then hit “enter” The format is: fnInt( expression, variable, smallest # the of input value interval, largest # of the input value interval) fnInt( expression, variable, smallest # the of input value interval, largest # of the input value interval) then hit “enter” “math” (gets you into a list of functions)

20. Numerical Integral Example: Find the integral of for the input variable interval x = [1, 5]. fnIint( -0.5x² + x + 10, x, 1, 5) = 31.333 = 31.333 Find the integral of the same function but for the interval x = [1, 8] x = [1, 8] = 16.333 = 16.333 Area above the x-axis is positive, area below the x-axis is negative!! negative!!

21. Your turn: 4. Compute the integral using the “fnInt” function for: On the interval [1, 4] You should get the same answer as you did for problem #3. I like the graphical display better. problem #3. I like the graphical display better.