1. Calculus BC Unit 2 Day 3 Finish Parametric Function Calculus Start Vector Value Function Calculus

2. 2010 BC Exam Question #3 Part b (CALCULATOR ACTIVE) A particle is moving along a curve so that its position at time t is, where and is not explicitly given. Both x and y are measured in meters, and t is measured in seconds. It is known that. (b) There is a point with x-coordinate 5 through which the particle passes twice. Find each of the following. (i) the two values of t when that occurs (ii) the slopes of the lines tangent to the particle’s path at that point (iii) The y-coordinate of that point, given

3. SOLUTION2010 BC Exam Question #3 Part d Calculator Suggestion: Put x(t) in y 1 and dy/dt in y 2 then type the following on the home screen of your calculator:

4. SOLUTION2010 BC Exam Question #3 Part d Calculator Suggestion: Put x(t) in y 1 and dy/dt in y 2 then type the following on the home screen of your calculator:

5. SOLUTION2010 BC Exam Question #3 Part d Calculator Suggestion: Based on prior suggestion dy/dt should be in y 2 then type the following on the home screen of your calculator:

6. Next Objective Calculate the length of a parametric curve

7. The length of a parametrized curve: (Notice the similarity to the distance formula.)

8. Example 2: Find the length of the following curve Graph it:

9. 2010 BC Exam Question #3 Part b (CALCULATOR ACTIVE) A particle is moving along a curve so that its position at time t is, where and is not explicitly given. Both x and y are measured in meters, and t is measured in seconds. It is known that. (b) Find the total distance traveled by the particle for seconds.

10. 2010 BC Exam Question #3 Part b (CALCULATOR ACTIVE) SOLUTION A particle is moving along a curve so that its position at time t is, where and is not explicitly given. Both x and y are measured in meters, and t is measured in seconds. It is known that. (b) Find the total distance traveled by the particle for seconds.

11. Special Functions Polar Parametric Vector-Value Functions

12. Before Vector-Value Functions… Vectors: have magnitude at direction Things expressed with vectors: Force, displacement, velocity. Vectors represented as rays A Start Point B Terminal Point

13. Equal vectors have the same length and direction (same slope). Determine why the following are not equal vectors, then draw an equal vector to u

14. A vector is in standard position if the initial point is at the origin. x y The component form of this vector is: The magnitude (length) ofis: 14

15. P Q (-3,4) (-5,2) The component form of is: v (-2,-2) Order matters! Terminal Point – Initial Point Order matters! Terminal Point – Initial Point What is the component form of ? 15

16. If then v is a unit vector. is the zero vector and has no direction. OTHER important terms 16

17. Vector Operations: (Add the components.) 17 v u u+v u + v is the resultant vector. (Parallelogram law of addition)

18. Vector Operations: Scalar Multiplication: Negative (opposite): (Subtract the components.) 18

19. Example Problem Let and. Find the (a)component form and (b) magnitude of the the following: a)Component form: b)Magnitude: 19

20. Practice 20

21. V T (-6,1) The component form of is: v (-2,-2) Order matters! Terminal Point – Initial Point Order matters! Terminal Point – Initial Point Slopes and Vectors The slope of is: (-2,5) Order does not matter

22. Even though order does not matter for slope calculations... To ensure direction and magnitude are preserved: Do not reduce or cancel signs in slope calculation It’s terminal – initial in slope calculation.

23. P Q (-3, 4) (5,-2) Example: Find the slope and component form of the vector. If the slope had been reduced then the component form would be wrong.