In chapter 1, we talked about parametric equations.

1. In chapter 1, we talked about parametric equations. Parametric equations can be used to describe motion that is not a function. If f and g have derivatives at t, then the parametrized curve also has a derivative at t.

2. The formula for finding the slope of a parametrized curve is: This makes sense if we think about canceling dt.

3. The formula for finding the slope of a parametrized curve is: We assume that the denominator is not zero.

4. To find the second derivative of a parametrized curve, we find the derivative of the first derivative: • Find the first derivative (dy/dx). 2. Find the derivative of dy/dx with respect to t. 3. Divide by dx/dt.

6. Find the first derivative (dy/dx).

7. 2. Find the derivative of dy/dx with respect to t. Quotient Rule

8. 3. Divide by dx/dt.

9. The equation for the length of a parametrized curve is similar to our previous “length of curve” equation:

11. This curve is:

12. Warning: Only some of this is review.

13. Quantities that we measure that have magnitude but not direction are called scalars. Quantities such as force, displacement or velocity that have direction as well as magnitude are represented by directed line segments. B terminal point The length is initial point A

14. B terminal point A initial point A vector is represented by a directed line segment. Vectors are equal if they have the same length and direction (same slope).

15. y A vector is in standard position if the initial point is at the origin. x The component form of this vector is:

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

17. The component form of (-3,4) P is: (-5,2) Q v (-2,-2)

18. Then v is a unit vector. If is the zero vector and has no direction.

19. Vector Operations: (Add the components.) (Subtract the components.)

20. Vector Operations: Scalar Multiplication: Negative (opposite):

21. u v u + v is the resultant vector. u+v (Parallelogram law of addition) v u

22. The angle between two vectors is given by: This comes from the law of cosines. See page 524 for the proof if you are interested.

23. The dot product (also called inner product) is defined as: Read “u dot v” This could be substituted in the formula for the angle between vectors to give:

24. Application: Example 7 A Boeing 727 airplane, flying due east at 500mph in still air, encounters a 70-mph tail wind acting in the direction of 60o north of east. The airplane holds its compass heading due east but, because of the wind, acquires a new ground speed and direction. What are they? N E

25. Application: Example 7 A Boeing 727 airplane, flying due east at 500mph in still air, encounters a 70-mph tail wind acting in the direction of 60o north of east. The airplane holds its compass heading due east but, because of the wind, acquires a new ground speed and direction. What are they? N E u

26. Application: Example 7 A Boeing 727 airplane, flying due east at 500mph in still air, encounters a 70-mph tail wind acting in the direction of 60o north of east. The airplane holds its compass heading due east but, because of the wind, acquires a new ground speed and direction. What are they? N v 60o E u

27. Application: Example 7 A Boeing 727 airplane, flying due east at 500mph in still air, encounters a 70-mph tail wind acting in the direction of 60o north of east. The airplane holds its compass heading due east but, because of the wind, acquires a new ground speed and direction. What are they? N We need to find the magnitude and direction of the resultant vectoru + v. v u+v E u

28. N The component forms of u and v are: v 70 u+v E 500 u Therefore: and:

29. N 538.4 6.5o E The new ground speed of the airplane is about 538.4 mph, and its new direction is about 6.5o north of east. p