1. 10.2: Vectors In A Plane Gabe Ren, Tommy Cwalina, Emily He, Eric Mi, Siddarth Narayan

2. We have used scalar quantities until now (Which have only defined magnitude) Vector- A quantity with defined magnitude AND direction. For purposes of this lesson, we will only be using two dimensions (xy plane) Introduction

3. ●Vector v is an ordered pair of real numbers, denoted in component form ○ a and b are called components of v ○ represented by arrow from origin to point (a, b) in xy coordinate plane ○ Magnitude of v-Absolute value of v, or √(a 2 +b 2 ) Fundamentals

4. In order to define direction of vector, we use what's called the direction angle o Denoted by Θ symbol o Range of [0, 360) (degrees) and [0, 2pi) (radians) ●Important-Direction angle is smallest angle formed with positive x-axis for nonzero vector v ●One frequent method you will use to find angle: arctan(b/a)=Θ (where b and a are the vertical and horizontal components of the vector respectively) Direction of Vectors

5. Bold letters (u and v) Lines above ( → u and → v) Angled brackets ( ) Note: Any two arrows with the same length and pointing in the same direction represent the same vector. Also known as equivalent vectors Representation of Vectors

6. If the vector v runs through initial point (a, b) and through terminal point (x, y), the vector representation is Head Minus Tail (HMT) Rule

7. ●If provided with the magnitude (A) and direction angle (Θ) of a vector, it can be represented by ○ Represent the vector and its components as a right triangle to help visualize The Vector Angles

8. Find the magnitude and the direction angle Θ of the vector v= The magnitude of v is |v|=√((-1-0) 2 +(√3-0) 2 )=2. Using triangle ratios, we see that the direction angle cos=-½ satisfies sin=3/2, so Θ=120° or 2/3 radians. Example 1-Finding Magnitude and Direction

9. Find the component form of a vector with magnitude 3 and direction angle 40°. The components of the vector, found using triangles, are x=3cos(40°) and y=3sin(40°). The vector is = Example 2-Finding Component Form

10. a) A river flows at 3 mph and a rower rows at 6 mph. What heading should the rower take to go straight across a river? b) Answer the same question if the river flows at 6 mph and the rower rows at 3 mph. Example 3-Finding Direction

11. a) We need to find the direction angle. We can use: sin(θ)=3/6=½ θ=30◦=π/6 radians b) sin(θ)=6/3. Since this is impossible (sin(θ)>1) we conclude no heading will work. This makes sense because the river flows faster than the rower rows, so the boat will be pushed downstream no matter what the heading. Example 3-Finding Direction (Cont’d)

12. Vector Addition/Subtraction o If vector u is and vector v is, then the sum u+v= o u-v=u+(-v) Scalar Multiplication o If vector u is and is multiplied by scalar k, the product ku= Opposite vector o The opposite of vector v is -v=(-1)v Vector Operations

13. Vector Addition/Subtraction (easier) o Tail-to-head Representation  Arrow from origin to (u 1, u 2 ) is representation of vector u  Arrow from (u 1, u 2 ) to (u 1 +v 1, u 2 +v 2 ) is representation of u+v o Parallelogram Representation  Representation of vectors u and v from origin form a parallelogram whose diagonal is u+v Vector Operation Graphs

14. Scalar Multiplication o Product ku can be represented by stretch (or shrink) of u by a factor of k.  If k>0, then ku points in the same direction as u  If k<0, then ku points in the opposite direction as u Vector Operation Graphs (Cont’d)

15. Let u= and v=. Find the following: a)2u+3v b)u-v c)|1/2u| a) 2u+3v=2 +3 = = b) u-v= - = = c) |1/2u|=|<-1/2, 3/2|=√((-1/2) 2 +(3/2) 2 )=1/2√10 Example 4-Performing Operations on Vectors

16. Let u, v, w be vectors and a, b be scalars Complete Operation Rules u+v=v+u(u+v)+w=u+(v+w)u+0=u u+(-u)=00u=01u=u1u=u a(bu)=(ab)ua(u+v)=au+av(a+b)u=au+bu(a+b)u=au+bu

17. Vector v/|v| has a magnitude 1 and is called the unit vector. o Represented as (As the magnitude is equal to 1) Unit Vector

18. Find a unit vector of. Using the formula for a unit vector (v/|v|), we get the following: /| | /√(2 2 +3 2 ) /√13 So a unit vector for is Example 5-Finding Unit Vectors

19. A Boeing 727 airplane, flying due east at 500 mph in still air, encounters a 70 mph tail wind acting in the direction 60° 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? Example 6-Finding Ground Speed and Direction

20. If u=the velocity of the airplane alone and v=the velocity of the tail wind, then |u|=500 and |v|=70. Speed and direction are the magnitude and direction of u+v. Let the positive x-axis represent east and the positive y-axis represent north, then the component forms of u and v are u= and v= =. Example 6-Finding Ground Speed and Direction (Cont’d)

21. u+v= |u+v|=√((535)^2+(35√3)^2)=538.4 =arctan((35√3)/535)=6.5 The ground speed is 538.4 mph and its new direction is 6.5 north of east. Example 6-Finding Ground Speed and Direction (Cont’d)

22. Remember common formulas and ratio for sine and cosine Whenever stuck, always try to put direction angle in terms of a trigonometric function Visualizing components and resultants on coordinate plane greatly helps (triangle, parallelogram) General Tips

23. Vectors are vital in determining the following: o Position o Velocity o Speed o Acceleration o Direction of motion Calculus and Vectors

24. Suppose a particle moves along a smooth curve so that its position at any time t is (x(t)), (y(t)), where x and y are differentiable functions of t. o Position Vector-r(t)= o Velocity Vector-v(t)= o Speed- Magnitude of v, or |v| (Because speed is scalar value) o Acceleration Vector-a(t)= o Direction of motion-v/|v| (Unit Vector) DIFFERENTIATE COMPONENT BY COMPONENT Calculus and Vectors Equations

25. A particle moves in the plane so that its position at any time t≥0 is given by (sin(t), t 2 /2). a) Find the position vector of the particle at time t. b) Find the velocity vector of the particle at time t. c) Find the acceleration of the particle at time t. d) Describe the position and motion of the particle at time t=6. Example 7-Doing Calculus Componentwise

26. a) The position vector, which has the same components as the position point, is. b) Differentiate each component of the position vector to get (since velocity is derivative of position). Example 7-Doing Calculus Componentwise (Cont’d)

27. c) Differentiate each component of the velocity vector to get. (since acceleration is derivative of velocity). d) The particle is at the point (sin(6), 18), with velocity and acceleration. Example 7-Doing Calculus Componentwise (Cont’d)

28. A particle moves in the plane with position vector r(t)=. Find the velocity and acceleration vectors. Velocity v(t)= =<3cos(3t), - 5sin(5t)). Acceleration a(t)= =. Example 8-Studying Planar Motion

29. A particle moves in an elliptical path so that its position at any time t≥0 is given by (4sin(t), 2cos(t)). a) Find the velocity and acceleration vectors. b) Find the velocity, acceleration, and speed of motion at t=π/4 c) Sketch the path of the particle and show the velocity vector at the point (4, 0). d) Does the particle travel clockwise or counterclockwise around the origin? Example 9-Studying Planar Motion

30. a) Velocity v(t)= = Acceleration a(t)= = b) Velocity v(pi/4)= = c) Graph parametrically, with x=4sin(t) and y=2cos(t). d) Graph parametrically, with x=4sin(t) and y=2cos(t). Example 9-Studying Planar Motion (Cont’d)

31. Suppose a particle moves along a path so that its velocity at any time t is v(t)=(v 1 (t), v 2 (t)), where v 1 and v 2 are integrable functions of t. o The displacement from t=a to t=b is given by the vector / ⌠ b ⌠ b \ \ ⌡ a(v 1 (t)dt), ⌡ a(v 2 (t)dt)/ o Preceding vector can be added to position at time t=a to get the position at time t=b Displacement of Vectors

32. Suppose a particle moves along a path so that its velocity at any time t is v(t)=(v 1 (t), v 2 (t)), where v 1 and v 2 are integrable functions of t. o The distance traveled from t=a to t=b is the ⌠ b ⌠ b ⌡ a(|v(t)|dt)= ⌡ a(√((v(t))^2+(v(t))^2)dt) Distance Traveled

33. A particle moves in the plane with velocity vector v(t)=(t-3cos(t), 2t-sin(t)). At t=0, the particle is at the point (1, 5). a) Find the position of the particle at t=4. b) What is the total distance traveled by the particle from t=0 to t=4? Example 10-Finding Displacement and Distance Traveled

34. a) Displacement= ⌠ 4 ⌠ 4 ⌡ 0(t-3πcos(πt)dt), ⌡ 0(2t-πsin(πt))dt)> = The particle is at the point (1+8, 5+16)=(9, 21) b) Distance traveled= ⌠ 4 ⌡ 0(√((t-3πcos(πt))^2+(2t- πsin(πt))^2)dt)=33.533 Example 10-Finding Displacement and Distance Traveled

35. Determine the path that the particle in Example 10 travels going from (1, 5) to (9, 21). The velocity vector and the position at t=0 combine to give us the vector equivalent of an initial value problem. The components of the position vector are simply found separately. Example 11-Finding the Path of the Particle

36. x’=t-3πcos(πt) x=t 2 /2-3sin(πt)+C x=t 2 /2-3sin(πt)+1, since x=1 when t=0 y’=2t-πsin(πt) y=t 2 +cos(πt)+C y=t 2 +cos(πt)+4, since y=4 when t=0 Graph the position parametrically from t=0 to t=4. Example 11-Finding the Path of the Particle (Cont’d)