10.2: Vectors In A Plane

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

2. Introduction • 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)

3. Fundamentals • Vector v is an ordered pair of real numbers, denoted in component form <a, b> • a and b are called components of v • <a, b> represented by arrow from origin to point (a, b) in xy coordinate plane • Magnitude of v-Absolute value of v, or √(a2+b2)

4. Direction of Vectors • In order to define direction of vector, we use what's called the direction angle • Denoted by Θ symbol • 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)

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

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

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

8. Example 1-Finding Magnitude and Direction Find the magnitude and the direction angle Θ of the vector v=<-1, √3> 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.

9. Example 2-Finding Component Form 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 <3cos(40°), 3sin(40°)>=<2.298, 1.928>

10. Example 3-Finding Direction 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.

11. Example 3-Finding Direction (Cont’d) 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.

12. Vector Operations • Vector Addition/Subtraction • If vector u is <a, b> and vector v is <x, y>, then the sum u+v=<a+x, b+y> • u-v=u+(-v) • Scalar Multiplication • If vector u is <a, b> and is multiplied by scalar k, the product ku=<ka, kb> • Opposite vector • The opposite of vector v is -v=(-1)v

13. Vector Operation Graphs • Vector Addition/Subtraction (easier) • Tail-to-head Representation • Arrow from origin to (u1, u2) is representation of vector u • Arrow from (u1, u2) to (u1+v1, u2+v2) is representation of u+v • Parallelogram Representation • Representation of vectors u and v from origin form a parallelogram whose diagonal is u+v

14. Vector Operation Graphs (Cont’d) • Scalar Multiplication • 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

15. Example 4-Performing Operations on Vectors Let u=<-1, 3> and v=<4, 7>. Find the following: a)2u+3v b)u-v c)|1/2u| a) 2u+3v=2<-1, 3>+3<4, 7>=<2(-1)+3(4), 2(3)+3(7)>=<10, 27> b) u-v=<-1, 3>-<4, 7>=<-1-4, 3-7>=<-5, -4> c) |1/2u|=|<-1/2, 3/2|=√((-1/2)2+(3/2)2)=1/2√10

16. Complete Operation Rules • Let u, v, w be vectors and a,b be scalars

17. Unit Vector • Vector v/|v| has a magnitude 1 and is called the unit vector. • Represented as <cos(Θ), sin(Θ)> (As the magnitude is equal to 1)

18. Example 5-Finding Unit Vectors Find a unit vector of <2, 3>. Using the formula for a unit vector (v/|v|), we get the following: <2, 3>/|<2, 3>| <2, 3>/√(22+32) <2, 3>/√13 <2/√13, 3/√13> So a unit vector for <2, 3> is <2/√13, 3/√13>

19. Example 6-Finding Ground Speed and Direction 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?

20. Example 6-Finding Ground Speed and Direction (Cont’d) 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= <500, 0> and v=<70cos(60°), 70sin(60°)>=<35, 35√3>.

21. Example 6-Finding Ground Speed and Direction (Cont’d) u+v=<535, 35√3> |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.

22. General Tips • 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)

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

24. Calculus and Vectors Equations • 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. • Position Vector-r(t)=<x(t), y(t)> • Velocity Vector-v(t)=<dx/dt, dy/dt> • Speed- Magnitude of v, or |v| (Because speed is scalar value) • Acceleration Vector-a(t)=<d2x/dt2, d2y/dt2> • Direction of motion-v/|v| (Unit Vector) DIFFERENTIATE COMPONENT BY COMPONENT

25. Example 7-Doing Calculus Componentwise A particle moves in the plane so that its position at any time t≥0 is given by (sin(t), t2/2). 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.

26. Example 7-Doing Calculus Componentwise (Cont’d) a) The position vector, which has the same components as the position point, is <sin(t), t2/2>. b) Differentiate each component of the position vector <sin(t), t2/2> to get <cos(t), t> (since velocity is derivative of position).

27. Example 7-Doing Calculus Componentwise (Cont’d) c) Differentiate each component of the velocity vector to get <-sin(t), 1>. (since acceleration is derivative of velocity). d) The particle is at the point (sin(6), 18), with velocity <cos(6), 6> and acceleration <-sin(6), 1>.

28. Example 8-Studying Planar Motion A particle moves in the plane with position vector r(t)=<sin(3t), cos(5t)>. Find the velocity and acceleration vectors. Velocity v(t)=<t’(sin(3t)), t’(cos(5t))>=<3cos(3t), -5sin(5t)). Acceleration a(t)=<t’’(sin(3t)), t’’(cos(5t))>=<-9in(3t), -25cos(3t)>.

29. Example 9-Studying Planar Motion 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?

30. Example 9-Studying Planar Motion (Cont’d) a) Velocity v(t)=<t’(4sin(t)), t’(2cos(t))>=<4cos(t), -2sin(t)> Acceleration a(t)=<t’(4cos(t)), t’(-2sin(t))>= <-4sin(t), -2cos(t)> b) Velocity v(pi/4)=<4cos(pi/4), -2sin(pi/4)>=<2(2)½, -(2)½> c) Graph parametrically, with x=4sin(t) and y=2cos(t). d) Graph parametrically, with x=4sin(t) and y=2cos(t).

31. Displacement of Vectors • Suppose a particle moves along a path so that its velocity at any time t is v(t)=(v1(t), v2(t)), where v1 and v2 are integrable functions of t. • The displacement from t=a to t=b is given by the vector /⌠b ⌠b \ \ ⌡a(v1(t)dt), ⌡a(v2(t)dt)/ • Preceding vector can be added to position at time t=a to get the position at time t=b

32. Distance Traveled • Suppose a particle moves along a path so that its velocity at any time t is v(t)=(v1(t), v2(t)), where v1 and v2 are integrable functions of t. • 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)

33. Example 10-Finding Displacement and Distance Traveled 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?

34. Example 10-Finding Displacement and Distance Traveled a) Displacement=⌠4 ⌠4 ⌡0(t-3πcos(πt)dt), ⌡0(2t-πsin(πt))dt)> =<8, 16> 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

35. Example 11-Finding the Path of the Particle 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.

36. Example 11-Finding the Path of the Particle (Cont’d) x’=t-3πcos(πt) x=t2/2-3sin(πt)+C x=t2/2-3sin(πt)+1, since x=1 when t=0 y’=2t-πsin(πt) y=t2+cos(πt)+C y=t2+cos(πt)+4, since y=4 when t=0 Graph the position <t2/2-3sin(πt)+1, t2+cos(πt)+4> parametrically from t=0 to t=4.