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VECTOR FUNCTIONS

13. VECTOR FUNCTIONS. VECTOR FUNCTIONS. 13.4 Motion in Space: Velocity and Acceleration. In this section, we will learn about: The motion of an object using tangent and normal vectors. MOTION IN SPACE: VELOCITY AND ACCELERATION.

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VECTOR FUNCTIONS

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  1. 13 VECTOR FUNCTIONS

  2. VECTOR FUNCTIONS 13.4 Motion in Space: Velocity and Acceleration In this section, we will learn about: The motion of an object using tangent and normal vectors.

  3. MOTION IN SPACE: VELOCITY AND ACCELERATION • Here, we show how the ideas of tangent and normal vectors and curvature can be used in physics to study: • The motion of an object, including its velocity and acceleration, along a space curve.

  4. VELOCITY AND ACCELERATION • In particular, we follow in the footsteps of Newton by using these methods to derive Kepler’s First Law of planetary motion.

  5. VELOCITY • Suppose a particle moves through space so that its position vector at time t is r(t).

  6. VELOCITY Vector 1 • Notice from the figure that, for small values of h, the vector approximates the direction of the particle moving along the curve r(t).

  7. VELOCITY • Its magnitude measures the size of the displacement vector per unit time.

  8. VELOCITY • The vector 1 gives the average velocity over a time interval of length h.

  9. VELOCITY VECTOR Equation 2 • Its limit is the velocity vector v(t) at time t :

  10. VELOCITY VECTOR • Thus, the velocity vector is also the tangent vector and points in the direction of the tangent line.

  11. SPEED • The speed of the particle at time tis the magnitude of the velocity vector, that is, |v(t)|.

  12. SPEED • This is appropriate because, from Equation 2 and from Equation 7 in Section 13.3, we have: • = rate of change of distance with respect to time

  13. ACCELERATION • As in the case of one-dimensional motion, the acceleration of the particle is defined as the derivative of the velocity: a(t) = v’(t) = r”(t)

  14. VELOCITY & ACCELERATION Example 1 • The position vector of an object moving in a plane is given by: r(t) = t3 i + t2 j • Find its velocity, speed, and acceleration when t = 1 and illustrate geometrically.

  15. VELOCITY & ACCELERATION Example 1 • The velocity and acceleration at time tare: • v(t) = r’(t) = 3t2i + 2t j • a(t) = r”(t) = 6t I +2 j

  16. VELOCITY & ACCELERATION Example 1 • The speed at t is:

  17. VELOCITY & ACCELERATION Example 1 • When t = 1, we have: v(1) = 3 i + 2 j a(1) = 6 i + 2 j|v(1)| =

  18. VELOCITY & ACCELERATION Example 1 • These velocity and acceleration vectors are shown here.

  19. VELOCITY & ACCELERATION Example 2 • Find the velocity, acceleration, and speed of a particle with position vector r(t) = ‹t2, et, tet›

  20. VELOCITY & ACCELERATION Example 2

  21. VELOCITY & ACCELERATION • The figure shows the path of the particle in Example 2 with the velocity and acceleration vectors when t = 1.

  22. VELOCITY & ACCELERATION • The vector integrals that were introduced in Section 13.2 can be used to find position vectors when velocity or acceleration vectors are known, as in the next example.

  23. VELOCITY & ACCELERATION Example 3 • A moving particle starts at an initial position r(0) = ‹1, 0, 0›with initial velocity v(0) = i – j +k • Its acceleration is a(t) = 4t i + 6t j + k • Find its velocity and position at time t.

  24. VELOCITY & ACCELERATION Example 3 • Since a(t) = v’(t), we have: • v(t) = ∫a(t) dt =∫ (4t i + 6t j + k) dt • =2t2i + 3t2j + t k + C

  25. VELOCITY & ACCELERATION Example 3 • To determine the value of the constant vector C, we use the fact that v(0) = i –j + k • The preceding equation gives v(0) = C. • So, C = i – j +k

  26. VELOCITY & ACCELERATION Example 3 • It follows: • v(t) = 2t2i + 3t2j + t k + i – j + k • = (2t2 + 1) i + (3t2 – 1) j + (t + 1) k

  27. VELOCITY & ACCELERATION Example 3 • Since v(t) = r’(t), we have: • r(t) = ∫v(t) dt • = ∫ [(2t2 + 1) i + (3t2 – 1) j + (t + 1) k] dt • = (⅔t3 + t) i + (t3 – t) j + (½t2 + t) k + D

  28. VELOCITY & ACCELERATION Example 3 • Putting t = 0, we find that D = r(0) = i. • So,the position at time t is given by: • r(t) = (⅔t3 + t + 1) i + (t3 – t)j + (½t2 + t) k

  29. VELOCITY & ACCELERATION • The expression for r(t) that we obtained in Example 3 was used to plot the path of the particle here for 0 ≤t ≤ 3.

  30. VELOCITY & ACCELERATION • In general, vector integrals allow us to recover: • Velocity, when acceleration is known • Position, when velocity is known

  31. VELOCITY & ACCELERATION • If the force that acts on a particle is known, then the acceleration can be found from Newton’s Second Law of Motion.

  32. VELOCITY & ACCELERATION • The vector version of this law states that if, at any time t, a force F(t) acts on an object of mass m producing an acceleration a(t), then • F(t) = ma(t)

  33. VELOCITY & ACCELERATION Example 4 • An object with mass m that moves in a circular path with constant angular speed ω has position vector r(t) = a cos ωt i + a sin ωt j • Find the force acting on the object and show that it is directed toward the origin.

  34. VELOCITY & ACCELERATION Example 4 • To find the force, we first need to know the acceleration: • v(t) = r’(t) = –aω sin ωt i + aω cos ωt j • a(t) = v’(t) = –aω2cos ωt i – aω2sin ωt j

  35. VELOCITY & ACCELERATION Example 4 • Therefore, Newton’s Second Law gives the force as: • F(t) = ma(t) • = –mω2 (a cos ωt i + a sin ωt j)

  36. VELOCITY & ACCELERATION Example 4 • Notice that: F(t) = –mω2r(t) • This shows that the force acts in the direction opposite to the radius vector r(t).

  37. VELOCITY & ACCELERATION Example 4 • Therefore, it points toward the origin.

  38. CENTRIPETAL FORCE Example 4 • Such a force is called a centripetal (center-seeking) force.

  39. VELOCITY & ACCELERATION Example 5 • A projectile is fired with: • Angle of elevation α • Initial velocity v0

  40. VELOCITY & ACCELERATION Example 5 • Assuming that air resistance is negligible and the only external force is due to gravity, find the position function r(t) of the projectile.

  41. VELOCITY & ACCELERATION Example 5 • What value of α maximizes the range (the horizontal distance traveled)?

  42. VELOCITY & ACCELERATION Example 5 • We set up the axes so that the projectile starts at the origin.

  43. VELOCITY & ACCELERATION Example 5 • As the force due to gravity acts downward, we have: F = ma = –mgjwhere g = |a|≈ 9.8 m/s2. • Therefore, a = –gj

  44. VELOCITY & ACCELERATION Example 5 • Since v(t) = a, we have: • v(t) = –gt j + C • where C = v(0) = v0. • Therefore, r’(t) = v(t) = –gt j + v0

  45. VELOCITY & ACCELERATION Example 5 • Integrating again, we obtain: • r(t) = –½ gt2j + t v0 + D • However, D = r(0) = 0

  46. VELOCITY & ACCELERATION E. g. 5—Equation 3 • So, the position vector of the projectile is given by: r(t) = –½gt2j + t v0

  47. VELOCITY & ACCELERATION Example 5 • If we write |v0| = v0 (the initial speed of the projectile), then • v0 = v0 cos αi + v0 sin αj • Equation 3 becomes: r(t) = (v0 cos α)t i + [(v0 sin α)t – ½gt2] j

  48. VELOCITY & ACCELERATION E. g. 5—Equations 4 • Therefore, the parametric equations of the trajectory are: x = (v0 cos α)t y = (v0 sin α)t – ½gt2

  49. VELOCITY & ACCELERATION Example 5 • If you eliminate t from Equations 4, you will see that y is a quadratic function of x.

  50. VELOCITY & ACCELERATION Example 5 • So, the path of the projectile is part of a parabola.

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