Vector-Valued Functions and Motion Along a Curve · seed 1 · A4, ink-friendly. The answer key prints on its own page for grown-ups.

Position, velocity, and the length of a path

Mathematics · Calculus & Analysis · ages 19-20
Name ______________________   Date ____________
  1. For r(t) = (3t, 4t, 0), what is the speed?

    Answer: ______________

  2. A particle has position r(t) = (t², t³, 0). At t = 0 the velocity is (0, 0, 0). Can the formula T = v / |v| be used to find the unit tangent at t = 0?

    Circle one:   True   False

  3. A particle has velocity v(t) = (2t, 3, 6t). What is its speed at t = 1?

    Answer: ______________

  4. If r(t) = (t, t²), what is its velocity vector?

    • (1, 2t)
    • (1, t²)
    • (2t, 1)
  5. A particle has position r(t) = (t², t³, 2t). What is the magnitude of its acceleration at t = 1? Give your answer to two decimal places.

    Answer: ______________

  6. A particle moves along a path with speed |v(t)| = 3t² from t = 0 to t = 2. The antiderivative of 3t² is t³. What is the distance traveled?

    Answer: ______________

  7. A particle has velocity v = (-3, 4, 0). A student writes the unit tangent as (-3, 4, 0) / 7. What is wrong, and what is correct?

    • The speed is 7, so the student is correct
    • The speed is 5, not 7, so the unit tangent is (-3/5, 4/5, 0)
    • The speed is 5, so the unit tangent is (3/5, 4/5, 0)
  8. For r(t) = (2t, 3t, 6t), what is the unit tangent vector?

    • (2, 3, 6)
    • (1/3, 1/3, 1/3)
    • (2/7, 3/7, 6/7)
  9. For r(t) = (2t, 3t, 6t) with 0 <= t <= 3, what is the arc length?

    Answer: ______________

  10. A particle moves with constant speed 5 from t = 0 to t = 2 along a curving path. Ana says the arc length is still 10. Is Ana right?

    Circle one:   True   False

LightMySky · lightmysky.comW1-mt_Xcjt6exMy3-s1

Answer key

For grown-ups. Fold this page away before handing over the rest.

Position, velocity, and the length of a path W1-mt_Xcjt6exMy3-s1

  1. 5 · Velocity is (3, 4, 0), and its length is 5.
  2. False · At t = 0 the speed is 0, so the denominator |v| is 0 and the formula is undefined.
  3. 7 · At t = 1 the velocity is (2, 3, 6), and its length is √(4 + 9 + 36) = 7.
  4. (1, 2t) · Differentiate each piece: t gives 1 and t² gives 2t.
  5. 6.32 · Velocity is (2t, 3t², 2), acceleration is (2, 6t, 0), and at t = 1 the magnitude is √40 ≈ 6.32.
  6. 8 · Distance traveled is the integral of speed. The antiderivative t³ evaluated from 0 to 2 gives 8.
  7. The speed is 5, not 7, so the unit tangent is (-3/5, 4/5, 0) · The speed is √(9 + 16) = 5, not 7. Dividing by 5 gives (-3/5, 4/5, 0).
  8. (2/7, 3/7, 6/7) · Velocity (2, 3, 6) has length 7, so divide each component by 7.
  9. 21 · Speed 7 is constant, so length is 7 times 3.
  10. True · Arc length is the integral of speed. Constant speed 5 over a time span of 2 gives 10, whatever the path's shape.
Worksheet · LightMySky