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9.6 Solving Motion Problems Using Parametric and Vector-Valued Functions

Syllabus
2020
Topic
9.6
Level
—

Learning objectives

—FUN-8.B—Determine values for positions and rates of change in problems involving planar motion• FUN-8.B Determine values for positions and rates of change in problems involving planar motion.• FUN-8.B.1 Derivatives can be used to determine velocity, speed, and acceleration for a particle moving along a curve in the plane defined using parametric or vector-valued functions.• FUN-8.B.2 For a particle in planar motion over an interval of time, the definite integral of the velocity vector represents the particle’s displacement (net change in position) over the interval of time, from which we might determine its position. The definite integral of speed represents the particle’s total distance traveled over the interval of time.• Enduring understanding FUN-8: Solving an initial value problem allows us to determine an expression for the position of a particle moving in the plane.

Match Each Motion Quantity to the Right Operation

For planar motion, r(t)=⟨x(t),y(t)⟩\mathbf r(t)=\langle x(t),y(t)\rangle is position. Differentiate once for velocity and twice for acceleration; take the magnitude of velocity for speed. Integrate velocity to obtain displacement, or integrate speed to obtain total distance traveled.

\mathbf v(t)=\mathbf r'(t)=\langle x'(t),y'(t)\rangle,\qquad \text{speed}=|\mathbf v(t)|=\sqrt{[x'(t)]^2+[y'(t)]^2},\qquad \mathbf a(t)=\mathbf v'(t)

\text{displacement on }[a,b]=\int_a^b\mathbf v(t),dt=\mathbf r(b)-\mathbf r(a),\qquad \text{distance}=\int_a^b|\mathbf v(t)|,dt

Example: a particle's position in meters is r(t)=⟨t2+1,t3−3t⟩\mathbf r(t)=\langle t^2+1,t^3-3t\rangle, with tt in seconds. Then v(t)=⟨2t,3t2−3⟩\mathbf v(t)=\langle2t,3t^2-3\rangle m/s and a(t)=⟨2,6t⟩\mathbf a(t)=\langle2,6t\rangle m/s2^2. At t=1t=1, its position is ⟨2,−2⟩\langle2,-2\rangle m, velocity is ⟨2,0⟩\langle2,0\rangle m/s, speed is 22+02=2\sqrt{2^2+0^2}=2 m/s, and acceleration is ⟨2,6⟩\langle2,6\rangle m/s2^2. From t=1t=1 to t=2t=2, displacement is r(2)−r(1)=⟨5,2⟩−⟨2,−2⟩=⟨3,4⟩\mathbf r(2)-\mathbf r(1)=\langle5,2\rangle-\langle2,-2\rangle=\langle3,4\rangle m.

Displacement is the net position change, so opposite motions can cancel; total distance cannot cancel because speed is nonnegative. In the example, ∥⟨3,4⟩∥=5\|\langle3,4\rangle\|=5 m is the magnitude of displacement, not automatically the distance traveled along the curved path.