A particle traveling in the plane has position
r ( t ) = ⟨ x ( t ) , y ( t ) ⟩ \mathbf{r}(t) = \langle x(t), y(t)\rangle r ( t ) = ⟨ x ( t ) , y ( t )⟩ at time
t t t . Its velocity and acceleration are the first and second derivatives of position, and each is a vector.
Velocity, speed, and acceleration Motion in the plane
The velocity
v \mathbf{v} v , the acceleration
a \mathbf{a} a , and the speed
∣ v ∣ |\mathbf{v}| ∣ v ∣ of a particle at
r ( t ) \mathbf{r}(t) r ( t ) are
v ( t ) = ⟨ x ′ ( t ) , y ′ ( t ) ⟩ , a ( t ) = ⟨ x ′ ′ ( t ) , y ′ ′ ( t ) ⟩ , ∣ v ( t ) ∣ = x ′ ( t ) 2 + y ′ ( t ) 2 . \begin{aligned}
\mathbf{v}(t) &= \big\langle x'(t),\, y'(t)\big\rangle, \\[4pt]
\mathbf{a}(t) &= \big\langle x''(t),\, y''(t)\big\rangle, \\[4pt]
|\mathbf{v}(t)| &= \sqrt{x'(t)^2 + y'(t)^2}.
\end{aligned} v ( t ) a ( t ) ∣ v ( t ) ∣ = ⟨ x ′ ( t ) , y ′ ( t ) ⟩ , = ⟨ x ′′ ( t ) , y ′′ ( t ) ⟩ , = x ′ ( t ) 2 + y ′ ( t ) 2 . When the particle isn't at rest, the velocity points the way it's traveling, along the tangent to its path. Speed is a number, the length of the velocity vector. Each component tells its own story. The particle travels right when
x ′ > 0 x' > 0 x ′ > 0 and left when
x ′ < 0 x' < 0 x ′ < 0 , up when
y ′ > 0 y' > 0 y ′ > 0 and down when
y ′ < 0 y' < 0 y ′ < 0 .
Since speed is
x ′ 2 + y ′ 2 \sqrt{x'^2 + y'^2} x ′2 + y ′2 , the chain rule gives its derivative:
d d t ∣ v ( t ) ∣ = x ′ ( t ) x ′ ′ ( t ) + y ′ ( t ) y ′ ′ ( t ) ∣ v ( t ) ∣ . \frac{d}{dt}|\mathbf{v}(t)| = \frac{x'(t)\,x''(t) + y'(t)\,y''(t)}{|\mathbf{v}(t)|}. d t d ∣ v ( t ) ∣ = ∣ v ( t ) ∣ x ′ ( t ) x ′′ ( t ) + y ′ ( t ) y ′′ ( t ) . So where the particle isn't at rest, its speed is increasing when
x ′ x ′ ′ + y ′ y ′ ′ x'x'' + y'y'' x ′ x ′′ + y ′ y ′′ is positive and decreasing when it's negative.
Interactive: Parametric Curves and Velocity The cycloid below is the path of a point on the rim of a wheel of radius
1 1 1 rolling along the
x x x -axis. Its velocity,
⟨ 1 − cos t , sin t ⟩ \langle 1 - \cos t, \sin t\rangle ⟨ 1 − cos t , sin t ⟩ , is the zero vector at
t = 0 t = 0 t = 0 ,
2 π 2\pi 2 π , and
4 π 4\pi 4 π , where the path has a sharp point. The speed is greatest at the top of each arch.
Example 1
A particle's position is
r ( t ) = ⟨ t 3 − 12 t , 3 t 2 − 6 t ⟩ \mathbf{r}(t) = \left\langle t^3 - 12t,\; 3t^2 - 6t\right\rangle r ( t ) = ⟨ t 3 − 12 t , 3 t 2 − 6 t ⟩ for
t ≥ 0 t \ge 0 t ≥ 0 . (a) Find the velocity, speed, and acceleration at
t = 1 t = 1 t = 1 and at
t = 2 t = 2 t = 2 , and describe the direction of travel at each time. (b) Decide whether the speed is increasing or decreasing at each time. (c) Show that the particle is never at rest.
v ( t ) = ⟨ 3 t 2 − 12 , 6 t − 6 ⟩ , a ( t ) = ⟨ 6 t , 6 ⟩ . \begin{aligned}
\mathbf{v}(t) &= \left\langle 3t^2 - 12,\; 6t - 6\right\rangle, \\[4pt]
\mathbf{a}(t) &= \langle 6t,\; 6\rangle.
\end{aligned} v ( t ) a ( t ) = ⟨ 3 t 2 − 12 , 6 t − 6 ⟩ , = ⟨ 6 t , 6 ⟩ . (a) At
t = 1 t = 1 t = 1 :
v ( 1 ) = ⟨ − 9 , 0 ⟩ \mathbf{v}(1) = \langle -9, 0\rangle v ( 1 ) = ⟨ − 9 , 0 ⟩ and
a ( 1 ) = ⟨ 6 , 6 ⟩ \mathbf{a}(1) = \langle 6, 6\rangle a ( 1 ) = ⟨ 6 , 6 ⟩ . The particle is traveling straight to the left. At
t = 2 t = 2 t = 2 :
v ( 2 ) = ⟨ 0 , 6 ⟩ \mathbf{v}(2) = \langle 0, 6\rangle v ( 2 ) = ⟨ 0 , 6 ⟩ and
a ( 2 ) = ⟨ 12 , 6 ⟩ \mathbf{a}(2) = \langle 12, 6\rangle a ( 2 ) = ⟨ 12 , 6 ⟩ . The particle is traveling straight up. The speeds are
∣ v ( 1 ) ∣ = ( − 9 ) 2 + 0 2 = 9 , ∣ v ( 2 ) ∣ = 0 2 + 6 2 = 6. \begin{aligned}
|\mathbf{v}(1)| &= \sqrt{(-9)^2 + 0^2} = 9, \\[4pt]
|\mathbf{v}(2)| &= \sqrt{0^2 + 6^2} = 6.
\end{aligned} ∣ v ( 1 ) ∣ ∣ v ( 2 ) ∣ = ( − 9 ) 2 + 0 2 = 9 , = 0 2 + 6 2 = 6. (b) Evaluate
x ′ x ′ ′ + y ′ y ′ ′ x'x'' + y'y'' x ′ x ′′ + y ′ y ′′ at each time:
t = 1 : ( − 9 ) ( 6 ) + 0 ( 6 ) = − 54 < 0 , t = 2 : 0 ( 12 ) + 6 ( 6 ) = 36 > 0. \begin{aligned}
t = 1&: \; (-9)(6) + 0(6) = -54 < 0, \\[4pt]
t = 2&: \; 0(12) + 6(6) = 36 > 0.
\end{aligned} t = 1 t = 2 : ( − 9 ) ( 6 ) + 0 ( 6 ) = − 54 < 0 , : 0 ( 12 ) + 6 ( 6 ) = 36 > 0. The speed is decreasing at
t = 1 t = 1 t = 1 and increasing at
t = 2 t = 2 t = 2 .
(c) At rest means both components of
v \mathbf{v} v are zero. For
t ≥ 0 t \ge 0 t ≥ 0 ,
x ′ ( t ) = 0 x'(t) = 0 x ′ ( t ) = 0 only at
t = 2 t = 2 t = 2 , and
y ′ ( 2 ) = 6 ≠ 0 y'(2) = 6 \ne 0 y ′ ( 2 ) = 6 = 0 . So the velocity is never the zero vector.
The path for 0 ≤ t ≤ 2.6, with the velocity at t = 1 (straight left) and at t = 2 (straight up). Both arrows are drawn at a third of their length. Position, displacement, and distance Integrating the velocity undoes the derivative, one component at a time.
The length of the displacement vector is the straight-line distance from the starting point to the end. The total distance follows the path, so it's at least as long.
Example 2
A calculator is allowed. A robot travels on a factory floor with velocity
v ( t ) = ⟨ 3 cos 0.8 t , 2 − 0.5 t ⟩ \mathbf{v}(t) = \langle 3\cos 0.8t,\; 2 - 0.5t\rangle v ( t ) = ⟨ 3 cos 0.8 t , 2 − 0.5 t ⟩ meters per second for
0 ≤ t ≤ 6 0 \le t \le 6 0 ≤ t ≤ 6 seconds. It starts at
( 2 , − 1 ) (2, -1) ( 2 , − 1 ) , in meters, when
t = 0 t = 0 t = 0 .
(a) Find the robot's speed at
t = 4 t = 4 t = 4 , and decide whether its speed is increasing or decreasing then. (b) Find the robot's position at
t = 6 t = 6 t = 6 and its straight-line distance from where it started. (c) Find the total distance the robot travels over
0 ≤ t ≤ 6 0 \le t \le 6 0 ≤ t ≤ 6 .
(a) At
t = 4 t = 4 t = 4 , the velocity components are
x ′ ( 4 ) = 3 cos 3.2 x'(4) = 3\cos 3.2 x ′ ( 4 ) = 3 cos 3.2 , about
− 2.995 -2.995 − 2.995 , and
y ′ ( 4 ) = 0 y'(4) = 0 y ′ ( 4 ) = 0 . The speed is
( 3 cos 3.2 ) 2 + 0 2 ≈ 2.995 \sqrt{(3\cos 3.2)^2 + 0^2} \approx 2.995 ( 3 cos 3.2 ) 2 + 0 2 ≈ 2.995 meters per second. The acceleration components are
x ′ ′ ( 4 ) = − 2.4 sin 3.2 x''(4) = -2.4\sin 3.2 x ′′ ( 4 ) = − 2.4 sin 3.2 , about 0.140, and
y ′ ′ ( 4 ) = − 0.5 y''(4) = -0.5 y ′′ ( 4 ) = − 0.5 . Then
x ′ x ′ ′ + y ′ y ′ ′ ≈ ( − 2.995 ) ( 0.140 ) + 0 ≈ − 0.42 < 0 , \begin{aligned}
x'x'' + y'y'' &\approx (-2.995)(0.140) + 0 \\
&\approx -0.42 < 0,
\end{aligned} x ′ x ′′ + y ′ y ′′ ≈ ( − 2.995 ) ( 0.140 ) + 0 ≈ − 0.42 < 0 , so the speed is decreasing at
t = 4 t = 4 t = 4 .
(b) Add the change in each coordinate to the start:
x ( 6 ) = 2 + ∫ 0 6 3 cos 0.8 t d t ≈ 2 − 3.736 = − 1.736 , y ( 6 ) = − 1 + ∫ 0 6 ( 2 − 0.5 t ) d t = − 1 + 3 = 2. \begin{aligned}
x(6) &= 2 + \int_{0}^{6} 3\cos 0.8t\,dt \\
&\approx 2 - 3.736 = -1.736, \\[6pt]
y(6) &= -1 + \int_{0}^{6} (2 - 0.5t)\,dt \\
&= -1 + 3 = 2.
\end{aligned} x ( 6 ) y ( 6 ) = 2 + ∫ 0 6 3 cos 0.8 t d t ≈ 2 − 3.736 = − 1.736 , = − 1 + ∫ 0 6 ( 2 − 0.5 t ) d t = − 1 + 3 = 2. The robot is at about
( − 1.736 , 2 ) (-1.736, 2) ( − 1.736 , 2 ) . The displacement is about
⟨ − 3.736 , 3 ⟩ \langle -3.736, 3\rangle ⟨ − 3.736 , 3 ⟩ , so the robot ends about
3.736 2 + 3 2 ≈ 4.791 \sqrt{3.736^2 + 3^2} \approx 4.791 3.73 6 2 + 3 2 ≈ 4.791 meters from where it started.
(c) The total distance is the integral of the speed:
∫ 0 6 9 cos 2 0.8 t + ( 2 − 0.5 t ) 2 d t ≈ 13.117 meters . \begin{aligned}
&\int_{0}^{6} \sqrt{9\cos^2 0.8t + (2 - 0.5t)^2}\,dt \\
&\qquad \approx 13.117 \text{ meters}.
\end{aligned} ∫ 0 6 9 cos 2 0.8 t + ( 2 − 0.5 t ) 2 d t ≈ 13.117 meters .
The robot's path from (2, −1) to about (−1.736, 2), with its velocity at t = 4. The straight segment is the displacement, about 4.791 meters; the path is about 13.117 meters.