A particle moving in the plane has a position vector
r ( t ) = ⟨ x ( t ) , y ( t ) ⟩ \mathbf{r}(t) = \langle x(t), y(t)\rangle r ( t ) = ⟨ x ( t ) , y ( t )⟩ t t t vector-valued . Calculus acts on it one component at a time:
v ( t ) = r ′ ( t ) = ⟨ x ′ ( t ) , y ′ ( t ) ⟩ a ( t ) = v ′ ( t ) = ⟨ x ′ ′ ( t ) , y ′ ′ ( t ) ⟩ \mathbf{v}(t) = \mathbf{r}'(t) = \langle x'(t), y'(t)\rangle \qquad\qquad \mathbf{a}(t) = \mathbf{v}'(t) = \langle x''(t), y''(t)\rangle v ( t ) = r ′ ( t ) = ⟨ x ′ ( t ) , y ′ ( t )⟩ a ( t ) = v ′ ( t ) = ⟨ x ′′ ( t ) , y ′′ ( t )⟩ The speed is the length of the velocity vector:
∣ v ( t ) ∣ = x ′ ( t ) 2 + y ′ ( t ) 2 \lvert \mathbf{v}(t) \rvert = \sqrt{x'(t)^2 + y'(t)^2} ∣ v ( t )∣ = x ′ ( t ) 2 + y ′ ( t ) 2
Over a short time h h h r ( t ) \mathbf{r}(t) r ( t ) r ( t + h ) \mathbf{r}(t + h) r ( t + h ) r ( t + h ) − r ( t ) \mathbf{r}(t + h) - \mathbf{r}(t) r ( t + h ) − r ( t ) h h h h → 0 h \to 0 h → 0 The velocity vector points where the particle is headed, and its
length is how fast it is going. Acceleration points toward the side the path
bends to, whenever the path bends.
Velocity and acceleration on an elliptical path
A particle travels counterclockwise around the ellipse x = 3 cos t, y = 2 sin t. At t = π/4, near (2.1, 1.4), the velocity arrow v points up and to the left, along the path, and the acceleration arrow a points straight at the origin, the center of the ellipse.
v
a
-4 -2 2 4 -3 -2 -1 1 2 3 x y
Velocity and acceleration on an elliptical path
Example 1 — Velocity, speed and acceleration
A particle has position r ( t ) = ⟨ t 2 , t 3 ⟩ \mathbf{r}(t) = \langle t^2, t^3 \rangle r ( t ) = ⟨ t 2 , t 3 ⟩ t = 1 t = 1 t = 1
v ( t ) = ⟨ 2 t , 3 t 2 ⟩ \mathbf{v}(t) = \langle 2t, 3t^2 \rangle v ( t ) = ⟨ 2 t , 3 t 2 ⟩ v ( 1 ) = ⟨ 2 , 3 ⟩ \mathbf{v}(1) = \langle 2, 3 \rangle v ( 1 ) = ⟨ 2 , 3 ⟩ 4 + 9 = 13 ≈ 3.61 \sqrt{4 + 9} = \sqrt{13} \approx 3.61 4 + 9 = 13 ≈ 3.61 a ( t ) = ⟨ 2 , 6 t ⟩ \mathbf{a}(t) = \langle 2, 6t \rangle a ( t ) = ⟨ 2 , 6 t ⟩ a ( 1 ) = ⟨ 2 , 6 ⟩ \mathbf{a}(1) = \langle 2, 6 \rangle a ( 1 ) = ⟨ 2 , 6 ⟩
Integration runs the other way, one component at a time. Starting from
r ( 0 ) \mathbf{r}(0) r ( 0 )
r ( t ) = r ( 0 ) + ∫ 0 t v ( u ) d u \mathbf{r}(t) = \mathbf{r}(0) + \int_0^t \mathbf{v}(u)\,du r ( t ) = r ( 0 ) + ∫ 0 t v ( u ) d u The integral of velocity is the displacement , the arrow from start to
finish. The distance traveled is the length of the path, the integral of
speed:
distance = ∫ a b x ′ ( t ) 2 + y ′ ( t ) 2 d t \text{distance} = \int_a^b \sqrt{x'(t)^2 + y'(t)^2}\,dt distance = ∫ a b x ′ ( t ) 2 + y ′ ( t ) 2 d t Example 2 — The ellipse in the graph
A particle moves with r ( t ) = ⟨ 3 cos t , 2 sin t ⟩ \mathbf{r}(t) = \langle 3\cos t, 2\sin t \rangle r ( t ) = ⟨ 3 cos t , 2 sin t ⟩ t = π 4 t = \tfrac{\pi}{4} t = 4 π
v ( t ) = ⟨ − 3 sin t , 2 cos t ⟩ \mathbf{v}(t) = \langle -3\sin t, 2\cos t \rangle v ( t ) = ⟨ − 3 sin t , 2 cos t ⟩ a ( t ) = ⟨ − 3 cos t , − 2 sin t ⟩ = − r ( t ) \mathbf{a}(t) = \langle -3\cos t, -2\sin t \rangle = -\mathbf{r}(t) a ( t ) = ⟨ − 3 cos t , − 2 sin t ⟩ = − r ( t ) t = π 4 t = \tfrac{\pi}{4} t = 4 π v ≈ ⟨ − 2.12 , 1.41 ⟩ \mathbf{v} \approx \langle -2.12, 1.41 \rangle v ≈ ⟨ − 2.12 , 1.41 ⟩ 4.5 + 2 ≈ 2.55 \sqrt{4.5 + 2} \approx 2.55 4.5 + 2 ≈ 2.55 a = − r \mathbf{a} = -\mathbf{r} a = − r
Example 3 — Position from velocity
A particle has v ( t ) = ⟨ 2 t , cos t ⟩ \mathbf{v}(t) = \langle 2t, \cos t \rangle v ( t ) = ⟨ 2 t , cos t ⟩ r ( 0 ) = ⟨ 1 , 3 ⟩ \mathbf{r}(0) = \langle 1, 3 \rangle r ( 0 ) = ⟨ 1 , 3 ⟩ r ( t ) \mathbf{r}(t) r ( t )
Integrate each component from 0 0 0 t t t
r ( t ) = ⟨ 1 + t 2 , 3 + sin t ⟩ \mathbf{r}(t) = \left\langle 1 + t^2,\ 3 + \sin t \right\rangle r ( t ) = ⟨ 1 + t 2 , 3 + sin t ⟩ Example 4 — Displacement and distance
For r ( t ) = ⟨ cos t , sin t ⟩ \mathbf{r}(t) = \langle \cos t, \sin t \rangle r ( t ) = ⟨ cos t , sin t ⟩ 0 ≤ t ≤ 2 π 0 \le t \le 2\pi 0 ≤ t ≤ 2 π
The particle goes once around the unit circle and returns to its start, so the
displacement is ⟨ 0 , 0 ⟩ \langle 0, 0 \rangle ⟨ 0 , 0 ⟩ sin 2 t + cos 2 t = 1 \sqrt{\sin^2 t + \cos^2 t} = 1 sin 2 t + cos 2 t = 1 ∫ 0 2 π 1 d t = 2 π \int_0^{2\pi} 1\,dt = 2\pi ∫ 0 2 π 1 d t = 2 π
Example 5 — A projectile
A ball is thrown with r ( 0 ) = ⟨ 0 , 0 ⟩ \mathbf{r}(0) = \langle 0, 0 \rangle r ( 0 ) = ⟨ 0 , 0 ⟩ v ( 0 ) = ⟨ 10 , 20 ⟩ \mathbf{v}(0) = \langle 10, 20 \rangle v ( 0 ) = ⟨ 10 , 20 ⟩ a ( t ) = ⟨ 0 , − 9.8 ⟩ \mathbf{a}(t) = \langle 0, -9.8 \rangle a ( t ) = ⟨ 0 , − 9.8 ⟩ r ( t ) \mathbf{r}(t) r ( t )
v ( t ) = ⟨ 10 , 20 − 9.8 t ⟩ \mathbf{v}(t) = \langle 10, 20 - 9.8t \rangle v ( t ) = ⟨ 10 , 20 − 9.8 t ⟩
r ( t ) = ⟨ 10 t , 20 t − 4.9 t 2 ⟩ \mathbf{r}(t) = \left\langle 10t,\ 20t - 4.9t^2 \right\rangle r ( t ) = ⟨ 10 t , 20 t − 4.9 t 2 ⟩ These are the parametric equations of a projectile, now derived from the
acceleration.
Common mistake
Adding components to get speed. The speed of ⟨ 3 , 4 ⟩ \langle 3, 4 \rangle ⟨ 3 , 4 ⟩ 9 + 16 = 5 \sqrt{9 + 16} = 5 9 + 16 = 5 3 + 4 = 7 3 + 4 = 7 3 + 4 = 7
Common mistake
Confusing displacement with distance. Integrating velocity gives the net
change in position, which can be zero after a round trip. Distance integrates
speed, which is never negative.
Common mistake
Calling a particle at rest when one component stops. At rest means both
components of velocity are 0 0 0 x ′ ( t ) = 0 x'(t) = 0 x ′ ( t ) = 0
For r ( t ) = ⟨ 3 t , t 2 ⟩ \mathbf{r}(t) = \langle 3t, t^2 \rangle r ( t ) = ⟨ 3 t , t 2 ⟩ v ( 2 ) \mathbf{v}(2) v ( 2 ) t = 2 t = 2 t = 2 a ( t ) \mathbf{a}(t) a ( t )
Answer
⟨ 3 , 4 ⟩ \langle 3, 4 \rangle ⟨ 3 , 4 ⟩ 5 5 5 ⟨ 0 , 2 ⟩ \langle 0, 2 \rangle ⟨ 0 , 2 ⟩
Full solution
v ( t ) = ⟨ 3 , 2 t ⟩ \mathbf{v}(t) = \langle 3, 2t \rangle v ( t ) = ⟨ 3 , 2 t ⟩ v ( 2 ) = ⟨ 3 , 4 ⟩ \mathbf{v}(2) = \langle 3, 4 \rangle v ( 2 ) = ⟨ 3 , 4 ⟩ 5 5 5 a ( t ) = ⟨ 0 , 2 ⟩ \mathbf{a}(t) = \langle 0, 2 \rangle a ( t ) = ⟨ 0 , 2 ⟩
Show that r ( t ) = ⟨ cos 2 t , sin 2 t ⟩ \mathbf{r}(t) = \langle \cos 2t, \sin 2t \rangle r ( t ) = ⟨ cos 2 t , sin 2 t ⟩
Answer
The speed is always 2 2 2
Full solution
v ( t ) = ⟨ − 2 sin 2 t , 2 cos 2 t ⟩ \mathbf{v}(t) = \langle -2\sin 2t, 2\cos 2t \rangle v ( t ) = ⟨ − 2 sin 2 t , 2 cos 2 t ⟩ 4 sin 2 2 t + 4 cos 2 2 t = 2 \sqrt{4\sin^2 2t + 4\cos^2 2t} = 2 4 sin 2 2 t + 4 cos 2 2 t = 2
For r ( t ) = ⟨ e t , e − t ⟩ \mathbf{r}(t) = \langle e^t, e^{-t} \rangle r ( t ) = ⟨ e t , e − t ⟩ t = 0 t = 0 t = 0
Answer
⟨ 1 , − 1 ⟩ \langle 1, -1 \rangle ⟨ 1 , − 1 ⟩ 2 \sqrt{2} 2
Full solution
v ( t ) = ⟨ e t , − e − t ⟩ \mathbf{v}(t) = \langle e^t, -e^{-t} \rangle v ( t ) = ⟨ e t , − e − t ⟩ v ( 0 ) = ⟨ 1 , − 1 ⟩ \mathbf{v}(0) = \langle 1, -1 \rangle v ( 0 ) = ⟨ 1 , − 1 ⟩ 2 \sqrt{2} 2
A particle has v ( t ) = ⟨ 4 t , 3 ⟩ \mathbf{v}(t) = \langle 4t, 3 \rangle v ( t ) = ⟨ 4 t , 3 ⟩ r ( 0 ) = ⟨ 0 , 1 ⟩ \mathbf{r}(0) = \langle 0, 1 \rangle r ( 0 ) = ⟨ 0 , 1 ⟩ r ( 2 ) \mathbf{r}(2) r ( 2 )
Answer
⟨ 8 , 7 ⟩ \langle 8, 7 \rangle ⟨ 8 , 7 ⟩
Full solution
r ( t ) = ⟨ 2 t 2 , 1 + 3 t ⟩ \mathbf{r}(t) = \langle 2t^2, 1 + 3t \rangle r ( t ) = ⟨ 2 t 2 , 1 + 3 t ⟩ r ( 2 ) = ⟨ 8 , 7 ⟩ \mathbf{r}(2) = \langle 8, 7 \rangle r ( 2 ) = ⟨ 8 , 7 ⟩
A particle has a ( t ) = ⟨ 6 t , 2 ⟩ \mathbf{a}(t) = \langle 6t, 2 \rangle a ( t ) = ⟨ 6 t , 2 ⟩ v ( 0 ) = ⟨ 1 , 0 ⟩ \mathbf{v}(0) = \langle 1, 0 \rangle v ( 0 ) = ⟨ 1 , 0 ⟩ r ( 0 ) = ⟨ 0 , 0 ⟩ \mathbf{r}(0) = \langle 0, 0 \rangle r ( 0 ) = ⟨ 0 , 0 ⟩ r ( t ) \mathbf{r}(t) r ( t )
Answer
⟨ t 3 + t , t 2 ⟩ \langle t^3 + t, t^2 \rangle ⟨ t 3 + t , t 2 ⟩
Full solution
v ( t ) = ⟨ 3 t 2 + 1 , 2 t ⟩ \mathbf{v}(t) = \langle 3t^2 + 1, 2t \rangle v ( t ) = ⟨ 3 t 2 + 1 , 2 t ⟩ r ( t ) = ⟨ t 3 + t , t 2 ⟩ \mathbf{r}(t) = \langle t^3 + t, t^2 \rangle r ( t ) = ⟨ t 3 + t , t 2 ⟩
Find the distance traveled by r ( t ) = ⟨ t 2 , 2 3 t 3 ⟩ \mathbf{r}(t) = \langle t^2, \tfrac{2}{3}t^3 \rangle r ( t ) = ⟨ t 2 , 3 2 t 3 ⟩ 0 ≤ t ≤ 1 0 \le t \le 1 0 ≤ t ≤ 1
Answer
2 3 ( 2 2 − 1 ) ≈ 1.22 \tfrac{2}{3}\left(2\sqrt{2} - 1\right) \approx 1.22 3 2 ( 2 2 − 1 ) ≈ 1.22
Full solution
The speed is 4 t 2 + 4 t 4 = 2 t 1 + t 2 \sqrt{4t^2 + 4t^4} = 2t\sqrt{1 + t^2} 4 t 2 + 4 t 4 = 2 t 1 + t 2 ∫ 0 1 2 t 1 + t 2 d t = 2 3 ( 2 3 / 2 − 1 ) \int_0^1 2t\sqrt{1 + t^2}\,dt = \tfrac{2}{3}\left(2^{3/2} - 1\right) ∫ 0 1 2 t 1 + t 2 d t = 3 2 ( 2 3/2 − 1 )
A particle has v ( t ) = ⟨ 1 , 2 t ⟩ \mathbf{v}(t) = \langle 1, 2t \rangle v ( t ) = ⟨ 1 , 2 t ⟩ t = 0 t = 0 t = 0 t = 3 t = 3 t = 3
Answer
⟨ 3 , 9 ⟩ \langle 3, 9 \rangle ⟨ 3 , 9 ⟩
Full solution
∫ 0 3 1 d t = 3 \int_0^3 1\,dt = 3 ∫ 0 3 1 d t = 3 ∫ 0 3 2 t d t = 9 \int_0^3 2t\,dt = 9 ∫ 0 3 2 t d t = 9
When is the particle r ( t ) = ⟨ t 3 − 3 t , t 2 − 2 t ⟩ \mathbf{r}(t) = \langle t^3 - 3t, t^2 - 2t \rangle r ( t ) = ⟨ t 3 − 3 t , t 2 − 2 t ⟩
Answer
At t = 1 t = 1 t = 1
Full solution
v ( t ) = ⟨ 3 t 2 − 3 , 2 t − 2 ⟩ \mathbf{v}(t) = \langle 3t^2 - 3, 2t - 2 \rangle v ( t ) = ⟨ 3 t 2 − 3 , 2 t − 2 ⟩ 0 0 0 t = ± 1 t = \pm 1 t = ± 1 t = 1 t = 1 t = 1 t = 1 t = 1 t = 1 0 0 0
For r ( t ) = ⟨ ln t , t 2 ⟩ \mathbf{r}(t) = \langle \ln t, t^2 \rangle r ( t ) = ⟨ ln t , t 2 ⟩ t > 0 t > 0 t > 0 t = 1 t = 1 t = 1
Answer
Right and up, with velocity ⟨ 1 , 2 ⟩ \langle 1, 2 \rangle ⟨ 1 , 2 ⟩
Full solution
v ( t ) = ⟨ 1 t , 2 t ⟩ \mathbf{v}(t) = \left\langle \tfrac{1}{t}, 2t \right\rangle v ( t ) = ⟨ t 1 , 2 t ⟩ v ( 1 ) = ⟨ 1 , 2 ⟩ \mathbf{v}(1) = \langle 1, 2 \rangle v ( 1 ) = ⟨ 1 , 2 ⟩
A student says the speed of r ( t ) = ⟨ 3 t , 4 t ⟩ \mathbf{r}(t) = \langle 3t, 4t \rangle r ( t ) = ⟨ 3 t , 4 t ⟩ 3 + 4 = 7 3 + 4 = 7 3 + 4 = 7
Hint
Speed is the length of the velocity vector.
Answer
The components were added instead of combined by the Pythagorean theorem. The speed is 5 5 5
Full solution
v ( t ) = ⟨ 3 , 4 ⟩ \mathbf{v}(t) = \langle 3, 4 \rangle v ( t ) = ⟨ 3 , 4 ⟩ 3 2 + 4 2 = 5 \sqrt{3^2 + 4^2} = 5 3 2 + 4 2 = 5 3 3 3 4 4 4 5 5 5