Fundamental Relationships Vector-valued functions describe motion in the plane through position, velocity, acceleration, and speed.
Position function: r ( t ) = ⟨ x ( t ) , y ( t ) ⟩ \mathbf{r}(t) = \langle x(t), y(t) \rangle r ( t ) = ⟨ x ( t ) , y ( t )⟩ Velocity vector: v ( t ) = r ′ ( t ) = ⟨ x ′ ( t ) , y ′ ( t ) ⟩ \mathbf{v}(t) = \mathbf{r}'(t) = \langle x'(t), y'(t) \rangle v ( t ) = r ′ ( t ) = ⟨ x ′ ( t ) , y ′ ( t )⟩ Acceleration vector: a ( t ) = v ′ ( t ) = r ′ ′ ( t ) = ⟨ x ′ ′ ( t ) , y ′ ′ ( t ) ⟩ \mathbf{a}(t) = \mathbf{v}'(t) = \mathbf{r}''(t) = \langle x''(t), y''(t) \rangle a ( t ) = v ′ ( t ) = r ′′ ( t ) = ⟨ x ′′ ( t ) , y ′′ ( t )⟩ Speed , which is a scalar, not a vector:
Speed = ∥ v ( t ) ∥ = [ x ′ ( t ) ] 2 + [ y ′ ( t ) ] 2 \text{Speed} = \|\mathbf{v}(t)\| = \sqrt{[x'(t)]^2 + [y'(t)]^2} Speed = ∥ v ( t ) ∥ = [ x ′ ( t ) ] 2 + [ y ′ ( t ) ] 2 Important: speed is the magnitude of velocity, so it isn’t a vector quantity.
Finding Velocity, Acceleration, and Speed Example 1: Circular Motion Given the position function
r ( t ) = ⟨ 3 sin ( t 4 ) , 3 cos ( t 4 ) ⟩ \mathbf{r}(t) = \left\langle 3\sin\left(\frac{t}{4}\right), 3\cos\left(\frac{t}{4}\right) \right\rangle r ( t ) = ⟨ 3 sin ( 4 t ) , 3 cos ( 4 t ) ⟩ Find the velocity vector by taking the derivative of each component:
v ( t ) = r ′ ( t ) = ⟨ 3 cos ( t 4 ) ⋅ 1 4 , − 3 sin ( t 4 ) ⋅ 1 4 ⟩ \mathbf{v}(t) = \mathbf{r}'(t) = \left\langle 3\cos\left(\frac{t}{4}\right) \cdot \frac{1}{4}, \, -3\sin\left(\frac{t}{4}\right) \cdot \frac{1}{4} \right\rangle v ( t ) = r ′ ( t ) = ⟨ 3 cos ( 4 t ) ⋅ 4 1 , − 3 sin ( 4 t ) ⋅ 4 1 ⟩ v ( t ) = ⟨ 3 4 cos ( t 4 ) , − 3 4 sin ( t 4 ) ⟩ \mathbf{v}(t) = \left\langle \frac{3}{4}\cos\left(\frac{t}{4}\right), \, -\frac{3}{4}\sin\left(\frac{t}{4}\right) \right\rangle v ( t ) = ⟨ 4 3 cos ( 4 t ) , − 4 3 sin ( 4 t ) ⟩ Find the acceleration vector by taking the derivative of velocity:
a ( t ) = v ′ ( t ) = ⟨ − 3 4 sin ( t 4 ) ⋅ 1 4 , − 3 4 cos ( t 4 ) ⋅ 1 4 ⟩ \mathbf{a}(t) = \mathbf{v}'(t) = \left\langle -\frac{3}{4}\sin\left(\frac{t}{4}\right) \cdot \frac{1}{4}, \, -\frac{3}{4}\cos\left(\frac{t}{4}\right) \cdot \frac{1}{4} \right\rangle a ( t ) = v ′ ( t ) = ⟨ − 4 3 sin ( 4 t ) ⋅ 4 1 , − 4 3 cos ( 4 t ) ⋅ 4 1 ⟩ a ( t ) = ⟨ − 3 16 sin ( t 4 ) , − 3 16 cos ( t 4 ) ⟩ \mathbf{a}(t) = \left\langle -\frac{3}{16}\sin\left(\frac{t}{4}\right), \, -\frac{3}{16}\cos\left(\frac{t}{4}\right) \right\rangle a ( t ) = ⟨ − 16 3 sin ( 4 t ) , − 16 3 cos ( 4 t ) ⟩ Find the speed:
Speed = [ 3 4 cos ( t 4 ) ] 2 + [ − 3 4 sin ( t 4 ) ] 2 \text{Speed} = \sqrt{\left[\frac{3}{4}\cos\left(\frac{t}{4}\right)\right]^2 + \left[-\frac{3}{4}\sin\left(\frac{t}{4}\right)\right]^2} Speed = [ 4 3 cos ( 4 t ) ] 2 + [ − 4 3 sin ( 4 t ) ] 2 = 9 16 cos 2 ( t 4 ) + 9 16 sin 2 ( t 4 ) = \sqrt{\frac{9}{16}\cos^2\left(\frac{t}{4}\right) + \frac{9}{16}\sin^2\left(\frac{t}{4}\right)} = 16 9 cos 2 ( 4 t ) + 16 9 sin 2 ( 4 t ) = 9 16 [ cos 2 ( t 4 ) + sin 2 ( t 4 ) ] = \sqrt{\frac{9}{16}\left[\cos^2\left(\frac{t}{4}\right) + \sin^2\left(\frac{t}{4}\right)\right]} = 16 9 [ cos 2 ( 4 t ) + sin 2 ( 4 t ) ] Using the Pythagorean identity
sin 2 ( θ ) + cos 2 ( θ ) = 1 \sin^2(\theta) + \cos^2(\theta) = 1 sin 2 ( θ ) + cos 2 ( θ ) = 1 :
= 9 16 ( 1 ) = 3 4 = \sqrt{\frac{9}{16}(1)} = \frac{3}{4} = 16 9 ( 1 ) = 4 3 The speed is constant at 3 4 \frac{3}{4} 4 3 . Speeding Up vs. Slowing Down Whether a particle’s speed increases or decreases depends on the relationship between velocity and acceleration.
In the plane, check the derivative of the speed. Speed is
[ x ′ ( t ) ] 2 + [ y ′ ( t ) ] 2 \sqrt{[x'(t)]^2 + [y'(t)]^2} [ x ′ ( t ) ] 2 + [ y ′ ( t ) ] 2 , so by the chain rule:
d d t ( speed ) = x ′ ( t ) x ′ ′ ( t ) + y ′ ( t ) y ′ ′ ( t ) [ x ′ ( t ) ] 2 + [ y ′ ( t ) ] 2 \frac{d}{dt}\left(\text{speed}\right) = \frac{x'(t)x''(t) + y'(t)y''(t)}{\sqrt{[x'(t)]^2 + [y'(t)]^2}} d t d ( speed ) = [ x ′ ( t ) ] 2 + [ y ′ ( t ) ] 2 x ′ ( t ) x ′′ ( t ) + y ′ ( t ) y ′′ ( t ) The bottom is positive whenever the particle is moving, so the top decides the sign.
Speeding up: x ′ ( t ) x ′ ′ ( t ) + y ′ ( t ) y ′ ′ ( t ) > 0 x'(t)x''(t) + y'(t)y''(t) > 0 x ′ ( t ) x ′′ ( t ) + y ′ ( t ) y ′′ ( t ) > 0 Slowing down: x ′ ( t ) x ′ ′ ( t ) + y ′ ( t ) y ′ ′ ( t ) < 0 x'(t)x''(t) + y'(t)y''(t) < 0 x ′ ( t ) x ′′ ( t ) + y ′ ( t ) y ′′ ( t ) < 0 Careful: the “same sign” rule from Topic 4.2 only works one component at a time. If
y ′ ( t ) y'(t) y ′ ( t ) and
y ′ ′ ( t ) y''(t) y ′′ ( t ) have the same sign, the
vertical motion is speeding up, but the overall speed can still be decreasing. You can also find the speed as a function of
t t t and check whether its derivative is positive or negative (a calculator works).
Example: a particle moves with velocity
v ( t ) = ⟨ 4 − t 2 , t + 1 ⟩ \mathbf{v}(t) = \langle 4 - t^2, \, t + 1 \rangle v ( t ) = ⟨ 4 − t 2 , t + 1 ⟩ . Is its speed increasing or decreasing at
t = 1 t = 1 t = 1 ?
a ( t ) = ⟨ − 2 t , 1 ⟩ v ( 1 ) = ⟨ 3 , 2 ⟩ a ( 1 ) = ⟨ − 2 , 1 ⟩ \mathbf{a}(t) = \langle -2t, \, 1 \rangle \qquad \mathbf{v}(1) = \langle 3, 2 \rangle \qquad \mathbf{a}(1) = \langle -2, 1 \rangle a ( t ) = ⟨ − 2 t , 1 ⟩ v ( 1 ) = ⟨ 3 , 2 ⟩ a ( 1 ) = ⟨ − 2 , 1 ⟩ x ′ ( 1 ) x ′ ′ ( 1 ) + y ′ ( 1 ) y ′ ′ ( 1 ) = ( 3 ) ( − 2 ) + ( 2 ) ( 1 ) = − 4 x'(1)x''(1) + y'(1)y''(1) = (3)(-2) + (2)(1) = -4 x ′ ( 1 ) x ′′ ( 1 ) + y ′ ( 1 ) y ′′ ( 1 ) = ( 3 ) ( − 2 ) + ( 2 ) ( 1 ) = − 4 That’s negative, so the speed is
decreasing at
t = 1 t = 1 t = 1 , even though the vertical velocity and vertical acceleration are both positive. (A calculator agrees: the derivative of the speed at
t = 1 t = 1 t = 1 is about
− 1.109 -1.109 − 1.109 .)
Working Between Position and Velocity Example 2: Basic Polynomial Motion Given the position
r ( t ) = ⟨ 2 t 3 + t , t 2 − 1 ⟩ \mathbf{r}(t) = \langle 2t^3 + t, \, t^2 - 1 \rangle r ( t ) = ⟨ 2 t 3 + t , t 2 − 1 ⟩ Velocity vector:
v ( t ) = ⟨ 6 t 2 + 1 , 2 t ⟩ \mathbf{v}(t) = \langle 6t^2 + 1, \, 2t \rangle v ( t ) = ⟨ 6 t 2 + 1 , 2 t ⟩ Acceleration vector:
a ( t ) = ⟨ 12 t , 2 ⟩ \mathbf{a}(t) = \langle 12t, \, 2 \rangle a ( t ) = ⟨ 12 t , 2 ⟩ Example 3: Speed with Exponential Components Given the position
r ( t ) = ⟨ 3 t , e − t 2 ⟩ \mathbf{r}(t) = \langle 3t, \, e^{-t^2} \rangle r ( t ) = ⟨ 3 t , e − t 2 ⟩ , find the speed at
t = 2 t = 2 t = 2 .
Find velocity:
v ( t ) = ⟨ 3 , − 2 t e − t 2 ⟩ \mathbf{v}(t) = \langle 3, \, -2te^{-t^2} \rangle v ( t ) = ⟨ 3 , − 2 t e − t 2 ⟩ Calculate the speed at t = 2 t = 2 t = 2 : Speed = ( 3 ) 2 + [ − 2 ( 2 ) e − 2 2 ] 2 = 9 + [ − 4 e − 4 ] 2 = 9 + 16 e − 8 \text{Speed} = \sqrt{(3)^2 + \left[-2(2)e^{-2^2}\right]^2} = \sqrt{9 + \left[-4e^{-4}\right]^2} = \sqrt{9 + 16e^{-8}} Speed = ( 3 ) 2 + [ − 2 ( 2 ) e − 2 2 ] 2 = 9 + [ − 4 e − 4 ] 2 = 9 + 16 e − 8 ≈ 9 + 0.00537 ≈ 3.001 \approx \sqrt{9 + 0.00537} \approx 3.001 ≈ 9 + 0.00537 ≈ 3.001 Total Distance Traveled The total distance a particle travels from
t = a t = a t = a to
t = b t = b t = b is:
Distance = ∫ a b ∥ v ( t ) ∥ d t = ∫ a b [ x ′ ( t ) ] 2 + [ y ′ ( t ) ] 2 d t \text{Distance} = \int_a^b \|\mathbf{v}(t)\| \, dt = \int_a^b \sqrt{[x'(t)]^2 + [y'(t)]^2} \, dt Distance = ∫ a b ∥ v ( t ) ∥ d t = ∫ a b [ x ′ ( t ) ] 2 + [ y ′ ( t ) ] 2 d t This integral usually needs a calculator.
Example 4: Distance Calculation Given the velocity
v ( t ) = ⟨ 3 t + 2 , 6 ⟩ \mathbf{v}(t) = \langle 3t + 2, \, 6 \rangle v ( t ) = ⟨ 3 t + 2 , 6 ⟩ from
t = 0 t = 0 t = 0 to
t = 4 t = 4 t = 4 .
Set up the integral:
Distance = ∫ 0 4 ( 3 t + 2 ) 2 + ( 6 ) 2 d t = ∫ 0 4 ( 3 t + 2 ) 2 + 36 d t \text{Distance} = \int_0^4 \sqrt{(3t + 2)^2 + (6)^2} \, dt = \int_0^4 \sqrt{(3t + 2)^2 + 36} \, dt Distance = ∫ 0 4 ( 3 t + 2 ) 2 + ( 6 ) 2 d t = ∫ 0 4 ( 3 t + 2 ) 2 + 36 d t Evaluating with a calculator gives approximately
40.968 40.968 40.968 .
Check: the speed grows from
40 ≈ 6.3 \sqrt{40} \approx 6.3 40 ≈ 6.3 at
t = 0 t = 0 t = 0 to
232 ≈ 15.2 \sqrt{232} \approx 15.2 232 ≈ 15.2 at
t = 4 t = 4 t = 4 , so the average speed is around
10 10 10 , and
4 × 10 = 40 4 \times 10 = 40 4 × 10 = 40 is close.
Finding Position from Velocity There are two methods for finding position when you’re given velocity and an initial condition.
Method 1: Direct Integration Given: v ( t ) = ⟨ 4 t + 1 , 7 ⟩ \mathbf{v}(t) = \langle 4t + 1, \, 7 \rangle v ( t ) = ⟨ 4 t + 1 , 7 ⟩ and
r ( 0 ) = ⟨ 2 , − 1 ⟩ \mathbf{r}(0) = \langle 2, -1 \rangle r ( 0 ) = ⟨ 2 , − 1 ⟩ . Find
r ( 3 ) \mathbf{r}(3) r ( 3 ) .
r ( 3 ) = r ( 0 ) + ∫ 0 3 v ( t ) d t \mathbf{r}(3) = \mathbf{r}(0) + \int_0^3 \mathbf{v}(t) \, dt r ( 3 ) = r ( 0 ) + ∫ 0 3 v ( t ) d t = ⟨ 2 , − 1 ⟩ + ∫ 0 3 ⟨ 4 t + 1 , 7 ⟩ d t = \langle 2, -1 \rangle + \int_0^3 \langle 4t + 1, \, 7 \rangle \, dt = ⟨ 2 , − 1 ⟩ + ∫ 0 3 ⟨ 4 t + 1 , 7 ⟩ d t = ⟨ 2 + ∫ 0 3 ( 4 t + 1 ) d t , − 1 + ∫ 0 3 7 d t ⟩ = \left\langle 2 + \int_0^3 (4t + 1) \, dt, \; -1 + \int_0^3 7 \, dt \right\rangle = ⟨ 2 + ∫ 0 3 ( 4 t + 1 ) d t , − 1 + ∫ 0 3 7 d t ⟩ = ⟨ 2 + [ 2 t 2 + t ] 0 3 , − 1 + [ 7 t ] 0 3 ⟩ = \left\langle 2 + [2t^2 + t]_0^3, \; -1 + [7t]_0^3 \right\rangle = ⟨ 2 + [ 2 t 2 + t ] 0 3 , − 1 + [ 7 t ] 0 3 ⟩ = ⟨ 2 + ( 18 + 3 ) , − 1 + 21 ⟩ = ⟨ 23 , 20 ⟩ = \langle 2 + (18 + 3), \; -1 + 21 \rangle = \langle 23, 20 \rangle = ⟨ 2 + ( 18 + 3 ) , − 1 + 21 ⟩ = ⟨ 23 , 20 ⟩ Method 2: Finding the General Position Function Find the general position function first, then evaluate it.
Step 1: Integrate velocity to get position with constants:
p ( t ) = ⟨ 2 t 2 + t + c 1 , 7 t + c 2 ⟩ \mathbf{p}(t) = \langle 2t^2 + t + c_1, \, 7t + c_2 \rangle p ( t ) = ⟨ 2 t 2 + t + c 1 , 7 t + c 2 ⟩ Step 2: Use the initial condition
r ( 0 ) = ⟨ 2 , − 1 ⟩ \mathbf{r}(0) = \langle 2, -1 \rangle r ( 0 ) = ⟨ 2 , − 1 ⟩ :
For the
x x x -component:
2 = 2 ( 0 ) 2 + 0 + c 1 2 = 2(0)^2 + 0 + c_1 2 = 2 ( 0 ) 2 + 0 + c 1 , so
c 1 = 2 c_1 = 2 c 1 = 2 For the
y y y -component:
− 1 = 7 ( 0 ) + c 2 -1 = 7(0) + c_2 − 1 = 7 ( 0 ) + c 2 , so
c 2 = − 1 c_2 = -1 c 2 = − 1 Step 3: Write the complete position function:
p ( t ) = ⟨ 2 t 2 + t + 2 , 7 t − 1 ⟩ \mathbf{p}(t) = \langle 2t^2 + t + 2, \, 7t - 1 \rangle p ( t ) = ⟨ 2 t 2 + t + 2 , 7 t − 1 ⟩ Step 4: Evaluate at
t = 3 t = 3 t = 3 :
p ( 3 ) = ⟨ 2 ( 9 ) + 3 + 2 , 7 ( 3 ) − 1 ⟩ = ⟨ 23 , 20 ⟩ \mathbf{p}(3) = \langle 2(9) + 3 + 2, \, 7(3) - 1 \rangle = \langle 23, 20 \rangle p ( 3 ) = ⟨ 2 ( 9 ) + 3 + 2 , 7 ( 3 ) − 1 ⟩ = ⟨ 23 , 20 ⟩ Displacement vs. Total Distance Method 1 above works because integrating velocity gives displacement , the net change in position. Displacement is a vector :
Displacement = ∫ a b v ( t ) d t = ⟨ ∫ a b x ′ ( t ) d t , ∫ a b y ′ ( t ) d t ⟩ = r ( b ) − r ( a ) \text{Displacement} = \int_a^b \mathbf{v}(t) \, dt = \left\langle \int_a^b x'(t) \, dt, \int_a^b y'(t) \, dt \right\rangle = \mathbf{r}(b) - \mathbf{r}(a) Displacement = ∫ a b v ( t ) d t = ⟨ ∫ a b x ′ ( t ) d t , ∫ a b y ′ ( t ) d t ⟩ = r ( b ) − r ( a ) Add the displacement to the starting position and you get the ending position:
r ( b ) = r ( a ) + ∫ a b v ( t ) d t \mathbf{r}(b) = \mathbf{r}(a) + \int_a^b \mathbf{v}(t) \, dt r ( b ) = r ( a ) + ∫ a b v ( t ) d t .
Total distance traveled is a
number , the integral of the speed,
∫ a b ∥ v ( t ) ∥ d t \int_a^b \|\mathbf{v}(t)\| \, dt ∫ a b ∥ v ( t ) ∥ d t . It counts every bit of the path, including curves and backtracking, so it’s always at least as big as the length of the displacement vector.
Example: a particle has velocity
v ( t ) = ⟨ − π sin ( π t ) , π cos ( π t ) ⟩ \mathbf{v}(t) = \langle -\pi\sin(\pi t), \, \pi\cos(\pi t) \rangle v ( t ) = ⟨ − π sin ( π t ) , π cos ( π t )⟩ for
0 ≤ t ≤ 1 0 \le t \le 1 0 ≤ t ≤ 1 . Find its displacement and the total distance it travels.
Displacement = ⟨ ∫ 0 1 − π sin ( π t ) d t , ∫ 0 1 π cos ( π t ) d t ⟩ = ⟨ [ cos ( π t ) ] 0 1 , [ sin ( π t ) ] 0 1 ⟩ = ⟨ − 2 , 0 ⟩ \text{Displacement} = \left\langle \int_0^1 -\pi\sin(\pi t) \, dt, \int_0^1 \pi\cos(\pi t) \, dt \right\rangle = \left\langle \big[\cos(\pi t)\big]_0^1, \big[\sin(\pi t)\big]_0^1 \right\rangle = \langle -2, 0 \rangle Displacement = ⟨ ∫ 0 1 − π sin ( π t ) d t , ∫ 0 1 π cos ( π t ) d t ⟩ = ⟨ [ cos ( π t ) ] 0 1 , [ sin ( π t ) ] 0 1 ⟩ = ⟨ − 2 , 0 ⟩ The speed is
π 2 sin 2 ( π t ) + π 2 cos 2 ( π t ) = π \sqrt{\pi^2\sin^2(\pi t) + \pi^2\cos^2(\pi t)} = \pi π 2 sin 2 ( π t ) + π 2 cos 2 ( π t ) = π , so
Total distance = ∫ 0 1 π d t = π ≈ 3.142 \text{Total distance} = \int_0^1 \pi \, dt = \pi \approx 3.142 Total distance = ∫ 0 1 π d t = π ≈ 3.142 The particle ends up only
2 2 2 units from where it started, but it travels
π \pi π units to get there. If it starts at
( 1 , 0 ) (1, 0) ( 1 , 0 ) , it goes halfway around the unit circle and ends at
( − 1 , 0 ) (-1, 0) ( − 1 , 0 ) .
Direction of Motion You can figure out which way a particle is moving by looking at the signs of the velocity components.
Moving right: x ′ ( t ) > 0 x'(t) > 0 x ′ ( t ) > 0 , positive horizontal velocity
Moving left: x ′ ( t ) < 0 x'(t) < 0 x ′ ( t ) < 0 , negative horizontal velocity
Moving up: y ′ ( t ) > 0 y'(t) > 0 y ′ ( t ) > 0 , positive vertical velocity
Moving down: y ′ ( t ) < 0 y'(t) < 0 y ′ ( t ) < 0 , negative vertical velocity
Example 5: Determining Direction Given: r ( t ) = ⟨ 4 e − t / 3 , t − 2 t + 3 ⟩ \mathbf{r}(t) = \left\langle 4e^{-t/3}, \, \frac{t - 2}{t + 3} \right\rangle r ( t ) = ⟨ 4 e − t /3 , t + 3 t − 2 ⟩ for
t ≥ 0 t \geq 0 t ≥ 0 Velocity:
v ( t ) = ⟨ − 4 3 e − t / 3 , ( t + 3 ) ( 1 ) − ( t − 2 ) ( 1 ) ( t + 3 ) 2 ⟩ = ⟨ − 4 3 e − t / 3 , 5 ( t + 3 ) 2 ⟩ \mathbf{v}(t) = \left\langle -\frac{4}{3}e^{-t/3}, \; \frac{(t + 3)(1) - (t - 2)(1)}{(t + 3)^2} \right\rangle = \left\langle -\frac{4}{3}e^{-t/3}, \; \frac{5}{(t + 3)^2} \right\rangle v ( t ) = ⟨ − 3 4 e − t /3 , ( t + 3 ) 2 ( t + 3 ) ( 1 ) − ( t − 2 ) ( 1 ) ⟩ = ⟨ − 3 4 e − t /3 , ( t + 3 ) 2 5 ⟩ When is the particle moving right?
Solve
− 4 3 e − t / 3 > 0 -\frac{4}{3}e^{-t/3} > 0 − 3 4 e − t /3 > 0 . Since
e − t / 3 e^{-t/3} e − t /3 is always positive and we’re multiplying it by
− 4 3 -\frac{4}{3} − 3 4 , which is negative, this expression is always negative.
So the particle is
never moving right for
t ≥ 0 t \geq 0 t ≥ 0 . It’s always moving left.
When is the particle moving up?
Solve
5 ( t + 3 ) 2 > 0 \frac{5}{(t + 3)^2} > 0 ( t + 3 ) 2 5 > 0 . The numerator is positive and the denominator is always positive for
t ≥ 0 t \geq 0 t ≥ 0 .
So the particle is
always moving up for
t ≥ 0 t \geq 0 t ≥ 0 .
Key Takeaways Velocity is the derivative of position: v ( t ) = r ′ ( t ) \mathbf{v}(t) = \mathbf{r}'(t) v ( t ) = r ′ ( t ) Acceleration is the derivative of velocity: a ( t ) = v ′ ( t ) = r ′ ′ ( t ) \mathbf{a}(t) = \mathbf{v}'(t) = \mathbf{r}''(t) a ( t ) = v ′ ( t ) = r ′′ ( t ) Speed is the magnitude of velocity: Speed = ∥ v ( t ) ∥ = [ x ′ ( t ) ] 2 + [ y ′ ( t ) ] 2 \text{Speed} = \|\mathbf{v}(t)\| = \sqrt{[x'(t)]^2 + [y'(t)]^2} Speed = ∥ v ( t ) ∥ = [ x ′ ( t ) ] 2 + [ y ′ ( t ) ] 2 Speed is a scalar, not a vector Pythagorean identity: sin 2 ( θ ) + cos 2 ( θ ) = 1 \sin^2(\theta) + \cos^2(\theta) = 1 sin 2 ( θ ) + cos 2 ( θ ) = 1 Speeding up happens when x ′ ( t ) x ′ ′ ( t ) + y ′ ( t ) y ′ ′ ( t ) > 0 x'(t)x''(t) + y'(t)y''(t) > 0 x ′ ( t ) x ′′ ( t ) + y ′ ( t ) y ′′ ( t ) > 0 Slowing down happens when x ′ ( t ) x ′ ′ ( t ) + y ′ ( t ) y ′ ′ ( t ) < 0 x'(t)x''(t) + y'(t)y''(t) < 0 x ′ ( t ) x ′′ ( t ) + y ′ ( t ) y ′′ ( t ) < 0 Total distance: ∫ a b ∥ v ( t ) ∥ d t \int_a^b \|\mathbf{v}(t)\| \, dt ∫ a b ∥ v ( t ) ∥ d t Displacement: ∫ a b v ( t ) d t \int_a^b \mathbf{v}(t) \, dt ∫ a b v ( t ) d t r ( b ) − r ( a ) \mathbf{r}(b) - \mathbf{r}(a) r ( b ) − r ( a ) Moving right or left is determined by the sign of x ′ ( t ) x'(t) x ′ ( t ) Moving up or down is determined by the sign of y ′ ( t ) y'(t) y ′ ( t ) 
