In AB you studied a particle moving back and forth along a line. Now the particle is free to move anywhere in the plane: a drone, a hockey puck, a ball in flight. Its position is a vector-valued function , and everything you learned about motion along a line still works, one component at a time. This topic appears on almost every AP Calculus BC exam, usually as a calculator-active free-response question.
For a particle with position r ⃗ ( t ) = ⟨ x ( t ) , y ( t ) ⟩ \vec{r}(t) = \langle x(t),\ y(t) \rangle r ( t ) = ⟨ x ( t ) , y ( t )⟩ t t t
Quantity Formula What it tells you Position r ⃗ ( t ) = ⟨ x ( t ) , y ( t ) ⟩ \vec{r}(t) = \langle x(t),\ y(t) \rangle r ( t ) = ⟨ x ( t ) , y ( t )⟩ where the particle is Velocity v ⃗ ( t ) = r ⃗ ′ ( t ) = ⟨ x ′ ( t ) , y ′ ( t ) ⟩ \vec{v}(t) = \vec{r}\,'(t) = \langle x'(t),\ y'(t) \rangle v ( t ) = r ′ ( t ) = ⟨ x ′ ( t ) , y ′ ( t )⟩ direction of motion and how fast; tangent to the path Acceleration a ⃗ ( t ) = v ⃗ ′ ( t ) = ⟨ x ′ ′ ( t ) , y ′ ′ ( t ) ⟩ \vec{a}(t) = \vec{v}\,'(t) = \langle x''(t),\ y''(t) \rangle a ( t ) = v ′ ( t ) = ⟨ x ′′ ( t ) , y ′′ ( t )⟩ how the velocity is changing Speed ∣ v ⃗ ( t ) ∣ = ( x ′ ( t ) ) 2 + ( y ′ ( t ) ) 2 \lvert \vec{v}(t) \rvert = \sqrt{(x'(t))^2 + (y'(t))^2} ∣ v ( t )∣ = ( x ′ ( t ) ) 2 + ( y ′ ( t ) ) 2 how fast, as a single number ≥ 0 \ge 0 ≥ 0
Velocity is a vector ; speed is a number , the length (magnitude) of the velocity vector.
Each component tells you about one direction:
x ′ ( t ) > 0 x'(t) \gt 0 x ′ ( t ) > 0 right . x ′ ( t ) < 0 x'(t) \lt 0 x ′ ( t ) < 0 left .y ′ ( t ) > 0 y'(t) \gt 0 y ′ ( t ) > 0 up . y ′ ( t ) < 0 y'(t) \lt 0 y ′ ( t ) < 0 down .If x ′ ( t ) = 0 x'(t) = 0 x ′ ( t ) = 0 y ′ ( t ) ≠ 0 y'(t) \ne 0 y ′ ( t ) = 0
The particle is at rest only when x ′ ( t ) = 0 x'(t) = 0 x ′ ( t ) = 0 and y ′ ( t ) = 0 y'(t) = 0 y ′ ( t ) = 0
The path x = t cubed - 3t, y = t squared for t from -2 to 2. Velocity vectors are drawn tangent to the path: v(-1) = <0, -2> at (2, 1) points straight down, v(0) = <-3, 0> at the origin points left, and v(1) = <0, 2> at (-2, 1) points straight up.
−3
−2
−1
1
2
3
−1
1
2
3
4
t = −1
t = 0
t = 1
v(−1) = ⟨0, −2⟩
v(0) = ⟨−3, 0⟩
v(1) = ⟨0, 2⟩
t = −2
t = 2
x
y
For r ⃗ ( t ) = ⟨ t 3 − 3 t , t 2 ⟩ \vec{r}(t) = \langle t^3 - 3t,\ t^2 \rangle r ( t ) = ⟨ t 3 − 3 t , t 2 ⟩ Integrate each component and add the starting position:
r ⃗ ( b ) = r ⃗ ( a ) + ∫ a b v ⃗ ( t ) d t that is, x ( b ) = x ( a ) + ∫ a b x ′ ( t ) d t , y ( b ) = y ( a ) + ∫ a b y ′ ( t ) d t \vec{r}(b) = \vec{r}(a) + \int_a^b \vec{v}(t)\, dt
\qquad\text{that is,}\qquad
x(b) = x(a) + \int_a^b x'(t)\, dt, \quad y(b) = y(a) + \int_a^b y'(t)\, dt r ( b ) = r ( a ) + ∫ a b v ( t ) d t that is, x ( b ) = x ( a ) + ∫ a b x ′ ( t ) d t , y ( b ) = y ( a ) + ∫ a b y ′ ( t ) d t The integral ∫ a b v ⃗ ( t ) d t \int_a^b \vec{v}(t)\, dt ∫ a b v ( t ) d t displacement vector.
Total distance is the integral of speed (the same integral as parametric arc length ):
distance travelled from t = a to t = b = ∫ a b ( x ′ ( t ) ) 2 + ( y ′ ( t ) ) 2 d t \text{distance travelled from } t = a \text{ to } t = b = \int_a^b \sqrt{(x'(t))^2 + (y'(t))^2}\ dt distance travelled from t = a to t = b = ∫ a b ( x ′ ( t ) ) 2 + ( y ′ ( t ) ) 2 d t Motion questions are usually calculator active. Write each setup (the integral or the expression) before the number, give decimals to 3 3 3 radian mode. Common requests: speed at a time, the time when the particle moves in some direction, the position at a later time, and the total distance travelled.
A particle has position r ⃗ ( t ) = ⟨ t 3 − 3 t , t 2 ⟩ \vec{r}(t) = \langle t^3 - 3t,\ t^2 \rangle r ( t ) = ⟨ t 3 − 3 t , t 2 ⟩ t = 2 t = 2 t = 2 t ≥ 0 t \ge 0 t ≥ 0
Solution. Differentiate each component:
v ⃗ ( t ) = ⟨ 3 t 2 − 3 , 2 t ⟩ , a ⃗ ( t ) = ⟨ 6 t , 2 ⟩ \vec{v}(t) = \langle 3t^2 - 3,\ 2t \rangle, \qquad \vec{a}(t) = \langle 6t,\ 2 \rangle v ( t ) = ⟨ 3 t 2 − 3 , 2 t ⟩ , a ( t ) = ⟨ 6 t , 2 ⟩ At t = 2 t = 2 t = 2
v ⃗ ( 2 ) = ⟨ 9 , 4 ⟩ , a ⃗ ( 2 ) = ⟨ 12 , 2 ⟩ , speed = 81 + 16 = 97 ≈ 9.849 \vec{v}(2) = \langle 9, 4 \rangle, \qquad \vec{a}(2) = \langle 12, 2 \rangle, \qquad \text{speed} = \sqrt{81 + 16} = \sqrt{97} \approx 9.849 v ( 2 ) = ⟨ 9 , 4 ⟩ , a ( 2 ) = ⟨ 12 , 2 ⟩ , speed = 81 + 16 = 97 ≈ 9.849 Left: x ′ ( t ) = 3 t 2 − 3 = 3 ( t − 1 ) ( t + 1 ) < 0 x'(t) = 3t^2 - 3 = 3(t - 1)(t + 1) \lt 0 x ′ ( t ) = 3 t 2 − 3 = 3 ( t − 1 ) ( t + 1 ) < 0 − 1 < t < 1 -1 \lt t \lt 1 − 1 < t < 1 t ≥ 0 t \ge 0 t ≥ 0 0 ≤ t < 1 0 \le t \lt 1 0 ≤ t < 1
Up: y ′ ( t ) = 2 t > 0 y'(t) = 2t \gt 0 y ′ ( t ) = 2 t > 0 t > 0 t \gt 0 t > 0 t > 0 t \gt 0 t > 0
At t = 1 t = 1 t = 1 ⟨ 0 , 2 ⟩ \langle 0, 2 \rangle ⟨ 0 , 2 ⟩ y ′ ( t ) = 0 y'(t) = 0 y ′ ( t ) = 0 t = 0 t = 0 t = 0 x ′ ( 0 ) = − 3 ≠ 0 x'(0) = -3 \ne 0 x ′ ( 0 ) = − 3 = 0
A particle has velocity v ⃗ ( t ) = ⟨ 2 t + 1 , π cos ( π t ) ⟩ \vec{v}(t) = \langle 2t + 1,\ \pi\cos(\pi t) \rangle v ( t ) = ⟨ 2 t + 1 , π cos ( π t )⟩ r ⃗ ( 0 ) = ⟨ 3 , − 1 ⟩ \vec{r}(0) = \langle 3, -1 \rangle r ( 0 ) = ⟨ 3 , − 1 ⟩ t = 2 t = 2 t = 2
Solution.
x ( 2 ) = 3 + ∫ 0 2 ( 2 t + 1 ) d t = 3 + [ t 2 + t ] 0 2 = 3 + 6 = 9 x(2) = 3 + \int_0^2 (2t + 1)\, dt = 3 + \Big[ t^2 + t \Big]_0^2 = 3 + 6 = 9 x ( 2 ) = 3 + ∫ 0 2 ( 2 t + 1 ) d t = 3 + [ t 2 + t ] 0 2 = 3 + 6 = 9 y ( 2 ) = − 1 + ∫ 0 2 π cos ( π t ) d t = − 1 + [ sin ( π t ) ] 0 2 = − 1 + 0 = − 1 y(2) = -1 + \int_0^2 \pi\cos(\pi t)\, dt = -1 + \Big[ \sin(\pi t) \Big]_0^2 = -1 + 0 = -1 y ( 2 ) = − 1 + ∫ 0 2 π cos ( π t ) d t = − 1 + [ sin ( π t ) ] 0 2 = − 1 + 0 = − 1 So r ⃗ ( 2 ) = ⟨ 9 , − 1 ⟩ \vec{r}(2) = \langle 9, -1 \rangle r ( 2 ) = ⟨ 9 , − 1 ⟩ y y y
A particle has position r ⃗ ( t ) = ⟨ 3 t 2 , 2 t 3 ⟩ \vec{r}(t) = \langle 3t^2,\ 2t^3 \rangle r ( t ) = ⟨ 3 t 2 , 2 t 3 ⟩ t = 0 t = 0 t = 0 t = 2 2 t = 2\sqrt{2} t = 2 2
Solution. v ⃗ ( t ) = ⟨ 6 t , 6 t 2 ⟩ \vec{v}(t) = \langle 6t,\ 6t^2 \rangle v ( t ) = ⟨ 6 t , 6 t 2 ⟩ t ≥ 0 t \ge 0 t ≥ 0
speed = 36 t 2 + 36 t 4 = 6 t 1 + t 2 \text{speed} = \sqrt{36t^2 + 36t^4} = 6t\sqrt{1 + t^2} speed = 36 t 2 + 36 t 4 = 6 t 1 + t 2 Let u = 1 + t 2 u = 1 + t^2 u = 1 + t 2 d u = 2 t d t du = 2t\, dt d u = 2 t d t u u u 1 1 1 9 9 9
∫ 0 2 2 6 t 1 + t 2 d t = 3 ∫ 1 9 u d u = 2 [ u 3 / 2 ] 1 9 = 2 ( 27 − 1 ) = 52 \int_0^{2\sqrt{2}} 6t\sqrt{1 + t^2}\, dt = 3\int_1^9 \sqrt{u}\, du = 2\Big[ u^{3/2} \Big]_1^9 = 2(27 - 1) = 52 ∫ 0 2 2 6 t 1 + t 2 d t = 3 ∫ 1 9 u d u = 2 [ u 3/2 ] 1 9 = 2 ( 27 − 1 ) = 52 The particle goes from ( 0 , 0 ) (0, 0) ( 0 , 0 ) ( 24 , 32 2 ) \left( 24,\ 32\sqrt{2} \right) ( 24 , 32 2 ) 3 ( 2 2 ) 2 = 24 3(2\sqrt{2})^2 = 24 3 ( 2 2 ) 2 = 24 2 ( 2 2 ) 3 = 32 2 2(2\sqrt{2})^3 = 32\sqrt{2} 2 ( 2 2 ) 3 = 32 2
24 2 + ( 32 2 ) 2 = 576 + 2048 = 2624 = 8 41 ≈ 51.225 \sqrt{24^2 + \left( 32\sqrt{2} \right)^2} = \sqrt{576 + 2048} = \sqrt{2624} = 8\sqrt{41} \approx 51.225 2 4 2 + ( 32 2 ) 2 = 576 + 2048 = 2624 = 8 41 ≈ 51.225 Slightly less than 52 52 52
A particle moves in the plane with velocity v ⃗ ( t ) = ⟨ e 0.5 t − 2 , sin ( t 2 ) ⟩ \vec{v}(t) = \left\langle e^{0.5t} - 2,\ \sin(t^2) \right\rangle v ( t ) = ⟨ e 0.5 t − 2 , sin ( t 2 ) ⟩ 0 ≤ t ≤ 3 0 \le t \le 3 0 ≤ t ≤ 3 r ⃗ ( 0 ) = ⟨ 1 , 3 ⟩ \vec{r}(0) = \langle 1, 3 \rangle r ( 0 ) = ⟨ 1 , 3 ⟩
(a) Find the speed of the particle at t = 2 t = 2 t = 2 t = 2 t = 2 t = 2
(b) On what interval is the particle moving left?
(c) Find the position of the particle at t = 3 t = 3 t = 3
(d) Find the total distance travelled from t = 0 t = 0 t = 0 t = 3 t = 3 t = 3
Solution. (a)
∣ v ⃗ ( 2 ) ∣ = ( e − 2 ) 2 + ( sin 4 ) 2 ≈ 1.043 \lvert \vec{v}(2) \rvert = \sqrt{(e - 2)^2 + (\sin 4)^2} \approx 1.043 ∣ v ( 2 )∣ = ( e − 2 ) 2 + ( sin 4 ) 2 ≈ 1.043 a ⃗ ( t ) = ⟨ 0.5 e 0.5 t , 2 t cos ( t 2 ) ⟩ \vec{a}(t) = \left\langle 0.5e^{0.5t},\ 2t\cos(t^2) \right\rangle a ( t ) = ⟨ 0.5 e 0.5 t , 2 t cos ( t 2 ) ⟩ a ⃗ ( 2 ) = ⟨ 0.5 e , 4 cos 4 ⟩ ≈ ⟨ 1.359 , − 2.615 ⟩ \vec{a}(2) = \langle 0.5e,\ 4\cos 4 \rangle \approx \langle 1.359, -2.615 \rangle a ( 2 ) = ⟨ 0.5 e , 4 cos 4 ⟩ ≈ ⟨ 1.359 , − 2.615 ⟩
(b) Moving left means x ′ ( t ) = e 0.5 t − 2 < 0 x'(t) = e^{0.5t} - 2 \lt 0 x ′ ( t ) = e 0.5 t − 2 < 0 e 0.5 t < 2 e^{0.5t} \lt 2 e 0.5 t < 2 t < 2 ln 2 ≈ 1.386 t \lt 2\ln 2 \approx 1.386 t < 2 ln 2 ≈ 1.386 0 ≤ t < 1.386 0 \le t \lt 1.386 0 ≤ t < 1.386
(c)
x ( 3 ) = 1 + ∫ 0 3 ( e 0.5 t − 2 ) d t ≈ 1.963 , y ( 3 ) = 3 + ∫ 0 3 sin ( t 2 ) d t ≈ 3.774 x(3) = 1 + \int_0^3 \left( e^{0.5t} - 2 \right) dt \approx 1.963, \qquad y(3) = 3 + \int_0^3 \sin(t^2)\, dt \approx 3.774 x ( 3 ) = 1 + ∫ 0 3 ( e 0.5 t − 2 ) d t ≈ 1.963 , y ( 3 ) = 3 + ∫ 0 3 sin ( t 2 ) d t ≈ 3.774 So r ⃗ ( 3 ) ≈ ⟨ 1.963 , 3.774 ⟩ \vec{r}(3) \approx \langle 1.963, 3.774 \rangle r ( 3 ) ≈ ⟨ 1.963 , 3.774 ⟩ x ( 3 ) = 2 e 1.5 − 7 x(3) = 2e^{1.5} - 7 x ( 3 ) = 2 e 1.5 − 7
(d)
∫ 0 3 ( e 0.5 t − 2 ) 2 + sin 2 ( t 2 ) d t ≈ 3.347 \int_0^3 \sqrt{\left( e^{0.5t} - 2 \right)^2 + \sin^2(t^2)}\ dt \approx 3.347 ∫ 0 3 ( e 0.5 t − 2 ) 2 + sin 2 ( t 2 ) d t ≈ 3.347 Giving a vector when the question asks for speed. Speed is the number ( x ′ ) 2 + ( y ′ ) 2 \sqrt{(x')^2 + (y')^2} ( x ′ ) 2 + ( y ′ ) 2 ⟨ x ′ , y ′ ⟩ \langle x', y' \rangle ⟨ x ′ , y ′ ⟩ x ′ + y ′ x' + y' x ′ + y ′
Calling the particle “at rest” when only one component is zero. If x ′ ( t ) = 0 x'(t) = 0 x ′ ( t ) = 0 y ′ ( t ) ≠ 0 y'(t) \ne 0 y ′ ( t ) = 0 both components equal to 0 0 0
Forgetting the initial position. ∫ 0 3 x ′ ( t ) d t \int_0^3 x'(t)\, dt ∫ 0 3 x ′ ( t ) d t x x x x ( 0 ) x(0) x ( 0 ) x ( 3 ) x(3) x ( 3 )
Using displacement for distance. “Total distance travelled” is ∫ speed d t \int \text{speed}\, dt ∫ speed d t ∫ v ⃗ d t \int \vec{v}\, dt ∫ v d t
Mixing up “left/right” and “up/down”. Left and right come from the sign of x ′ ( t ) x'(t) x ′ ( t ) y ′ ( t ) y'(t) y ′ ( t ) y ( t ) y(t) y ( t ) x x x
Rounding too early, or degree mode. Store intermediate values in your calculator and round only the final answer to 3 3 3
1. (Warm-up) A particle has position r ⃗ ( t ) = ⟨ t 2 , 4 t − 1 ⟩ \vec{r}(t) = \langle t^2,\ 4t - 1 \rangle r ( t ) = ⟨ t 2 , 4 t − 1 ⟩ t = 1 t = 1 t = 1
Solution v ⃗ ( t ) = ⟨ 2 t , 4 ⟩ \vec{v}(t) = \langle 2t, 4 \rangle v ( t ) = ⟨ 2 t , 4 ⟩ a ⃗ ( t ) = ⟨ 2 , 0 ⟩ \vec{a}(t) = \langle 2, 0 \rangle a ( t ) = ⟨ 2 , 0 ⟩ t = 1 t = 1 t = 1
v ⃗ ( 1 ) = ⟨ 2 , 4 ⟩ , a ⃗ ( 1 ) = ⟨ 2 , 0 ⟩ , speed = 4 + 16 = 2 5 ≈ 4.472 \vec{v}(1) = \langle 2, 4 \rangle, \qquad \vec{a}(1) = \langle 2, 0 \rangle, \qquad \text{speed} = \sqrt{4 + 16} = 2\sqrt{5} \approx 4.472 v ( 1 ) = ⟨ 2 , 4 ⟩ , a ( 1 ) = ⟨ 2 , 0 ⟩ , speed = 4 + 16 = 2 5 ≈ 4.472 2. (Warm-up) Show that a particle with position r ⃗ ( t ) = ⟨ cos t , sin t ⟩ \vec{r}(t) = \langle \cos t,\ \sin t \rangle r ( t ) = ⟨ cos t , sin t ⟩ 1 1 1
Solution v ⃗ ( t ) = ⟨ − sin t , cos t ⟩ \vec{v}(t) = \langle -\sin t,\ \cos t \rangle v ( t ) = ⟨ − sin t , cos t ⟩
speed = sin 2 t + cos 2 t = 1 \text{speed} = \sqrt{\sin^2 t + \cos^2 t} = 1 speed = sin 2 t + cos 2 t = 1 The particle moves around the unit circle (counterclockwise) at constant speed 1 1 1
3. (Warm-up) A particle has velocity v ⃗ ( t ) = ⟨ 3 , − 2 t ⟩ \vec{v}(t) = \langle 3,\ -2t \rangle v ( t ) = ⟨ 3 , − 2 t ⟩ r ⃗ ( 0 ) = ⟨ 1 , 5 ⟩ \vec{r}(0) = \langle 1, 5 \rangle r ( 0 ) = ⟨ 1 , 5 ⟩ r ⃗ ( t ) \vec{r}(t) r ( t ) r ⃗ ( 2 ) \vec{r}(2) r ( 2 )
Solution r ⃗ ( t ) = ⟨ 1 + 3 t , 5 − t 2 ⟩ , r ⃗ ( 2 ) = ⟨ 7 , 1 ⟩ \vec{r}(t) = \langle 1 + 3t,\ 5 - t^2 \rangle, \qquad \vec{r}(2) = \langle 7,\ 1 \rangle r ( t ) = ⟨ 1 + 3 t , 5 − t 2 ⟩ , r ( 2 ) = ⟨ 7 , 1 ⟩ 4. (Core) A particle has position r ⃗ ( t ) = ⟨ t 3 − 12 t , t 2 − 2 t ⟩ \vec{r}(t) = \langle t^3 - 12t,\ t^2 - 2t \rangle r ( t ) = ⟨ t 3 − 12 t , t 2 − 2 t ⟩ 0 ≤ t ≤ 4 0 \le t \le 4 0 ≤ t ≤ 4
(a) When is it moving left? When is it moving down?
(b) Is the particle ever at rest? Explain.
Solution v ⃗ ( t ) = ⟨ 3 t 2 − 12 , 2 t − 2 ⟩ \vec{v}(t) = \langle 3t^2 - 12,\ 2t - 2 \rangle v ( t ) = ⟨ 3 t 2 − 12 , 2 t − 2 ⟩
(a) Left: 3 t 2 − 12 = 3 ( t − 2 ) ( t + 2 ) < 0 3t^2 - 12 = 3(t - 2)(t + 2) \lt 0 3 t 2 − 12 = 3 ( t − 2 ) ( t + 2 ) < 0 0 ≤ t < 2 0 \le t \lt 2 0 ≤ t < 2 2 t − 2 < 0 2t - 2 \lt 0 2 t − 2 < 0 0 ≤ t < 1 0 \le t \lt 1 0 ≤ t < 1
(b) x ′ ( t ) = 0 x'(t) = 0 x ′ ( t ) = 0 t = 2 t = 2 t = 2 y ′ ( 2 ) = 2 ≠ 0 y'(2) = 2 \ne 0 y ′ ( 2 ) = 2 = 0 0 0 0
5. (Core) A particle has position r ⃗ ( t ) = ⟨ e t + e − t , 2 t ⟩ \vec{r}(t) = \left\langle e^t + e^{-t},\ 2t \right\rangle r ( t ) = ⟨ e t + e − t , 2 t ⟩ t = 0 t = 0 t = 0 t = ln 2 t = \ln 2 t = ln 2
Solution v ⃗ ( t ) = ⟨ e t − e − t , 2 ⟩ \vec{v}(t) = \left\langle e^t - e^{-t},\ 2 \right\rangle v ( t ) = ⟨ e t − e − t , 2 ⟩
( e t − e − t ) 2 + 4 = e 2 t − 2 + e − 2 t + 4 = e 2 t + 2 + e − 2 t = ( e t + e − t ) 2 (e^t - e^{-t})^2 + 4 = e^{2t} - 2 + e^{-2t} + 4 = e^{2t} + 2 + e^{-2t} = (e^t + e^{-t})^2 ( e t − e − t ) 2 + 4 = e 2 t − 2 + e − 2 t + 4 = e 2 t + 2 + e − 2 t = ( e t + e − t ) 2 So the speed is e t + e − t e^t + e^{-t} e t + e − t
∫ 0 ln 2 ( e t + e − t ) d t = [ e t − e − t ] 0 ln 2 = ( 2 − 1 2 ) − 0 = 3 2 \int_0^{\ln 2} (e^t + e^{-t})\, dt = \Big[ e^t - e^{-t} \Big]_0^{\ln 2} = \left( 2 - \frac{1}{2} \right) - 0 = \frac{3}{2} ∫ 0 l n 2 ( e t + e − t ) d t = [ e t − e − t ] 0 l n 2 = ( 2 − 2 1 ) − 0 = 2 3 6. (Core) A particle has velocity v ⃗ ( t ) = ⟨ 1 t + 1 , 2 t ⟩ \vec{v}(t) = \left\langle \dfrac{1}{t + 1},\ 2t \right\rangle v ( t ) = ⟨ t + 1 1 , 2 t ⟩ t ≥ 0 t \ge 0 t ≥ 0 r ⃗ ( 0 ) = ⟨ 0 , 4 ⟩ \vec{r}(0) = \langle 0, 4 \rangle r ( 0 ) = ⟨ 0 , 4 ⟩ r ⃗ ( 3 ) \vec{r}(3) r ( 3 )
Solution x ( 3 ) = 0 + ∫ 0 3 1 t + 1 d t = ln 4 , y ( 3 ) = 4 + ∫ 0 3 2 t d t = 4 + 9 = 13 x(3) = 0 + \int_0^3 \frac{1}{t + 1}\, dt = \ln 4, \qquad y(3) = 4 + \int_0^3 2t\, dt = 4 + 9 = 13 x ( 3 ) = 0 + ∫ 0 3 t + 1 1 d t = ln 4 , y ( 3 ) = 4 + ∫ 0 3 2 t d t = 4 + 9 = 13 So r ⃗ ( 3 ) = ⟨ ln 4 , 13 ⟩ \vec{r}(3) = \langle \ln 4,\ 13 \rangle r ( 3 ) = ⟨ ln 4 , 13 ⟩
7. (Core) (Calculator active.) A particle has velocity v ⃗ ( t ) = ⟨ cos ( t 2 ) , 2 e − 0.5 t ⟩ \vec{v}(t) = \left\langle \cos(t^2),\ 2e^{-0.5t} \right\rangle v ( t ) = ⟨ cos ( t 2 ) , 2 e − 0.5 t ⟩ 0 ≤ t ≤ 2 0 \le t \le 2 0 ≤ t ≤ 2 r ⃗ ( 0 ) = ⟨ 2 , 1 ⟩ \vec{r}(0) = \langle 2, 1 \rangle r ( 0 ) = ⟨ 2 , 1 ⟩
(a) Find the speed at t = 1 t = 1 t = 1
(b) Find r ⃗ ( 2 ) \vec{r}(2) r ( 2 )
(c) Find the total distance travelled from t = 0 t = 0 t = 0 t = 2 t = 2 t = 2
Solution (a) cos 2 ( 1 ) + 4 e − 1 ≈ 1.328 \sqrt{\cos^2(1) + 4e^{-1}} \approx 1.328 cos 2 ( 1 ) + 4 e − 1 ≈ 1.328
(b)
x ( 2 ) = 2 + ∫ 0 2 cos ( t 2 ) d t ≈ 2.461 , y ( 2 ) = 1 + ∫ 0 2 2 e − 0.5 t d t ≈ 3.528 x(2) = 2 + \int_0^2 \cos(t^2)\, dt \approx 2.461, \qquad y(2) = 1 + \int_0^2 2e^{-0.5t}\, dt \approx 3.528 x ( 2 ) = 2 + ∫ 0 2 cos ( t 2 ) d t ≈ 2.461 , y ( 2 ) = 1 + ∫ 0 2 2 e − 0.5 t d t ≈ 3.528 So r ⃗ ( 2 ) ≈ ⟨ 2.461 , 3.528 ⟩ \vec{r}(2) \approx \langle 2.461, 3.528 \rangle r ( 2 ) ≈ ⟨ 2.461 , 3.528 ⟩ y ( 2 ) = 1 + [ − 4 e − 0.5 t ] 0 2 = 5 − 4 e − 1 y(2) = 1 + \Big[ -4e^{-0.5t} \Big]_0^2 = 5 - 4e^{-1} y ( 2 ) = 1 + [ − 4 e − 0.5 t ] 0 2 = 5 − 4 e − 1
(c)
∫ 0 2 cos 2 ( t 2 ) + 4 e − t d t ≈ 2.987 \int_0^2 \sqrt{\cos^2(t^2) + 4e^{-t}}\ dt \approx 2.987 ∫ 0 2 cos 2 ( t 2 ) + 4 e − t d t ≈ 2.987 8. (Challenge) A particle has position r ⃗ ( t ) = ⟨ t 2 , t 3 3 − 4 t ⟩ \vec{r}(t) = \left\langle t^2,\ \dfrac{t^3}{3} - 4t \right\rangle r ( t ) = ⟨ t 2 , 3 t 3 − 4 t ⟩ t ≥ 0 t \ge 0 t ≥ 0
Solution v ⃗ ( t ) = ⟨ 2 t , t 2 − 4 ⟩ \vec{v}(t) = \langle 2t,\ t^2 - 4 \rangle v ( t ) = ⟨ 2 t , t 2 − 4 ⟩
S ( t ) = ( 2 t ) 2 + ( t 2 − 4 ) 2 = t 4 − 4 t 2 + 16 , S ′ ( t ) = 4 t 3 − 8 t = 4 t ( t 2 − 2 ) S(t) = (2t)^2 + (t^2 - 4)^2 = t^4 - 4t^2 + 16, \qquad S'(t) = 4t^3 - 8t = 4t(t^2 - 2) S ( t ) = ( 2 t ) 2 + ( t 2 − 4 ) 2 = t 4 − 4 t 2 + 16 , S ′ ( t ) = 4 t 3 − 8 t = 4 t ( t 2 − 2 ) For t ≥ 0 t \ge 0 t ≥ 0 S ′ ( t ) = 0 S'(t) = 0 S ′ ( t ) = 0 t = 0 t = 0 t = 0 t = 2 t = \sqrt{2} t = 2 S ′ ( t ) < 0 S'(t) \lt 0 S ′ ( t ) < 0 ( 0 , 2 ) (0, \sqrt{2}) ( 0 , 2 ) S ′ ( t ) > 0 S'(t) \gt 0 S ′ ( t ) > 0 t > 2 t \gt \sqrt{2} t > 2 S S S t = 2 t = \sqrt{2} t = 2
speed = S ( 2 ) = 4 − 8 + 16 = 12 = 2 3 ≈ 3.464 \text{speed} = \sqrt{S(\sqrt{2})} = \sqrt{4 - 8 + 16} = \sqrt{12} = 2\sqrt{3} \approx 3.464 speed = S ( 2 ) = 4 − 8 + 16 = 12 = 2 3 ≈ 3.464 (Compare: at t = 0 t = 0 t = 0 4 4 4
9. (Challenge) A particle has velocity v ⃗ ( t ) = ⟨ 4 t − 4 , 3 t 2 − 3 ⟩ \vec{v}(t) = \langle 4t - 4,\ 3t^2 - 3 \rangle v ( t ) = ⟨ 4 t − 4 , 3 t 2 − 3 ⟩ t ≥ 0 t \ge 0 t ≥ 0 r ⃗ ( 0 ) = ⟨ 1 , 2 ⟩ \vec{r}(0) = \langle 1, 2 \rangle r ( 0 ) = ⟨ 1 , 2 ⟩
(a) At what time is the particle at rest? Where is it then?
(b) (Calculator active.) Find the total distance travelled from t = 0 t = 0 t = 0 t = 2 t = 2 t = 2 r ⃗ ( 0 ) \vec{r}(0) r ( 0 ) r ⃗ ( 2 ) \vec{r}(2) r ( 2 )
Solution (a) 4 t − 4 = 0 4t - 4 = 0 4 t − 4 = 0 t = 1 t = 1 t = 1 3 t 2 − 3 = 0 3t^2 - 3 = 0 3 t 2 − 3 = 0 t = ± 1 t = \pm 1 t = ± 1 0 0 0 t = 1 t = 1 t = 1
r ⃗ ( t ) = ⟨ 1 + 2 t 2 − 4 t , 2 + t 3 − 3 t ⟩ , r ⃗ ( 1 ) = ⟨ − 1 , 0 ⟩ \vec{r}(t) = \langle 1 + 2t^2 - 4t,\ 2 + t^3 - 3t \rangle, \qquad \vec{r}(1) = \langle -1,\ 0 \rangle r ( t ) = ⟨ 1 + 2 t 2 − 4 t , 2 + t 3 − 3 t ⟩ , r ( 1 ) = ⟨ − 1 , 0 ⟩ (b)
∫ 0 2 ( 4 t − 4 ) 2 + ( 3 t 2 − 3 ) 2 d t ≈ 7.314 \int_0^2 \sqrt{(4t - 4)^2 + (3t^2 - 3)^2}\ dt \approx 7.314 ∫ 0 2 ( 4 t − 4 ) 2 + ( 3 t 2 − 3 ) 2 d t ≈ 7.314 r ⃗ ( 2 ) = ⟨ 1 + 8 − 8 , 2 + 8 − 6 ⟩ = ⟨ 1 , 4 ⟩ \vec{r}(2) = \langle 1 + 8 - 8,\ 2 + 8 - 6 \rangle = \langle 1, 4 \rangle r ( 2 ) = ⟨ 1 + 8 − 8 , 2 + 8 − 6 ⟩ = ⟨ 1 , 4 ⟩ ( 1 , 2 ) (1, 2) ( 1 , 2 ) ( 1 , 4 ) (1, 4) ( 1 , 4 ) 2 2 2 ( − 1 , 0 ) (-1, 0) ( − 1 , 0 ) ( 1 , 4 ) (1, 4) ( 1 , 4 )
