Acceleration Is a Vector: The Cycloid Accelerates at the Instant It Stops
I Key FormulasKey Insight: Zero velocity does not mean zero acceleration. At the cusp the cycloid's velocity changes from downward, through 0 β \vec{0} 0 a β ( 0 ) = β¨ 0 , 1 β© \vec{a}(0) = \langle 0, 1 \rangle a ( 0 ) = β¨ 0 , 1 β©
II The Cycloid Never Stops Acceleratinga β = β¨ sin β‘ t , cos β‘ t β© \vec{a} = \langle \sin t, \cos t \rangle a = β¨ sin t , cos t β© 1 1 1 t = 0 t = 0 t = 0 v β = 0 β \vec{v} = \vec{0} v = 0
βΆ play
t β1.20
t = 0 (cusp)
III From Ξ v β / Ξ t \Delta\vec{v}/\Delta t Ξ v /Ξ t a β \vec{a} a On the unit circle the speed never changes, yet Ξ v β / Ξ t \Delta\vec{v}/\Delta t Ξ v /Ξ t
IV Speed Is Not the Derivative of the Distanceβ£ d r β / d t β£ |d\vec{r}/dt| β£ d r / d t β£ d β£ r β β£ / d t = r β β
v β / β£ r β β£ d|\vec{r}|/dt = \vec{r} \cdot \vec{v}/|\vec{r}| d β£ r β£/ d t = r β
v /β£ r β£ r β \vec{r} r
Path
unit circle β¨cos t, sin tβ©
cycloid β¨t β sin t, 1 β cos tβ©
line β¨t β 2, 1β©
t 1.00
V Quiz Questions
Problem 1 Β· Differentiate Twice
Given: the cycloid r β ( t ) = β¨ t β sin β‘ t , Β 1 β cos β‘ t β© \vec{r}(t) = \langle t - \sin t,\ 1 - \cos t \rangle r ( t ) = β¨ t β sin t , Β 1 β cos t β© find the acceleration a β ( Ο 3 ) \vec{a}\left(\tfrac{\pi}{3}\right) a ( 3 Ο β )
a β¨ 1 2 , Β 3 2 β© \left\langle \tfrac{1}{2},\ \tfrac{\sqrt{3}}{2} \right\rangle β¨ 2 1 β , Β 2 3 β β β© b β¨ β 3 2 , Β 1 2 β© \left\langle -\tfrac{\sqrt{3}}{2},\ \tfrac{1}{2} \right\rangle β¨ β 2 3 β β , Β 2 1 β β© c β¨ 3 2 , Β 1 2 β© \left\langle \tfrac{\sqrt{3}}{2},\ \tfrac{1}{2} \right\rangle β¨ 2 3 β β , Β 2 1 β β© d β¨ 3 2 , Β β 1 2 β© \left\langle \tfrac{\sqrt{3}}{2},\ -\tfrac{1}{2} \right\rangle β¨ 2 3 β β , Β β 2 1 β β©
β
Correct! a β = β¨ sin β‘ t , cos β‘ t β© \vec{a} = \langle \sin t, \cos t \rangle a = β¨ sin t , cos t β© , and at
t = Ο 3 t = \tfrac{\pi}{3} t = 3 Ο β that is
β¨ 3 2 , 1 2 β© \left\langle \tfrac{\sqrt{3}}{2}, \tfrac{1}{2} \right\rangle β¨ 2 3 β β , 2 1 β β© .
β That is the velocity. β¨ 1 β cos β‘ t , sin β‘ t β© \langle 1 - \cos t, \sin t \rangle β¨ 1 β cos t , sin t β© at
t = Ο 3 t = \tfrac{\pi}{3} t = 3 Ο β is
v β \vec{v} v ; the acceleration needs one more derivative.
β Check the first component's sign. d d t ( 1 β cos β‘ t ) = + sin β‘ t \frac{d}{dt}(1 - \cos t) = +\sin t d t d β ( 1 β cos t ) = + sin t , because the derivative of
β cos β‘ t -\cos t β cos t is
+ sin β‘ t +\sin t + sin t .
β Check the second component's sign. d d t sin β‘ t = + cos β‘ t \frac{d}{dt}\sin t = +\cos t d t d β sin t = + cos t , and
cos β‘ Ο 3 = + 1 2 \cos\tfrac{\pi}{3} = +\tfrac{1}{2} cos 3 Ο β = + 2 1 β .
β Not quite. Differentiate
v β = β¨ 1 β cos β‘ t , sin β‘ t β© \vec{v} = \langle 1 - \cos t, \sin t \rangle v = β¨ 1 β cos t , sin t β© component by component.
Show solution
First derivative (the velocity), component by component:
v β ( t ) = β¨ 1 β cos β‘ t , Β sin β‘ t β© \vec{v}(t) = \langle 1 - \cos t,\ \sin t \rangle v ( t ) = β¨ 1 β cos t , Β sin t β© Second derivative (the acceleration):
a β ( t ) = β¨ d d t ( 1 β cos β‘ t ) , Β d d t sin β‘ t β© = β¨ sin β‘ t , Β cos β‘ t β© \vec{a}(t) = \left\langle \tfrac{d}{dt}(1 - \cos t),\ \tfrac{d}{dt}\sin t \right\rangle = \langle \sin t,\ \cos t \rangle a ( t ) = β¨ d t d β ( 1 β cos t ) , Β d t d β sin t β© = β¨ sin t , Β cos t β© At t = Ο 3 t = \tfrac{\pi}{3} t = 3 Ο β
a β ( Ο 3 ) = β¨ sin β‘ Ο 3 , Β cos β‘ Ο 3 β© = β¨ 3 2 , Β 1 2 β© \vec{a}\left(\tfrac{\pi}{3}\right) = \left\langle \sin\tfrac{\pi}{3},\ \cos\tfrac{\pi}{3} \right\rangle = \left\langle \tfrac{\sqrt{3}}{2},\ \tfrac{1}{2} \right\rangle a ( 3 Ο β ) = β¨ sin 3 Ο β , Β cos 3 Ο β β© = β¨ 2 3 β β , Β 2 1 β β© Its length is 3 4 + 1 4 = 1 \sqrt{\tfrac{3}{4} + \tfrac{1}{4}} = 1 4 3 β + 4 1 β β = 1 t t t
Problem 2 Β· Stopped but Accelerating
Given: the same cycloid, at t = 2 Ο t = 2\pi t = 2 Ο find a β ( 2 Ο ) \vec{a}(2\pi) a ( 2 Ο )
a β¨ 0 , Β 0 β© \langle 0,\ 0 \rangle β¨ 0 , Β 0 β© b β¨ 0 , Β β 1 β© \langle 0,\ -1 \rangle β¨ 0 , Β β 1 β© c β¨ 1 , Β 0 β© \langle 1,\ 0 \rangle β¨ 1 , Β 0 β© d β¨ 0 , Β 1 β© \langle 0,\ 1 \rangle β¨ 0 , Β 1 β©
β
Correct! β¨ sin β‘ 2 Ο , cos β‘ 2 Ο β© = β¨ 0 , 1 β© \langle \sin 2\pi, \cos 2\pi \rangle = \langle 0, 1 \rangle β¨ sin 2 Ο , cos 2 Ο β© = β¨ 0 , 1 β© : the point is stopped, yet it accelerates straight up.
β That is the velocity, not the acceleration. v β ( 2 Ο ) = 0 β \vec{v}(2\pi) = \vec{0} v ( 2 Ο ) = 0 , but the velocity is changing from downward to upward, so its derivative is not zero.
β Check the direction. Just before
t = 2 Ο t = 2\pi t = 2 Ο the velocity points down, just after it points up; the change
Ξ v β \Delta\vec{v} Ξ v points up.
β Not quite. Evaluate
a β ( t ) = β¨ sin β‘ t , cos β‘ t β© \vec{a}(t) = \langle \sin t, \cos t \rangle a ( t ) = β¨ sin t , cos t β© at
t = 2 Ο t = 2\pi t = 2 Ο .
Show solution
The velocity at the cusp is zero:
v β ( 2 Ο ) = β¨ 1 β cos β‘ 2 Ο , Β sin β‘ 2 Ο β© = β¨ 0 , Β 0 β© \vec{v}(2\pi) = \langle 1 - \cos 2\pi,\ \sin 2\pi \rangle = \langle 0,\ 0 \rangle v ( 2 Ο ) = β¨ 1 β cos 2 Ο , Β sin 2 Ο β© = β¨ 0 , Β 0 β© The acceleration is not:
a β ( 2 Ο ) = β¨ sin β‘ 2 Ο , Β cos β‘ 2 Ο β© = β¨ 0 , Β 1 β© \vec{a}(2\pi) = \langle \sin 2\pi,\ \cos 2\pi \rangle = \langle 0,\ 1 \rangle a ( 2 Ο ) = β¨ sin 2 Ο , Β cos 2 Ο β© = β¨ 0 , Β 1 β© For t t t 2 Ο 2\pi 2 Ο v β β β¨ 0 , sin β‘ t β© \vec{v} \approx \langle 0, \sin t \rangle v β β¨ 0 , sin t β© sin β‘ t < 0 \sin t \lt 0 sin t < 0 t t t sin β‘ t > 0 \sin t \gt 0 sin t > 0 0 β \vec{0} 0 a β \vec{a} a
Problem 3 Β· Speed vs. d β£ r β β£ / d t d|\vec{r}|/dt d β£ r β£/ d t
Given: r β ( t ) = β¨ t , Β 3 β© \vec{r}(t) = \langle t,\ 3 \rangle r ( t ) = β¨ t , Β 3 β© y = 3 y = 3 y = 3 find the speed and d β£ r β β£ d t \dfrac{d|\vec{r}|}{dt} d t d β£ r β£ β t = 4 t = 4 t = 4
What is the speed β£ v β ( 4 ) β£ |\vec{v}(4)| β£ v ( 4 ) β£
a 5 5 5 b 1 1 1 c 4 5 \tfrac{4}{5} 5 4 β d 4 4 4
What is d β£ r β β£ d t \dfrac{d|\vec{r}|}{dt} d t d β£ r β£ β t = 4 t = 4 t = 4
a 1 1 1 b 3 5 \tfrac{3}{5} 5 3 β c 4 5 \tfrac{4}{5} 5 4 β d 4 4 4
β
Correct! Speed
1 1 1 , but the distance from the origin grows only at rate
4 5 \tfrac{4}{5} 5 4 β : part of the velocity is perpendicular to
r β \vec{r} r .
β That is d β£ r β β£ / d t d|\vec{r}|/dt d β£ r β£/ d t The speed is the length of
v β = β¨ 1 , 0 β© \vec{v} = \langle 1, 0 \rangle v = β¨ 1 , 0 β© .
β That is the speed. d β£ r β β£ / d t d|\vec{r}|/dt d β£ r β£/ d t counts only the part of
v β \vec{v} v along
r β \vec{r} r :
r β β
v β / β£ r β β£ \vec{r} \cdot \vec{v}/|\vec{r}| r β
v /β£ r β£ .
β Check the speed. Differentiate
r β \vec{r} r first:
v β = β¨ 1 , 0 β© \vec{v} = \langle 1, 0 \rangle v = β¨ 1 , 0 β© , then take its length.
β Check d β£ r β β£ / d t d|\vec{r}|/dt d β£ r β£/ d t Use
x β x β² + y β y β² x 2 + y 2 \dfrac{x\,x' + y\,y'}{\sqrt{x^2 + y^2}} x 2 + y 2 β x x β² + y y β² β with
x = 4 , Β y = 3 , Β x β² = 1 , Β y β² = 0 x = 4,\ y = 3,\ x' = 1,\ y' = 0 x = 4 , Β y = 3 , Β x β² = 1 , Β y β² = 0 .
Show solution
Speed: v β ( t ) = β¨ 1 , 0 β© \vec{v}(t) = \langle 1, 0 \rangle v ( t ) = β¨ 1 , 0 β© β£ v β β£ = 1 |\vec{v}| = 1 β£ v β£ = 1 t t t
Derivative of the length: at t = 4 t = 4 t = 4 r β = β¨ 4 , 3 β© \vec{r} = \langle 4, 3 \rangle r = β¨ 4 , 3 β© β£ r β β£ = 16 + 9 = 5 |\vec{r}| = \sqrt{16 + 9} = 5 β£ r β£ = 16 + 9 β = 5
d β£ r β β£ d t = r β β
v β β£ r β β£ = 4 β
1 + 3 β
0 5 = 4 5 \frac{d|\vec{r}|}{dt} = \frac{\vec{r} \cdot \vec{v}}{|\vec{r}|} = \frac{4 \cdot 1 + 3 \cdot 0}{5} = \frac{4}{5} d t d β£ r β£ β = β£ r β£ r β
v β = 5 4 β
1 + 3 β
0 β = 5 4 β Check directly: β£ r β β£ = t 2 + 9 |\vec{r}| = \sqrt{t^2 + 9} β£ r β£ = t 2 + 9 β d β£ r β β£ d t = t t 2 + 9 = 4 5 \dfrac{d|\vec{r}|}{dt} = \dfrac{t}{\sqrt{t^2 + 9}} = \dfrac{4}{5} d t d β£ r β£ β = t 2 + 9 β t β = 5 4 β t = 4 t = 4 t = 4
The two differ because v β \vec{v} v r β \vec{r} r 4 5 = cos β‘ ΞΈ \tfrac{4}{5} = \cos\theta 5 4 β = cos ΞΈ ΞΈ \theta ΞΈ
Problem 4 Β· A Faster, Wider Turn
Given: r β ( t ) = β¨ 2 cos β‘ 3 t , Β 2 sin β‘ 3 t β© \vec{r}(t) = \langle 2\cos 3t,\ 2\sin 3t \rangle r ( t ) = β¨ 2 cos 3 t , Β 2 sin 3 t β© 2 2 2 find the magnitude of its acceleration β£ a β β£ |\vec{a}| β£ a β£
β
Correct! a β = β¨ β 18 cos β‘ 3 t , β 18 sin β‘ 3 t β© = β 9 β r β \vec{a} = \langle -18\cos 3t, -18\sin 3t \rangle = -9\,\vec{r} a = β¨ β 18 cos 3 t , β 18 sin 3 t β© = β 9 r , of length
18 18 18 , pointing to the center.
β Constant speed is not zero acceleration. The velocity keeps turning, so
d v β / d t β 0 β d\vec{v}/dt \neq \vec{0} d v / d t ξ = 0 , exactly as on the unit circle.
β That is the speed. β£ v β β£ = β£ β¨ β 6 sin β‘ 3 t , 6 cos β‘ 3 t β© β£ = 6 |\vec{v}| = |\langle -6\sin 3t, 6\cos 3t \rangle| = 6 β£ v β£ = β£ β¨ β 6 sin 3 t , 6 cos 3 t β© β£ = 6 ; differentiate once more.
β The radius is missing. The chain rule gives the factor
3 3 3 twice, but the
2 2 2 in front stays:
2 β
3 β
3 2 \cdot 3 \cdot 3 2 β
3 β
3 .
β Not quite. Differentiate twice, using the chain rule on
3 t 3t 3 t each time.
Show solution
v β ( t ) = β¨ β 6 sin β‘ 3 t , Β 6 cos β‘ 3 t β© , β£ v β β£ = 6 \vec{v}(t) = \langle -6\sin 3t,\ 6\cos 3t \rangle, \qquad |\vec{v}| = 6 v ( t ) = β¨ β 6 sin 3 t , Β 6 cos 3 t β© , β£ v β£ = 6 a β ( t ) = β¨ β 18 cos β‘ 3 t , Β β 18 sin β‘ 3 t β© = β 9 β r β ( t ) , β£ a β β£ = 18 \vec{a}(t) = \langle -18\cos 3t,\ -18\sin 3t \rangle = -9\,\vec{r}(t), \qquad |\vec{a}| = 18 a ( t ) = β¨ β 18 cos 3 t , Β β 18 sin 3 t β© = β 9 r ( t ) , β£ a β£ = 18 Also β£ r β β£ = 2 |\vec{r}| = 2 β£ r β£ = 2 d β£ r β β£ d t = 0 \dfrac{d|\vec{r}|}{dt} = 0 d t d β£ r β£ β = 0 6 6 6 v β \vec{v} v a β β
v β = 108 cos β‘ 3 t sin β‘ 3 t β 108 sin β‘ 3 t cos β‘ 3 t = 0 \vec{a} \cdot \vec{v} = 108\cos 3t\sin 3t - 108\sin 3t\cos 3t = 0 a β
v = 108 cos 3 t sin 3 t β 108 sin 3 t cos 3 t = 0
Solved: 0 / 4