The Point on a Rolling Wheel Stops Dead: Velocity and Speed on the Cycloid
I Key FormulasKey Insight: At the bottom of the wheel the rotation carries the point backward at unit speed, β¨ β 1 , 0 β© \langle -1, 0 \rangle β¨ β 1 , 0 β© v β ( 0 ) = 0 β \vec{v}(0) = \vec{0} v ( 0 ) = 0 v β ( Ο ) = β¨ 2 , 0 β© \vec{v}(\pi) = \langle 2, 0 \rangle v ( Ο ) = β¨ 2 , 0 β©
II Velocity as a Limit of Ξr/ΞtAt t = Ο / 2 t = \pi/2 t = Ο /2 Ξ r β / Ξ t \Delta\vec{r}/\Delta t Ξ r /Ξ t v β ( Ο / 2 ) = β¨ 1 , 1 β© \vec{v}(\pi/2) = \langle 1, 1 \rangle v ( Ο /2 ) = β¨ 1 , 1 β© Ξ t β 0 \Delta t \to 0 Ξ t β 0
III Rolling Plus RotationThe velocity is the wheel's forward roll β¨ 1 , 0 β© \langle 1, 0 \rangle β¨ 1 , 0 β© β¨ β cos β‘ t , sin β‘ t β© \langle -\cos t, \sin t \rangle β¨ β cos t , sin t β©
π‘ The rotation part β¨ β cos β‘ t , sin β‘ t β© \langle -\cos t, \sin t \rangle β¨ β cos t , sin t β© 1 1 1
IV Speed Along the ArchThe length of the velocity arrow on the arch is the height of the speed graph β£ v β β£ = 2 β 2 cos β‘ t |\vec{v}| = \sqrt{2 - 2\cos t} β£ v β£ = 2 β 2 cos t β
V Quiz Questions
Problem 1 Β· Differentiate the Position Vector
Given: the cycloid r β ( t ) = β¨ t β sin β‘ t , β
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 , 1 β cos t β© find the velocity v β ( Ο / 3 ) \vec{v}(\pi/3) v ( Ο /3 )
a β¨ 1 2 , β
β 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 1 2 , β
β 3 2 β© \left\langle -\tfrac{1}{2},\; \tfrac{\sqrt{3}}{2} \right\rangle β¨ β 2 1 β , 2 3 β β β© c β¨ 1 2 , β
β 3 2 β© \left\langle \tfrac{1}{2},\; \tfrac{\sqrt{3}}{2} \right\rangle β¨ 2 1 β , 2 3 β β β© d β¨ Ο 3 β 3 2 , β
β 1 2 β© \left\langle \tfrac{\pi}{3} - \tfrac{\sqrt{3}}{2},\; \tfrac{1}{2} \right\rangle β¨ 3 Ο β β 2 3 β β , 2 1 β β©
β
Correct! v β ( t ) = β¨ 1 β cos β‘ t , sin β‘ t β© \vec{v}(t) = \langle 1 - \cos t, \sin t \rangle v ( t ) = β¨ 1 β cos t , sin t β© , and
cos β‘ Ο 3 = 1 2 \cos\frac{\pi}{3} = \frac{1}{2} cos 3 Ο β = 2 1 β ,
sin β‘ Ο 3 = 3 2 \sin\frac{\pi}{3} = \frac{\sqrt{3}}{2} sin 3 Ο β = 2 3 β β : the point is climbing the rising side of the arch.
β Check the sign. The derivative of
β cos β‘ t -\cos t β cos t is
+ sin β‘ t +\sin t + sin t , not
β sin β‘ t -\sin t β sin t .
β Not quite. You kept only the rotation part
β¨ β cos β‘ t , sin β‘ t β© \langle -\cos t, \sin t \rangle β¨ β cos t , sin t β© : the derivative of
t t t contributes the rolling part
β¨ 1 , 0 β© \langle 1, 0 \rangle β¨ 1 , 0 β© .
β Not quite. That is the position
r β ( Ο / 3 ) \vec{r}(\pi/3) r ( Ο /3 ) ; the velocity is its derivative
d r β d t \frac{d\vec{r}}{dt} d t d r β .
β Not quite. Differentiate each component, then substitute
t = Ο / 3 t = \pi/3 t = Ο /3 .
Show solution
Differentiate one component at a time:
d d t ( t β sin β‘ t ) = 1 β cos β‘ t , d d t ( 1 β cos β‘ t ) = 0 + sin β‘ t \frac{d}{dt}(t - \sin t) = 1 - \cos t, \qquad \frac{d}{dt}(1 - \cos t) = 0 + \sin t d t d β ( t β sin t ) = 1 β cos t , d t d β ( 1 β cos t ) = 0 + sin t so v β ( t ) = β¨ 1 β cos β‘ t , β
β sin β‘ t β© \vec{v}(t) = \langle 1 - \cos t,\; \sin t \rangle v ( t ) = β¨ 1 β cos t , sin t β© t = Ο / 3 t = \pi/3 t = Ο /3
v β ( Ο / 3 ) = β¨ 1 β 1 2 , β
β 3 2 β© = β¨ 1 2 , β
β 3 2 β© \vec{v}(\pi/3) = \left\langle 1 - \tfrac{1}{2},\; \tfrac{\sqrt{3}}{2} \right\rangle = \left\langle \tfrac{1}{2},\; \tfrac{\sqrt{3}}{2} \right\rangle v ( Ο /3 ) = β¨ 1 β 2 1 β , 2 3 β β β© = β¨ 2 1 β , 2 3 β β β©
Problem 2 Β· Speed Is the Length, Not the Sum
Given: the cycloid velocity v β ( t ) = β¨ 1 β cos β‘ t , β
β sin β‘ t β© \vec{v}(t) = \langle 1 - \cos t,\; \sin t \rangle v ( t ) = β¨ 1 β cos t , sin t β© find the speed β£ v β β£ |\vec{v}| β£ v β£ t = 2 Ο / 3 t = 2\pi/3 t = 2 Ο /3
a β£ v β β£ = 1 |\vec{v}| = 1 β£ v β£ = 1 b β£ v β β£ = 3 + 3 2 |\vec{v}| = \tfrac{3 + \sqrt{3}}{2} β£ v β£ = 2 3 + 3 β β c β£ v β β£ = 3 |\vec{v}| = 3 β£ v β£ = 3 d β£ v β β£ = 3 |\vec{v}| = \sqrt{3} β£ v β£ = 3 β
β
Correct! cos β‘ 2 Ο 3 = β 1 2 \cos\frac{2\pi}{3} = -\frac{1}{2} cos 3 2 Ο β = β 2 1 β , so
β£ v β β£ = 2 + 1 = 3 β 1.73 |\vec{v}| = \sqrt{2 + 1} = \sqrt{3} \approx 1.73 β£ v β£ = 2 + 1 β = 3 β β 1.73 : still climbing toward the top of the arch, the point already moves faster than the wheel.
β Check the cosine. cos β‘ 2 Ο 3 = β 1 2 \cos\frac{2\pi}{3} = -\frac{1}{2} cos 3 2 Ο β = β 2 1 β , not
+ 1 2 +\frac{1}{2} + 2 1 β , so
2 β 2 cos β‘ t = 2 + 1 2 - 2\cos t = 2 + 1 2 β 2 cos t = 2 + 1 .
β Not quite. You added the components
3 2 + 3 2 \frac{3}{2} + \frac{\sqrt{3}}{2} 2 3 β + 2 3 β β . The speed is the length of the vector: square, add, then take the square root.
β Almost. 2 β 2 cos β‘ t = 3 2 - 2\cos t = 3 2 β 2 cos t = 3 is the squared length; take the square root.
β Not quite. Use
β£ v β β£ = 2 β 2 cos β‘ t |\vec{v}| = \sqrt{2 - 2\cos t} β£ v β£ = 2 β 2 cos t β .
Show solution
At t = 2 Ο / 3 t = 2\pi/3 t = 2 Ο /3 cos β‘ t = β 1 2 \cos t = -\tfrac{1}{2} cos t = β 2 1 β sin β‘ t = 3 2 \sin t = \tfrac{\sqrt{3}}{2} sin t = 2 3 β β v β = β¨ 3 2 , β
β 3 2 β© \vec{v} = \left\langle \tfrac{3}{2},\; \tfrac{\sqrt{3}}{2} \right\rangle v = β¨ 2 3 β , 2 3 β β β©
β£ v β β£ = ( 3 2 ) 2 + ( 3 2 ) 2 = 9 4 + 3 4 = 3 |\vec{v}| = \sqrt{\left(\tfrac{3}{2}\right)^2 + \left(\tfrac{\sqrt{3}}{2}\right)^2} = \sqrt{\tfrac{9}{4} + \tfrac{3}{4}} = \sqrt{3} β£ v β£ = ( 2 3 β ) 2 + ( 2 3 β β ) 2 β = 4 9 β + 4 3 β β = 3 β The shortcut agrees: 2 β 2 cos β‘ t = 2 β 2 ( β 1 2 ) = 3 \sqrt{2 - 2\cos t} = \sqrt{2 - 2\left(-\tfrac{1}{2}\right)} = \sqrt{3} 2 β 2 cos t β = 2 β 2 ( β 2 1 β ) β = 3 β
Problem 3 Β· Matching the Wheel's Speed
Given: the unit cycloid, whose wheel rolls at speed 1 1 1 find the times 0 < t < 2 Ο 0 < t < 2\pi 0 < t < 2 Ο 1 1 1
When is the speed equal to 1?
a t = Ο 2 t = \tfrac{\pi}{2} t = 2 Ο β t = 3 Ο 2 t = \tfrac{3\pi}{2} t = 2 3 Ο β b t = Ο 3 t = \tfrac{\pi}{3} t = 3 Ο β t = 5 Ο 3 t = \tfrac{5\pi}{3} t = 3 5 Ο β c t = 2 Ο 3 t = \tfrac{2\pi}{3} t = 3 2 Ο β t = 4 Ο 3 t = \tfrac{4\pi}{3} t = 3 4 Ο β d t = Ο 6 t = \tfrac{\pi}{6} t = 6 Ο β t = 5 Ο 6 t = \tfrac{5\pi}{6} t = 6 5 Ο β
What is the velocity at the later time?
a β¨ 1 2 , β
β 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¨ 1 2 , β
β 3 2 β© \left\langle \tfrac{1}{2},\; \tfrac{\sqrt{3}}{2} \right\rangle β¨ 2 1 β , 2 3 β β β© c β¨ 3 2 , β
β β 3 2 β© \left\langle \tfrac{3}{2},\; -\tfrac{\sqrt{3}}{2} \right\rangle β¨ 2 3 β , β 2 3 β β β© d β¨ β 1 2 , β
β β 3 2 β© \left\langle -\tfrac{1}{2},\; -\tfrac{\sqrt{3}}{2} \right\rangle β¨ β 2 1 β , β 2 3 β β β©
β
Correct! Speed
1 1 1 needs
cos β‘ t = 1 2 \cos t = \tfrac{1}{2} cos t = 2 1 β ; at
t = 5 Ο 3 t = \tfrac{5\pi}{3} t = 3 5 Ο β the point is dropping into the next cusp, moving right and steeply down.
β Not quite. At
t = Ο / 2 t = \pi/2 t = Ο /2 the speed is
2 β 0 = 2 \sqrt{2 - 0} = \sqrt{2} 2 β 0 β = 2 β . Set the whole speed
2 β 2 cos β‘ t \sqrt{2 - 2\cos t} 2 β 2 cos t β equal to
1 1 1 , not just the first component.
β Check the sign. 2 β 2 cos β‘ t = 1 2 - 2\cos t = 1 2 β 2 cos t = 1 gives
cos β‘ t = + 1 2 \cos t = +\tfrac{1}{2} cos t = + 2 1 β ; at
t = 2 Ο / 3 t = 2\pi/3 t = 2 Ο /3 the speed is
3 \sqrt{3} 3 β .
β Not quite. Those solve
sin β‘ t = 1 2 \sin t = \tfrac{1}{2} sin t = 2 1 β ; the condition is on
cos β‘ t \cos t cos t .
β Check the times. Square both sides of
2 β 2 cos β‘ t = 1 \sqrt{2 - 2\cos t} = 1 2 β 2 cos t β = 1 and solve for
cos β‘ t \cos t cos t .
β Check the velocity. At
t = 5 Ο / 3 t = 5\pi/3 t = 5 Ο /3 :
cos β‘ t = 1 2 \cos t = \tfrac{1}{2} cos t = 2 1 β and
sin β‘ t = β 3 2 \sin t = -\tfrac{\sqrt{3}}{2} sin t = β 2 3 β β ; substitute into
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© .
Show solution
Step 1: set the speed equal to 1 1 1
2 β 2 cos β‘ t = 1 β
β β β
β 2 β 2 cos β‘ t = 1 β
β β β
β cos β‘ t = 1 2 \sqrt{2 - 2\cos t} = 1 \;\Rightarrow\; 2 - 2\cos t = 1 \;\Rightarrow\; \cos t = \tfrac{1}{2} 2 β 2 cos t β = 1 β 2 β 2 cos t = 1 β cos t = 2 1 β On 0 < t < 2 Ο 0 < t < 2\pi 0 < t < 2 Ο t = Ο / 3 t = \pi/3 t = Ο /3 t = 5 Ο / 3 t = 5\pi/3 t = 5 Ο /3
Step 2: at t = 5 Ο / 3 t = 5\pi/3 t = 5 Ο /3 sin β‘ t = β 3 2 \sin t = -\tfrac{\sqrt{3}}{2} sin t = β 2 3 β β
v β ( 5 Ο / 3 ) = β¨ 1 β 1 2 , β
β β 3 2 β© = β¨ 1 2 , β
β β 3 2 β© \vec{v}(5\pi/3) = \left\langle 1 - \tfrac{1}{2},\; -\tfrac{\sqrt{3}}{2} \right\rangle = \left\langle \tfrac{1}{2},\; -\tfrac{\sqrt{3}}{2} \right\rangle v ( 5 Ο /3 ) = β¨ 1 β 2 1 β , β 2 3 β β β© = β¨ 2 1 β , β 2 3 β β β© Its length is 1 4 + 3 4 = 1 \sqrt{\tfrac{1}{4} + \tfrac{3}{4}} = 1 4 1 β + 4 3 β β = 1
Problem 4 Β· A Point Inside the Rim
Given: the same wheel (radius 1 1 1 1 2 \tfrac{1}{2} 2 1 β r β ( t ) = β¨ t β 1 2 sin β‘ t , β
β 1 β 1 2 cos β‘ t β© \vec{r}(t) = \left\langle t - \tfrac{1}{2}\sin t,\; 1 - \tfrac{1}{2}\cos t \right\rangle r ( t ) = β¨ t β 2 1 β sin t , 1 β 2 1 β cos t β© find its speed at its lowest point (t = 0 t = 0 t = 0 t = Ο t = \pi t = Ο
Speed at t = 0?
a β£ v β ( 0 ) β£ = 1 2 |\vec{v}(0)| = \tfrac{1}{2} β£ v ( 0 ) β£ = 2 1 β b β£ v β ( 0 ) β£ = 0 |\vec{v}(0)| = 0 β£ v ( 0 ) β£ = 0 c β£ v β ( 0 ) β£ = 1 |\vec{v}(0)| = 1 β£ v ( 0 ) β£ = 1 d β£ v β ( 0 ) β£ = 3 2 |\vec{v}(0)| = \tfrac{3}{2} β£ v ( 0 ) β£ = 2 3 β
Speed at t = Ο?
a β£ v β ( Ο ) β£ = 2 |\vec{v}(\pi)| = 2 β£ v ( Ο ) β£ = 2 b β£ v β ( Ο ) β£ = 1 |\vec{v}(\pi)| = 1 β£ v ( Ο ) β£ = 1 c β£ v β ( Ο ) β£ = 3 2 |\vec{v}(\pi)| = \tfrac{3}{2} β£ v ( Ο ) β£ = 2 3 β d β£ v β ( Ο ) β£ = 1 2 |\vec{v}(\pi)| = \tfrac{1}{2} β£ v ( Ο ) β£ = 2 1 β
β
Correct! The rotation part now has length
1 2 \tfrac{1}{2} 2 1 β , so it can only halve the forward roll at the bottom and add half at the top: a point inside the rim never stops.
β Not quite. Zero speed at the bottom belongs to the rim point, where the rotation part has length
1 1 1 . Here it has length
1 2 \tfrac{1}{2} 2 1 β and cannot cancel the roll.
β Check the bottom. v β ( t ) = β¨ 1 β 1 2 cos β‘ t , β
β 1 2 sin β‘ t β© \vec{v}(t) = \left\langle 1 - \tfrac{1}{2}\cos t,\; \tfrac{1}{2}\sin t \right\rangle v ( t ) = β¨ 1 β 2 1 β cos t , 2 1 β sin t β© ; substitute
t = 0 t = 0 t = 0 .
β Check the top. At
t = Ο t = \pi t = Ο ,
cos β‘ t = β 1 \cos t = -1 cos t = β 1 : the roll
β¨ 1 , 0 β© \langle 1, 0 \rangle β¨ 1 , 0 β© and the rotation
β¨ 1 2 , 0 β© \left\langle \tfrac{1}{2}, 0 \right\rangle β¨ 2 1 β , 0 β© add.
Show solution
Differentiate component by component:
v β ( t ) = β¨ 1 β 1 2 cos β‘ t , β
β 1 2 sin β‘ t β© = β¨ 1 , 0 β© + β¨ β 1 2 cos β‘ t , β
β 1 2 sin β‘ t β© \vec{v}(t) = \left\langle 1 - \tfrac{1}{2}\cos t,\; \tfrac{1}{2}\sin t \right\rangle = \langle 1, 0 \rangle + \left\langle -\tfrac{1}{2}\cos t,\; \tfrac{1}{2}\sin t \right\rangle v ( t ) = β¨ 1 β 2 1 β cos t , 2 1 β sin t β© = β¨ 1 , 0 β© + β¨ β 2 1 β cos t , 2 1 β sin t β© At t = 0 t = 0 t = 0 v β ( 0 ) = β¨ 1 β 1 2 , β
β 0 β© = β¨ 1 2 , 0 β© \vec{v}(0) = \left\langle 1 - \tfrac{1}{2},\; 0 \right\rangle = \left\langle \tfrac{1}{2}, 0 \right\rangle v ( 0 ) = β¨ 1 β 2 1 β , 0 β© = β¨ 2 1 β , 0 β© 1 2 \tfrac{1}{2} 2 1 β
At t = Ο t = \pi t = Ο v β ( Ο ) = β¨ 1 + 1 2 , β
β 0 β© = β¨ 3 2 , 0 β© \vec{v}(\pi) = \left\langle 1 + \tfrac{1}{2},\; 0 \right\rangle = \left\langle \tfrac{3}{2}, 0 \right\rangle v ( Ο ) = β¨ 1 + 2 1 β , 0 β© = β¨ 2 3 β , 0 β© 3 2 \tfrac{3}{2} 2 3 β
In general β£ v β β£ = 5 4 β cos β‘ t |\vec{v}| = \sqrt{\tfrac{5}{4} - \cos t} β£ v β£ = 4 5 β β cos t β 1 2 \tfrac{1}{2} 2 1 β 3 2 \tfrac{3}{2} 2 3 β
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