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Show that if v . v' = 0 (both vectors) then speed v is constant 
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#1
Mar1612, 12:20 AM

P: 25

1. The problem statement, all variables and given/known data
Show that for a particle moving with velocity v(t), if v . v'=0 then the speed v is constant. 2. Relevant equations 3. The attempt at a solution Let v = (v_{1},v_{2},...,v_{n}) and let v'= (v'_{1},v'_{2},...,v'_{n}) So, v . v'= v_{1}v'_{1} + v_{2}v'_{2} + ... + v_{n}v'_{n} = 0 This is the only way i can think of to go about it but i'm stuck here. What do i do to show v is constant when v.v'=0? 


#2
Mar1612, 12:28 AM

P: 42




#3
Mar1612, 12:31 AM

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Suppose you've got a particle with mass=1.
What would the derivative of its kinetic energy be? 


#4
Mar1612, 12:31 AM

P: 97

Show that if v . v' = 0 (both vectors) then speed v is constant
since v dot v'=0 then you have to consider two cases
when v=the zero vector or when v'=the zero vector 


#5
Mar1612, 12:48 AM

P: 25

when m =1 dT/dt= v.dv/dt i can simplify furthur, but why? 


#6
Mar1612, 12:54 AM

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So, you know that T'=v.v'
Agreed? 


#7
Mar1612, 01:00 AM

P: 42

either v=0 or v'=0. v=o means constant vel. and dv/dt=0 means vel. is not changing w.r.t. time. thus vel. is constt.



#8
Mar1612, 01:12 AM

P: 25




#9
Mar1612, 02:06 AM

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You can perfectly well have nonconstant vectors whose dot product is always equal to zero. The critical issue is that when those two vectors are related as a function and its derivative, THEN, we may prove that the SPEED (not the velocity!!), must be constant. 


#10
Mar1612, 02:08 AM

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Agreed? Furthermore, if T is constant, what can you say about whether the speed is constant or not? 


#11
Mar1612, 05:04 AM

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#12
Mar1612, 05:35 AM

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If 1/2*v^2 =constant, (where v^2 is the dot product of the velocity by itself, equalling the squared speed), then it follows the speed must be constant. Suppose that you move, with uniform speed along a circle. Neither your velocity or accelerations are constants, but the acceleration vector is always orthogonal to the velocity vector. the velocity vector in this case is tangential to the circle (changing all the time in its direction), while the acceleration vector is strictly centripetal (changing all the time in its direction). nonuniform speed along the circle would make the acceleration somewhat tangentially orientated as well, i.e, its dot product with the velocity would be..nonzero 


#13
Mar1712, 06:33 PM

P: 25

alright i understand the theory behind it. But assume we didn't know anything about how this question is related to kinetic energy, how would i begin to solve it?
Since it was a question in my math class and you are not expected to know kinetic energy. 


#14
Mar1712, 07:18 PM

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You don't need kinetic energy. There's nothing special about velocity here; I would take that v to mean any arbitrary vector quantity that satisfies [itex]\mathbf v \cdot \mathbf v ' = 0[/itex]. All you just need to do is to look at the derivative of the inner product of this vector v with itself.



#15
Mar1812, 08:36 AM

P: 22

Deleted sorry didn't read the rules:
Also, please DO NOT do someone's homework for them or post complete solutions to problems. Please give all the help you can, but DO NOT simply do the problem yourself and post the solution (at least not until the original poster has tried his/her very best). 


#16
Mar1812, 10:27 AM

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To solve the problem at hand, it helps to first prove that the time derivative of a unit vector is either zero or is orthogonal to the unit vector. 


#17
Mar1812, 10:31 AM

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P: 15,204

My second opinion is that your proof is incorrect. 


#18
Mar1812, 10:41 AM

P: 22




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