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Show that if v . v' = 0 (both vectors) then speed v is constant

by Keshroom
Tags: constant, speed, vectors
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Keshroom
#1
Mar16-12, 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 = (v1,v2,...,vn)
and let v'= (v'1,v'2,...,v'n)

So,

v . v'= v1v'1 + v2v'2 + ... + vnv'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?
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altamashghazi
#2
Mar16-12, 12:28 AM
P: 42
Quote Quote by Keshroom View Post
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 = (v1,v2,...,vn)
and let v'= (v'1,v'2,...,v'n)

So,

v . v'= v1v'1 + v2v'2 + ... + vnv'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?
what do u mean by v' and v
arildno
#3
Mar16-12, 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?

miglo
#4
Mar16-12, 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
Keshroom
#5
Mar16-12, 12:48 AM
P: 25
Quote Quote by arildno View Post
Suppose you've got a particle with mass=1.
What would the derivative of its kinetic energy be?
dT/dt = mv.dv/dt

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

Quote Quote by altamashghazi View Post
what do u mean by v' and v
v' is the derivative of v ( velocity )
arildno
#6
Mar16-12, 12:54 AM
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So, you know that T'=v.v'
Agreed?
altamashghazi
#7
Mar16-12, 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.
Keshroom
#8
Mar16-12, 01:12 AM
P: 25
Quote Quote by arildno View Post
So, you know that T'=v.v'
Agreed?
yep agreed
arildno
#9
Mar16-12, 02:06 AM
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Quote Quote by altamashghazi View Post
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.
This is incorrectly argued.
You can perfectly well have non-constant 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.
arildno
#10
Mar16-12, 02:08 AM
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Quote Quote by Keshroom View Post
yep agreed
But, it then follows that since T'=0 in your case, then it must also be true that T=constant.
Agreed?

Furthermore, if T is constant, what can you say about whether the speed is constant or not?
Keshroom
#11
Mar16-12, 05:04 AM
P: 25
Quote Quote by arildno View Post
But, it then follows that since T'=0 in your case, then it must also be true that T=constant.
Agreed?

Furthermore, if T is constant, what can you say about whether the speed is constant or not?
ok, so if T' is zero, that means acceleration =0 so therefore the speed will be constant. right?
arildno
#12
Mar16-12, 05:35 AM
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Quote Quote by Keshroom View Post
ok, so if T' is zero, that means acceleration =0 so therefore the speed will be constant. right?
No.

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). non-uniform speed along the circle would make the acceleration somewhat tangentially orientated as well, i.e, its dot product with the velocity would be..non-zero
Keshroom
#13
Mar17-12, 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.
D H
#14
Mar17-12, 07:18 PM
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P: 15,152
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.
blastoise
#15
Mar18-12, 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).
D H
#16
Mar18-12, 10:27 AM
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Quote Quote by D H View Post
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.
This is not the correct approach. This approach (and arildno's) proves that if the speed is constant then vv'=0. The problem at hand is the reverse problem, given that vv'=0, show that speed must be constant.

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.
D H
#17
Mar18-12, 10:31 AM
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Quote Quote by blastoise View Post
Case 2, there exists at lest one vn(t) that is not constant. Then v'n will not be zero at some t. Thus, (v1,v2,...,vn) dot (v'1,v'2,...,v'n) ≠ 0


Opinions?
My first opinion is that you should read our rules regarding solving people's homework for them.

My second opinion is that your proof is incorrect.
blastoise
#18
Mar18-12, 10:41 AM
P: 22
Quote Quote by D H View Post
My first opinion is that you should read our rules regarding solving people's homework for them.

My second opinion is that your proof is incorrect.
wasn't a proof but deleted it due to rules, but still would like a solution to this myself


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