# Show that velocity is perpendicular to rxv

1. Aug 24, 2011

### msslowlearner

1. The problem statement, all variables and given/known data
A particle moves so that its pos. vector is gn by r = cos wt i+sin wt j. show that the velocity v of the particle is perpendicular to and that rxv is a constant vector.

2. Relevant equations

3. The attempt at a solution

I've tried working this out assuming [B]v[/B] = xi + y j + zk and proceeding with the vector product, but i'm not able to solve for sin wt = 1. (rxv / mod r mod v)
Actually this problem is under the chapter differential vector calculus, but i dont know if perpendicularity can be found usin DVC. Please help.

2. Aug 24, 2011

### tiny-tim

welcome to pf!

hi msslowlearner! welcome to pf!

oooh, you're missing the obvious …

v = dr/dt …

so what is dr/dt ?

3. Aug 26, 2011

### msslowlearner

ah ahhh .. yes, i get it ..the tangent , i.e., the velocity ... how cud i hv missed it ?? anyways, tats the first part .. what about the second ? r x v is a constant vector??

4. Aug 26, 2011

### msslowlearner

how do you say a vector product is constant? i understand that the vector product is the area of the parallelogram formed by the 2 vectors r and v. but how is this constancy represented mathematically ?

5. Aug 26, 2011

### tiny-tim

hi msslowlearner!

(just got up :zzz: …)
hmm … you're understanding a visualisation of the https://www.physicsforums.com/library.php?do=view_item&itemid=85"

visualisations are there to help you, they are not what the mathematics "really is"

a vector product is mathematics, and to prove things about it, you usually need to use the mathematical definitions, and write out the mathematical equations
you just write out r x v mathematically …

from that, it should be obvious that it's constant

(if something is constant, it usually is obvious! ) …

so what do you get?

Last edited by a moderator: Apr 26, 2017
6. Aug 26, 2011

### HallsofIvy

Staff Emeritus
I think when tiny time says "write out r x v mathematically" he means "in terms of the components of r and v".

7. Aug 26, 2011

### msslowlearner

so, here we go..
r x v = (z sin wt)i + (-z cos wt)j+ (y cos wt-x sin wt)k

since rxv is a vector perpendcular to both r and v, v is perpendicular to rxv. right ?? tis one's from the definition itself ...

now, rxv is a fn(t). d/dt (rxv) = zw()r) +yw(-2sin wt)k ... hope i'm right so far... now i assumed that as change in t i.e., del t tends to zero, the k component becomes zero , and
d/dt(rxv=zw (r)

this is what i've arrived at.. how do i say rxv is constant ???

8. Aug 26, 2011

### tiny-tim

hi msslowlearner!
where does this come from??

(and what is "z" supposed to be?)

start again …

1) write out r

2) write out v

3) then multiply them …

what do you get for 1) 2) and 3) ?

9. Aug 26, 2011

### msslowlearner

1) r= cos wt i+sin wtj

2) i assumed v=xi+yj+zk

3) i used the matrix form to calculate r x v = (z sin wt)i + (-z cos wt)j+ (y cos wt-x sin wt)k

10. Aug 26, 2011

### msslowlearner

i mean am i getting the cross product atleast correct?? vectors do stump me.

11. Aug 26, 2011

### tiny-tim

no, use v = dr/dt = … ?

12. Aug 27, 2011

### msslowlearner

oh yes ... v=dr/dt ... it escaped my mind .. i got rxv = w, which is a constant ... so simple .. thanks tiny-tim :)

13. Aug 27, 2011

### HallsofIvy

Staff Emeritus
Good! What constant did you get? What $is$ r x v?

14. Aug 27, 2011

### tiny-tim

he said w !

(but he left out the direction)

15. Aug 27, 2011

### stallionx

r = cos wt i+sin wt j
dr/dt=v=w (-sin(wt)) i + w (cos(wt)) j

r dot product V = 0

rXv = i j k
cwt swt 0
-wswt -wcwt 0
i j k

cwt swt 0

w cwt cwt k + w swt swt k= w k

rxv = w

and is a constant

16. Aug 27, 2011

### tiny-tim

hi stallionx!

no harm done (since msslowlearner got the answer in post #12),

but pleeease don't give out full answers on this forum!

17. Aug 27, 2011

### Tomer

It's probably important to mention that $\vec{r}\times\vec{v} = w\hat{k}$. That is, it is a vector in the z direction... not just a number.

18. Aug 27, 2011

### stallionx

Hiii, thanks.

I am terribly sorry...