Dot products, show that r'.r''=0

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Homework Statement



r(t)=(x(t),y(t),z(t))
t has been chosen so that r'.r'=1
show that r'.r''=0

Homework Equations



v'.w'=|v||w|cos(theta)

The Attempt at a Solution



Clearly what is being described is circular motion about a unit circle. And using the equation for a unit circle its easy to show that r'.r''=0 Velocity is tangent to the circle, acceleration is inward and therefore the dot product of the two is zero.

What I am having trouble with is showing for the general solution.

r'.r'=|r'||r'|cos(theta) = 1
theta is zero, and therefore |r'| is 1 , this isn't particularly helpful.

r''.r'=|r''|cos(theta) (theta not zero)

Expanding in parametric form doesn't seem to help either.

r''.r'= x'(t)x''(t)+y'(t)y''(t)+z'(t)z''(t)

I am at a loss.
 

Answers and Replies

  • #2
834
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This is a common problem with a trick. What is the time derivative of the magnitude of the velocity?
 
  • #3
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Thanks for your quick reply.

The question is only given in general terms. And reads explicitly as:

Let r(t)=(x(t),y(t),z(t)) be the position vector along some curve for -infinity to +infinity.
Suppose t has been chosen so that 1=r'.r' for all t. Show that 0=r'.r''.
 
  • #4
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They've told you that the [itex]r' \cdot r' = 1[/itex], so the magnitude of the velocity is a constant. What is the time derivative of a constant?

Edit:
Alternatively, you recognize that this is circular motion and it must happen in some plane. You can use the results of 2d motion to say that

[tex]r(t) = r_0 e^{i \omega t}[/tex]

What are the time derivatives of this parameterization?
 
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  • #5
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Okay, the magnitude of the velocity is a constant.
Therefore the magnitude of the acceleration is zero, |r''|=0,

And certainly then r'.r'' =|r'||r''|cos(theta)=0 , because |r''|=0

Is this really all there is to it?

Also getting a bit tripped up because while in circular motion the magnitude of the velocity is constant but its direction is changing and there is an acceleration on the object.

Been Edited.
 
Last edited:
  • #6
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Clearly what is being described is circular motion about a unit circle.
That is incorrect. There are many solutions besides uniform circular motion. Constant velocity (moving along a straight line at a constant speed) is another type of motion that satisfies the relation r'·r' = 1.

Okay, the magnitude of the velocity is a constant.
Therefore the magnitude of the acceleration is zero, |r''|=0
That is also incorrect. Your uniform circular motion does not have the magnitude of acceleration equal to zero. The only type of motion that has magnitude of acceleration equal to zero is constant velocity (as opposed to constant speed). Uniform circular motion is one solution to this kind of motion, and the magnitude of acceleration is not zero for uniform circular motion.


Hint: What is d/dt (r'·r')?
Forget that r'·r' = 1 for a second. Just use the chain rule. Now add in the given fact that r'·r' = 1.
 
  • #7
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That is incorrect. There are many solutions besides uniform circular motion. Constant velocity (moving along a straight line at a constant speed) is another type of motion that satisfies the relation r'·r' = 1.
To be fair, there is the added condition that r'' is orthogonal to r'.


Hint: What is d/dt (r'·r')?
Forget that r'·r' = 1 for a second. Just use the chain rule. Now add in the given fact that r'·r' = 1.
Okay.

r(t) = ((x(t), y(t), z(t))

r't=((x'(t),y'(t),z'(t))

r't·r't=(x'(t)^2)+(y'(t)^2)+((z't)^2)

d/dt (r'·r') = 2(x'(t))(x''(t))+2(y'(t))(y''(t))+(z'(t))(z''(t))

d/dt (r'·r') = 2(r'(t))·(r''(t))

r'·r' = 1

0 = 2(r'(t))·(r''(t))

therefore, (r'(t))·(r''(t))=0 ?

The world is beginning to make sense again. Thanks.
 
  • #8
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Yeah, the trick I referred to is using the product rule directly.

[tex]\frac{d}{dt} (r' \cdot r') = (r'' \cdot r') + (r' \cdot r'') = 2 (r'' \cdot r')[/tex]

But we established that [itex]r' \cdot r' = 1[/itex], a constant, so [itex]2 (r'' \cdot r')[/itex] quantity must be equal to zero, and this can be done without breaking it down into components.
 

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