Help with Deriving d/dt |r(t)| = (1/|r(t)|) r(t)dot r'(t)

  • Thread starter bubbers
  • Start date
In summary, the conversation discussed the solution to a question involving the derivative of the magnitude of a vector. After taking the derivatives of both sides of the given equation using the chain rule, the correct solution was found to be d/dt|r(t)| = (1/|r(t)|) (r'(t) dot r(t)). The conversation also cleared up some misunderstandings about the chain rule and the assumption of constant magnitude.
  • #1
6
0

Homework Statement



If vector r(t) is not 0, show that d/dt |r(t)| = (1/|r(t)|) r(t)dot r'(t).

Homework Equations



The hint given was that |r(t)|^2 = r(t) dot r(t)

The Attempt at a Solution



the post I saw about this question said that you should take the derivatives of both sides of the hint equation using chain rule, so I got:

2|r(t)|(1)=r'(t) dot r(t) + r(t) dot r'(t)
which is equal to:
|r(t)|= r'(t) dot r(t)

aaaaand now I don't know what to do
 
Last edited:
Physics news on Phys.org
  • #2


bubbers said:
2|r(t)|(1)=r'(t) dot r(t) + r(t) dot r'(t)

This step is incorrect. You have the equivalent of f(g(x)) -- the chain rule says d/dx f(g(x)) = f'(g(x)) * g'(x).
 
  • #3


Okay, now I'm confused... in this situation, wouldn't f'(g(x)) be equivalent to 2|r(t)| and g'(x) be equivalent to 0 because |r(t)| is a length and the derivative of a number is 0? So then it would be 2|r(t)|(0)=r'(t) dot r(t) + r(t) dot r'(t) --> 0=r'(t) dot r(t) + r(t) dot r'(t)?
If this is right, then I still don't know where to go from here.
 
  • #4


bubbers said:
Okay, now I'm confused... in this situation, wouldn't f'(g(x)) be equivalent to 2|r(t)| and g'(x) be equivalent to 0 because |r(t)| is a length and the derivative of a number is 0? So then it would be 2|r(t)|(0)=r'(t) dot r(t) + r(t) dot r'(t) --> 0=r'(t) dot r(t) + r(t) dot r'(t)?
If this is right, then I still don't know where to go from here.

No. |r(t)| has t dependence. Accordingly, we can't assume the magnitude is constant.
 
  • #5


Oh man...I totally get it now...i feel kinda dumb :P
the derivative would just be:
2|r(t)|d/dt|r(t)|= r'(t) dot r(t) + r(t) dot r'(t)
2|r(t)|d/dt|r(t)|= 2(r'(t) dot r(t))
|r(t)|d/dt|r(t)|= r'(t) dot r(t)
d/dt|r(t)|=(1/|r(t)|) ( r'(t) dot r(t))

Thanks for letting me waste your time :)
 
  • #6


No problem. Glad to have helped. :)
 

1. What is the meaning of "d/dt" in the equation?

"d/dt" represents the derivative with respect to time, indicating that the equation is describing the rate of change of the function over time.

2. How do you derive the equation "d/dt |r(t)| = (1/|r(t)|) r(t)dot r'(t)"?

To derive this equation, we use the chain rule and the product rule of differentiation. First, we use the chain rule to differentiate the absolute value function |r(t)| with respect to time. Then, we use the product rule to differentiate the product of (1/|r(t)|) and r(t)dot r'(t), where r(t)dot represents the dot product of r(t) and r'(t). Combining these steps results in the final equation.

3. What is the physical significance of the equation?

This equation describes the instantaneous change in the magnitude of the position vector r(t) with respect to time. In other words, it represents the speed at which an object is moving in a particular direction. This can be useful in analyzing the motion of objects in space or in other physical systems.

4. How is the equation used in practical applications?

The equation can be used in various fields such as physics, engineering, and astronomy to analyze the motion of objects and determine their velocities and accelerations. It can also be utilized in optimization problems to find the maximum or minimum values of functions involving position vectors.

5. Are there any limitations to using this equation?

This equation assumes that the position vector r(t) is differentiable, meaning that it is continuous and has a well-defined derivative at all points. It also assumes that the position vector is non-zero, as the equation involves taking the reciprocal of the magnitude of the vector. Additionally, the equation may not be applicable in cases where the position vector is constantly changing direction, such as in circular or oscillatory motion.

Suggested for: Help with Deriving d/dt |r(t)| = (1/|r(t)|) r(t)dot r'(t)

Back
Top