Dot Product, Derivative, and Vector Valued Functions

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Homework Help Overview

The discussion revolves around proving properties of a differentiable vector valued function \(\vec{r}(t)\), specifically focusing on the derivative of its magnitude and the relationship between the dot product of the vector and its derivative.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of differentiating the magnitude of a vector, questioning whether the problem statement might contain a typo regarding the derivative of the magnitude versus the magnitude of the derivative.
  • Some participants discuss the relationship between the dot product of the vector and its derivative, considering how to express this in terms of magnitudes.
  • There is an attempt to relate the expressions involving the dot product to more familiar forms, such as using component functions.

Discussion Status

The discussion is ongoing, with participants sharing their thoughts and attempts to clarify the problem. Some guidance has been offered regarding the use of the derivative product rule and expressing the vector in component form, but no consensus has been reached.

Contextual Notes

Participants are navigating the complexities of vector calculus, particularly the nuances of differentiating scalar magnitudes that depend on time. There is an acknowledgment of potential ambiguities in the problem statement that may affect interpretation.

ourio
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Homework Statement


Prove that, if \vec{r}(t) is a differentiable vector valued function, then so is ||\vec{r}(t)||, and \vec{r}(t) \bullet \vec{r'}(t) = ||\vec{r}(t)|| ||\vec{r}(t)||'

Homework Equations


I know how to do a dot product, but what bothers me is the fact that the question involves the derivative of the magnitude of r. But the magnitude of a vector is a scalar, the derivative of which would be zero.

If it's a typo and really supposed to be the magnitude of the derivative, I'm still lost as to how to relate the dot product of the vectors with their magnitudes??

The Attempt at a Solution


I know that the unit tangent vector involves r' and ||r'||, but I'm not sure that this helps.
 
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ourio said:

Homework Statement


Prove that, if \vec{r}(t) is a differentiable vector valued function, then so is ||\vec{r}(t)||, and \vec{r}(t) \bullet \vec{r'}(t) = ||\vec{r}(t)|| ||\vec{r}(t)||'


Homework Equations


I know how to do a dot product, but what bothers me is the fact that the question involves the derivative of the magnitude of r. But the magnitude of a vector is a scalar, the derivative of which would be zero.
scalars can still be time dependent, and clearly have non-zero derivatives... the magnitude just represents the length of r, if the length is changing the derivative will be non-zero


ourio said:
If it's a typo and really supposed to be the magnitude of the derivative, I'm still lost as to how to relate the dot product of the vectors with their magnitudes??
|\vec{r}| = (\vec{r(t)} \bullet \vec{r(t)})^{1/2}
ourio said:

The Attempt at a Solution


I know that the unit tangent vector involves r' and ||r'||, but I'm not sure that this helps.
 
Okay, so if it really is the derivative of the magnitude, I can understand that. But, what does \<br /> (\vec{r(t)} \bullet \vec{r&#039;(t)})^{1/2}<br /> equal?

If you replace r with something like f(x)+g(x) and r' with f'(x)+g'(x) and do the dot product, you get \sqrt{f(x)f&#039;(x)+g(x)g&#039;(x)}. But how would that help?


I'm still not clear...
 
how about starting wth
|\vec{r(t)}|^2 = |\vec{r}||\vec{r}|= (\vec{r(t)} \bullet \vec{r(t)})

if i rememebr correctly, the derivative product rulle works just as normal, but if you're unsure, try writing everything out in component form

\vec{r(t)} =(x(t), y(t), z(t))
 
Thanks. I think I've got it now... it just took some time staring at the problem! Thanks again.
 

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