Are Orthogonal Vectors Proven by Derivative and Dot Product?

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The discussion centers on proving that if a vector v(t) has a constant magnitude, its derivative v˙(t) is orthogonal to v(t). The initial misunderstanding involved interpreting v as a constant vector, leading to a dot product of zero. However, it is clarified that v has a constant magnitude but is not constant itself. The conclusion reached is that the derivative of a vector with constant magnitude is indeed orthogonal to the vector. This understanding aligns with the properties of dot products and orthogonality in vector calculus.
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Homework Statement
Prove that if v(t) is any vector that depends on time, but v(t) has constant magnitude, then
v˙(t) is orthogonal to v(t)
Relevant Equations
Dot Product
I feel like this question is very straight forward and my explanation below summarizes the answer pretty well. Could someone confirm this or tell me if I am missing something?

We have V which is a vector, but the question states it is a constant. If I take the derivative of V, represented by V', a constant, then I get 0.
If I dot product these to values, the product is then 0. And it is known that when the dot product between two vectors is zero, they are orthogonal.
 
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quittingthecult said:
Homework Statement:: Prove that if v(t) is any vector that depends on time, but v(t) has constant magnitude, then
v˙(t) is orthogonal to v(t)
Relevant Equations:: Dot Product

I feel like this question is very straight forward and my explanation below summarizes the answer pretty well. Could someone confirm this or tell me if I am missing something?

We have V which is a vector, but the question states it is a constant. If I take the derivative of V, represented by V', a constant, then I get 0.
If I dot product these to values, the product is then 0. And it is known that when the dot product between two vectors is zero, they are orthogonal.
It says that ##\vec v## has constant magnitude; not that ##\vec v## is constant.
 
PeroK said:
It says that ##\vec v## has constant magnitude; not that ##\vec v## is constant.
Ah I misread the question. That makes a lot of sense. Thank you for catching that.
 
So what does it look like now?
 
BvU said:
So what does it look like now?
IMG_0311.jpg
This is the conclusion I came to.
 
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