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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The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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