Vector valued velocity and acceleration question

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The discussion focuses on the integration of vector-valued functions in multivariable calculus, specifically transitioning from acceleration to position. The user, Lee, questions the ability to recover initial motion from acceleration when certain components are constant. The consensus is that while integrating acceleration yields a velocity vector, the lack of information about initial conditions prevents the recovery of the exact position function. It is established that two integrals of the same function differ only by a constant vector, emphasizing the necessity of additional information for complete recovery of motion.

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leehufford
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Hello,

In my multivariable calc class we are differentiating and integrating position, velocity and acceleration vector valued functions. My question is this:

When we integrate a vector valued function from acceleration to position, the constant vector only changes the definition of the functions that were not zero for acceleration. For example if the position function is <1,0,0> and the acceleration function is <0,1,0> , you could never recover that initial x motion from integrating the acceleration function.

What am I missing here? I understand the constant from integration is in the form of a vector now but this doesn't help me see what's wrong. Thanks for reading,

Lee
 
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This doesn't make sense because the position in the y component and z component are constant and 0, so there is no acceleration in this direction.
 
Any Two integrals are equal up to a constant term. The fact that the constant after integrating acceleration is a vector (velocity) doesn't mean that that integral is "more determined". You cannot recover exact situation unless given more than just acceleration function.
 

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