Differentiating Vector v = ai + bxj: What Does it Mean?

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The discussion centers on differentiating the vector v = ai + bxj and the implications of vector direction versus magnitude during differentiation. Participants express confusion about applying the chain rule and whether to consider vector directions or just magnitudes, with some arguing that magnitudes alone can lead to incorrect conclusions, particularly in non-linear motion scenarios like centripetal motion. The importance of maintaining vector components and their directions in calculations is emphasized, as neglecting them can yield erroneous results. The conversation also touches on the role of polar coordinates in understanding vector behavior. Overall, the consensus leans towards recognizing that both direction and magnitude are crucial in vector differentiation.
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I have a vector v = a i +bxj

Now I want to find a.So I apply chain rule and differentiate dv/dx * dx/dt = bj * Vx = bj*ai ??

Now what does this mean? Is this allright? Should I not care about the vector directions when differentiating and just use the magnitudes a = ab , if yes then why? And what will be the direction of the acceleration then?
 
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Hmm...

d/dt = \mathbf{v}\cdot \nabla

<br /> \frac{d\mathbf{v}}{dt} = \frac{\partial v_x}{\partial x}\frac{dx}{dt} + \frac{\partial v_y}{\partial y}\frac{dy}{dt}<br />
and we know that

\frac{dx}{dt} = a, \frac{dy}{dt} = bx, \frac{\partial v_x}{\partial x} = 0, \frac{\partial v_y}{\partial y} = 0

and then you get...
 
I don't get what you have written. Please tell me the physics and not just the math. How can a be zero ?
 
Last edited:
Hold on a second I did it wrong, didn't I? No one corrected me yet...

\frac{dv_x}{dt} = \frac{\partial v_x}{\partial x}\frac{dx}{dt} + \frac{\partial v_x}{\partial y}\frac{dy}{dt}

\frac{dv_y}{dt} = \frac{\partial v_y}{\partial x}\frac{dx}{dt} + \frac{\partial v_y}{\partial y}\frac{dy}{dt}

Which is just applying the chain rule to the components of v separately. Now that should make sense, it didn't before because it was wrong.
 
a_x = 0

a_y = ab

I'm sorry about my stupidity.
 
just differentiate the magnitudes. derivatives does not effect the basis vectors. so you write the related basis vector after the derivation. and it will give you the direction of the acceleration.
 
torehan said:
just differentiate the magnitudes.

That's wrong. Consider the counterexample-- centripetal motion. The speed is constant in time, so by your method you would conclude that acceleration is 0, but it's not.

And in this problem, differentiating the magnitude would also give you the wrong answer.
 
that's not totally wrong. if you should use "the polar coordinate" system and then you can find the direction.
 
torehan said:
that's not totally wrong. if you should use "the polar coordinate" system and then you can find the direction.

No, the polar coordinate basis vectors are not constant.

Face it, once you leave 1 dimensional motion, you can not characterize position, velocity and acceleration by their magnitudes.
 
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atavistic said:
I have a vector v = a i +bxj

Now I want to find a.So I apply chain rule and differentiate dv/dx * dx/dt = bj * Vx = bj*ai ??

Now what does this mean? Is this allright? Should I not care about the vector directions when differentiating and just use the magnitudes a = ab , if yes then why? And what will be the direction of the acceleration then?

I'm not sure I understand, is a a constant? what do you understand by the vector a ?
 
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