Question

interesting vector/calculus thing

Answer 1

what is

\(\int \, \vec V \times \dfrac{d^2 \vec{ V}}{d t^2} \ dt\)




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where \(\vec V = \vec V (t)\)

and \(\times \) is the cross/vector product

Answer 2

@FaiqRaees
@ganeshie8
@phi

following today's masterclass

Answer 3

.

Answer 4

Had a quick headache about this very terse problem and sort of came up with one possibility. If V sands for the word "Vector" (as opposed to standing for velocity), then if that vector were displacement, and if the angle involved in the cross product were 90 degrees, then you may be close to circular motion ..
Clues would help, though, if any are available.

Answer 5

@osprey The question is very in-descriptive indeed and is probably oriented towards the people that know the circumstances of the question and discussion in the mentioned masterclass.

In my best guess it's a vector field, and \[\vec V \rightarrow f(x,y,z,t)\] or more generally \[\vec V \rightarrow f(x_i,t) = f(x_1, x_2, \dots x_n, t)\] and not just a function of time as that would be pointless to try and cross multiply. In any case then you'd have:
\[\vec V = V_x\hat i + V_y\hat j +V_z\hat k\] which you could just derive and multiply to get:
\[ \left( V_y \frac{\partial ^2 V_z}{\partial t^2} - V_z \frac{\partial ^2 V_y}{\partial t^2}\right) \hat i +
\left( V_z \frac{\partial ^2 V_x}{\partial t^2} - V_x \frac{\partial ^2 V_z}{\partial t^2}\right) \hat j +
\left( V_x \frac{\partial ^2 V_y}{\partial t^2} - V_y \frac{\partial ^2 V_x}{\partial t^2}\right) \hat k \]
which once solved for can be hard to integrate or not, depending on the way V looks like. In any case I have to say that this does remind me of some operators you can run into that are combination of vector fields and vector operators, specifically I've seen vector Laplacian being used like this:
\[ V\Delta K = V_i \frac{\partial ^2 K_i}{\partial x_i^2}\hat x_i \] For the question however, probably the curl operator would be more appropriate starting point. I assume the idea of the task is to try and show this integrand as some combination of such operators and then use their identities to simplify the integral.

But this is where I stop looking further since the question could also mean something totally different.

Answer 6

oh, since there is no edit button here, one important thing to mention is that even if you try to go and rewrite the original integrand over vector operators you're still in a world of trouble because all the partial derivatives are done by time and not coordinates as the vector operators do so it's not probably all that good of an idea anyhow.

Answer 7

\(\int \, \vec V \times \dfrac{d^2 \vec{ V}}{d t^2} \ dt\)

\(= \int \, \vec V \times \vec{ V''} \ dt\)

\(= \int \, \vec V \times (\vec {V'})' \ dt\)

thus setting it up for IBP

\(= \vec V \times \vec{ V'} - \int \, (\vec V)' \times \vec{ V'} \ dt\)


\(= \vec V \times \vec{ V'} - \int \, \vec{ V'} \times \vec{ V'} \ dt\)

\(= \vec V \times \vec{ V'} - \int \, 0 \ dt\)

\(= \vec V \times \vec {V'}\)