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if i have some function f(x,y), what is fxx(x,y) (i dont mean just a deriv(deriv(f(x,y))) explanation)? is it still "acceleration", like physics projectile motion <-> acceleration, or is there some other geometric/physical thing that is associated with explaining what fxx is?
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Hi cal;
Good to see you!
Means differentiate with respect to x twice.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Well it is the accelaration with respect to x and x only.
X'(y-Xβ)=0
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Thanks!
Hi bobbym!
Now... what does fxy(x,y) mean in terms of acceleration? i cant seem to imagine it
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Hi;
I am not quite understanding what you are saying. I treat acceleration quite simple. If you start with the distance formula for a falling body in earths gravitational field. Leaving out the metric system and the vector notation. The one discovered by Galileo.
Distance fallen = y for time t:
Velocity is the instantaneous rate of change of distance. That is the derivative.
Acceleration is the instantaneous rate of change of velocity. The second derivative. If you are given some distance formula f with respect to t then
is the acceleration.In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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second partial derivative of a given function Y=F(x) exist if there will be no sharp turn in the graph of F'(x)
let y=f(x,z)
then y'=f(x+h,z)-f(x,z)/h when lim'h'tends to 0 is partial derivative because x is variable
similerly find derivative of y treating z as variable because of
derivative is limiting process so for y'' repeat above process
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i dont mean acceleration. i know what acceleration is. i just dont get how fxy works. i know to take the derivative with respect to x and then y, but how do you show that in the real world? it seems weird, going in 2 different directions on a plane with scalars
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Hi cal;
I know you know. All math seems weird to me. I am amazed each time it works. I go,"amazing" whenever an answer checks out. If you are having no problems with differentiating with respect to x then y or y then x, I think you are there! Let the physicists lose their minds explaining why this works in the real world.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Thanks for the advice bobbym. i guess for the moment, i'll live with that
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Hi calccrypto;
I really didn't mean it that strong. Keep your eyes open for an answer that you understand. I am the worst in the world at a geometric interpretation of math. You should see the mess I am making of a problem concerning cylindrical coordinates.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Ok. When/If i find out, i'll post the answer here
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Okay, everything else going reasonably well?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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yeah. thanks for asking
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