How can I Integrate Acceleration to Reconstruct Velocity?

[spidey]

is this correct?

∫(dx²/dy²) dx = dx/dy

 

[CompuChip]

What do you mean by (dx^2/dy^2) ?
Is it the second derivative of x with respect to y? Is it the first derivative of x with respect to y, squared? Is x a function of y (this is a bit unconventional notation, then)

 

[spidey]

What do you mean by (dx^2/dy^2) ?
Is it the second derivative of x with respect to y? Is it the first derivative of x with respect to y, squared? Is x a function of y (this is a bit unconventional notation, then)

i am sorry i placed the square on x but it is.. ∫ (d²x/dy²) dx ...

 

[Tac-Tics]

is this correct?

∫(dx²/dy²) dx = dx/dy

Suppose \frac{dx}{dy} = x + 1.

Then \frac{d^2x}{dy^2} = 1

However, it is *not* true that \int 1 dx = x + 1. As your calc teacher will say it, \int 1 dx = x + C for some abuse of notation/constant C.

The problem with your statement is that the integral isn't the inverse of the derivative. Integrating a derivative of a function doesn't give you back the original function.

However, the reverse *is* true. Taking the derivative of an integral of a function DOES give you back the original function. This is the fundamental theorem of calculus: \frac{d}{dx}(\int f(x) dx) = f(x)

(Note that I sort of disregarded your use of the second derivative, since it is just an added layer of confusion to this particular problem).

 

[spidey]

Suppose \frac{dx}{dy} = x + 1.

Then \frac{d^2x}{dy^2} = 1

However, it is *not* true that \int 1 dx = x + 1. As your calc teacher will say it, \int 1 dx = x + C for some abuse of notation/constant C.

The problem with your statement is that the integral isn't the inverse of the derivative. Integrating a derivative of a function doesn't give you back the original function.

However, the reverse *is* true. Taking the derivative of an integral of a function DOES give you back the original function. This is the fundamental theorem of calculus: \frac{d}{dx}(\int f(x) dx) = f(x)

(Note that I sort of disregarded your use of the second derivative, since it is just an added layer of confusion to this particular problem).

so u r saying this is not correct..∫ (d²x/dy²) dx not= dx/dy..

my basic question is i want to integrate acceleration.so
∫ a dx = ∫ (dv/dt) dx since [ a=dv/dt]
so to solve the above integration i thought
∫ a dx = ∫ (dv/dt) dx = ∫ (d²x/dt²) dx since [a=d²x/dt² or v=dx/dt]
= dx/dt ( i guess this is answer)
if its not,so what is the answer for this?

 

[CompuChip]

Actually,
∫ a dt = dx/dt =: v
by precisely that the theorem mentioned by Tac-Tics: a = d/dt( dx/dt ) so integrating it gives the velocity dx/dt.
You might be able to rework the integral over x to an integral over t, by a variable substitution:
∫ a(x(t)) dx = ∫ a(t) v(t) dt = ∫ v'(t) v(t) dt
(try to work that out for yourself).

I am thinking that partial integration may help you forward in this problem, but I'm not quite sure... play around a bit

 

[spidey]

Actually,
∫ a dt = dx/dt =: v
by precisely that the theorem mentioned by Tac-Tics: a = d/dt( dx/dt ) so integrating it gives the velocity dx/dt.
You might be able to rework the integral over x to an integral over t, by a variable substitution:
∫ a(x(t)) dx = ∫ a(t) v(t) dt = ∫ v'(t) v(t) dt
(try to work that out for yourself).

I am thinking that partial integration may help you forward in this problem, but I'm not quite sure... play around a bit

I understand the first part ∫ a dt = dx/dt =: v
but i am confused with rest...
i want answer for ∫ a dx = ∫ (dv/dt) dx= ?
don't know how to continue..

 

[spidey]

Actually,
∫ a dt = dx/dt =: v
by precisely that the theorem mentioned by Tac-Tics: a = d/dt( dx/dt ) so integrating it gives the velocity dx/dt.
You might be able to rework the integral over x to an integral over t, by a variable substitution:
∫ a(x(t)) dx = ∫ a(t) v(t) dt = ∫ v'(t) v(t) dt
(try to work that out for yourself).

I am thinking that partial integration may help you forward in this problem, but I'm not quite sure... play around a bit

I tried using integration by parts..i don't know is this correct..
∫ a dx = ∫ a v dt

u = v du = dv
dv= a dt v=v

∫ u dv = uv - ∫ vdu

= v.v - ∫ v dv
= v² - v²/2 or = v² - x = (dx/dt)² - x
= v²/2
= (dx/dt)²/2

is this correct...please someone help me...

 

[Tac-Tics]

Maybe my explanation was a little much for what you're trying to do (though it is important to understand it eventually too).

If you integrate acceleration, the function you get is something that looks a LOT like the function for velocity. In fact, if you know the velocity at any point in time, you can reconstruct the function for velocity entirely.

 

[spidey]

Maybe my explanation was a little much for what you're trying to do (though it is important to understand it eventually too).

If you integrate acceleration, the function you get is something that looks a LOT like the function for velocity. In fact, if you know the velocity at any point in time, you can reconstruct the function for velocity entirely.

Now i am getting this value...

∫ a dx = ∫ a v dt

u = a du = da
dv= v dt v= a

∫ u dv = uv - ∫ vdu

= a.a - ∫ a da
= a² - a²/2
= a²/2
= (d²x/dt²)²/2

is this correct...