Question about integration

[Kal335]

Hi everyone,

I have a question about integration/derivation.

Here is my question.
You are certainly familiar with this equation.
x= 1/2 at²+ v₀t+x₀[MATH][/MATH]
So looking at v₀ you certainly know that i can derive the x in respect to t to get it like so :

dx/dt = v by multiplying each side by dt i can get dx = v × dt.

so my issue is the 1/2 at².
Now to get 1/2 at²,i try deriving twice x like this :
d²x/dt² = a then multiplying both side by dt² i end up with x = a × dt² / d² while i am trying to get back the 1/2 at².

Any idea on what i am doing wrong ? Thanks

 

[Harry_the_cat]

What is the question?

 

[Kal335]

What is the question?

I just updated it.

 

[HallsofIvy]

Hi everyone,

I have a question about integration/derivation.

Here is my question.
You are certainly familiar with this equation.
x= 1/2 at²+ v₀t+x₀[MATH][/MATH]
So looking at v₀ you certainly know that i can derive the x in respect to t to get it like so :

dx/dt = v by multiplying each side by dt i can get dx = v × dt.

No, that's not at all correct! Differentiating with respect to t you get at+ v₀.

so my issue is the 1/2 at².
Now to get 1/2 at²,i try deriving twice x like this :
d²x/dt² = a then multiplying both side by dt² i end up with x = a × dt² / d² while i am trying to get back the 1/2 at².

Any idea on what i am doing wrong ? Thanks

Well, first, you started by differentiating incorrectly! The first derivative is, as I said, at+ v₀.
The second thing is that, while dx/dt= f is the same as dx= f dt, d²x/dt²= f is NOT the same as "d²x= fd^t". Instead, use the fact that d²x/dt²= d/dt(dx/dt)= a so that, integrating once, dx/dt= at+ constant. Taking t= 0, since we are told that the speed at t= 0 is v₀, constant= v₀. dx/dt= at+ v₀ so that, integrating again, x= (a/2)t²+ v₀+ constant. That constant is the x₀ in you original equation.

 

[Kal335]

Thanks for replying.
Maybe i was not clear enough.Sorry.

When i said differentiating x with respect to t,i didn't mean the function : x(t)=1/2at²+v₀t+x₀ to get at+v₀t

I am not trying to differentiate the whole function, i was trying to express the term 'at²/2' in terms of dx.

Such as dx = v × dt for the velocity.

I hope you understand me.

 

[HallsofIvy]

Then, again, that is not a valid operation. First, the derivative, \(\displaystyle \frac{dx}{dt}\) is NOT a fraction! However, it is the limit of a fraction, \(\displaystyle \lim_{\Delta t\to 0}\frac{\Delta x}{\Delta t}\), so that is can be treated like a fraction. We can, through an "abuse of notation", write \(\displaystyle \frac{dx}{dt}=f(t)\) as \(\displaystyle dx= f(t)dt\).

The second derivative, however, is not a fraction and cannot be treated like a fraction: \(\displaystyle \frac{d^2x}{dt^2}= f(t)\) makes sense but \(\displaystyle d^2x= f(t)dt^2\) is meaningless!

 

[Kal335]

Then, again, that is not a valid operation. First, the derivative, \(\displaystyle \frac{dx}{dt}\) is NOT a fraction! However, it is the limit of a fraction, \(\displaystyle \lim_{\Delta t\to 0}\frac{\Delta x}{\Delta t}\), so that is can be treated like a fraction. We can, through an "abuse of notation", write \(\displaystyle \frac{dx}{dt}=f(t)\) as \(\displaystyle dx= f(t)dt\). The second derivative, however, is not a fraction and cannot be treated like a fraction: \(\displaystyle \frac{d^2x}{dt^2}= f(t)\) makes sense but \(\displaystyle d^2x= f(t)dt^2\) is meaningless! Thank you i found the answer i was looking for.