Second, Third (ect) Derivatives, and their relation to the function value

[clm222]

Hello, I was pondering acceleration's relation to velocity, when I came across a problem that is more of an issue of differentiation.
Given our variables

v=velocity
a=acceleration
α=slope of 'a'
t=time
(a_t1→t2)=Total acceleration from t₁ to t₂
(α_t1→t2)=Total slope of 'a' from t₁ to t₂

a=\frac{dv}{dt}

α=\frac{da}{dt}

would the two equations above imply that: α=\frac{{d^2}v}{d{t^2}}

Since I can find (assuming a has no slope) v(t) with: a_{{t_1}→{t_2}}=\int_{t_1}^{t_2} a\;dt

we can apply this to the slope of 'a' that:

α_{{t_1}→{t_2}}=\int_{t_1}^{t_2} α\;dt

would this also imply that the total slope of 'v' with reguard to the slope of 'a' from t₁ to t₂ be equal to:

\int\int_{t_1}^{t_2} α\; dt

I ask this in the differentiation page since it seems this applies generally to calculus. Converting to generall variables (v=y t=x) would:

f(x)=\int\int \frac{{d^2}y}{d{x^2}}\;dx

I'm asking to clairify. I also noticed that if you look at this algebriaclly you get (given that α=\frac{{d^2}y}{d{x^2}})
α=\frac{y}{x^2}\Rightarrow y=α{x^2}
but that gives a different answer then the integration, unless if I square x (which I only misunderstand because of the fact, that Leibnez's notation squares the 'd' instead of the 'y'). When I do square 'y', I get the same answer as when I integrate.

so basiclly what I am asking is:

to find the value of f(x), can i integrate all the way from whatever slope of f(x)?

can i express this algebriaclly by putting y to the power of the [number of slopes], and x to the same power respectivly?

if α=\frac{dy}{dx} and if β=\frac{dα}{dx}, does that automaticly mean that β=\frac{{d^2}y}{d{x^2}}
?
sorry if i made any silly asumptions in my math, I am new to calculus and math-interpreted-physics, thank you.

 

[haruspex]

Given our variables

v=velocity
a=acceleration
α=slope of 'a'
t=time
(a_t1→t2)=Total acceleration from t₁ to t₂
(α_t1→t2)=Total slope of 'a' from t₁ to t₂

a=\frac{dv}{dt}

α=\frac{da}{dt}

would the two equations above imply that: α=\frac{{d^2}v}{d{t^2}}<br />

<br /> Yes.<br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Since I can find (assuming a has no slope) v(t) with: a_{{t_1}→{t_2}}=\int_{t_1}^{t_2} a\;dt<br /> <br /> we can apply this to the slope of &#039;a&#039; that:<br /> <br /> α_{{t_1}→{t_2}}=\int_{t_1}^{t_2} α\;dt<br /> <br /> would this also imply that the total slope of &#039;v&#039; with regard to the slope of &#039;a&#039; from t₁ to t₂ be equal to:<br /> <br /> \int\int_{t_1}^{t_2} α\; dt </div> </div> </blockquote>Need to be a bit careful with the limits. Remember that the variable of integration has no meaning (strictly speaking) outside the integral. Using the same symbol in both places is a &#039;pun&#039;, i.e. they refer to different entities by the same name. Clearer is:<br /> \int_{u=t_1}^{t_2}\int_{t=t_1}^{u} α\; dt du<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I ask this in the differentiation page since it seems this applies generally to calculus. Converting to generall variables (v=y t=x) would:<br /> f(x)=\int\int \frac{{d^2}y}{d{x^2}}\;dx </div> </div> </blockquote>Yes, but with the same clarifications.<br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> I also noticed that if you look at this algebraically you get (given that α=\frac{{d^2}y}{d{x^2}})<br /> α=\frac{y}{x^2}\Rightarrow y=α{x^2} </div> </div> </blockquote>You&#039;ve lost me. Where did α=\frac{y}{x^2} come from?<br />

 

[HallsofIvy]

Hello, I was pondering acceleration's relation to velocity, when I came across a problem that is more of an issue of differentiation.
Given our variables

v=velocity
a=acceleration
α=slope of 'a'
t=time
(a_t1→t2)=Total acceleration from t₁ to t₂
(α_t1→t2)=Total slope of 'a' from t₁ to t₂

a=\frac{dv}{dt}

α=\frac{da}{dt}

would the two equations above imply that: α=\frac{{d^2}v}{d{t^2}}

Since I can find (assuming a has no slope) v(t) with: a_{{t_1}→{t_2}}=\int_{t_1}^{t_2} a\;dt

we can apply this to the slope of 'a' that:

α_{{t_1}→{t_2}}=\int_{t_1}^{t_2} α\;dt

would this also imply that the total slope of 'v' with reguard to the slope of 'a' from t₁ to t₂ be equal to:

\int\int_{t_1}^{t_2} α\; dt

I ask this in the differentiation page since it seems this applies generally to calculus. Converting to generall variables (v=y t=x) would:

f(x)=\int\int \frac{{d^2}y}{d{x^2}}\;dx

I'm asking to clairify. I also noticed that if you look at this algebriaclly you get (given that α=\frac{{d^2}y}{d{x^2}})
α=\frac{y}{x^2}\Rightarrow y=α{x^2}

NO, you cannot treat this as if it were algebra.

but that gives a different answer then the integration, unless if I square x (which I only misunderstand because of the fact, that Leibnez's notation squares the 'd' instead of the 'y'). When I do square 'y', I get the same answer as when I integrate.

It is incorrect to think of those as "squares" at all. They only indicate that this is a second derivative, have nothing to do with multiplying and cannot be cancelled

so basiclly what I am asking is:

to find the value of f(x), can i integrate all the way from whatever slope of f(x)?

I don't know what you mean by "integrate all the way" or "whatever slope"..

can i express this algebriaclly by putting y to the power of the [number of slopes], and x to the same power respectivly?

No, you cannnot solve differential equations "algebraically" and those superscripts are NOT powers.

if α=\frac{dy}{dx} and if β=\frac{dα}{dx}, does that automaticly mean that β=\frac{{d^2}y}{d{x^2}}

Yes, that is true but is not "algebraic". Rather, you are just saying that a "derivative of a derivative is a second derivative."

?
sorry if i made any silly asumptions in my math, I am new to calculus and math-interpreted-physics, thank you.

 

[clm222]

ok no algebra, can I still do algebriac rearrangment? (ie)
\int \frac{{x^2}}{x^{-2}}\;dx = \int \frac{x^2}{(\frac{1}{x^2})}\;dx = \int \frac{{x^2}{x^2}}{1}\;dx = \int x^{2+2}\;dx = \int x^4\;dx

 

[chiro]

ok no algebra, can I still do algebriac rearrangment? (ie)
\int \frac{{x^2}}{x^{-2}}\;dx = \int \frac{x^2}{(\frac{1}{x^2})}\;dx = \int \frac{{x^2}{x^2}}{1}\;dx = \int x^{2+2}\;dx = \int x^4\;dx

You can do this, but the thing to be careful about is how to do substitutions, how to treat your limits and differentials, and when to know if it's ok to move things in and out of the integral itself.

Some key tips for the above is whether something is a constant, or whether variables are independent of one another (say like if you are integrating xy^2 with respect to dx and y is independent, then you can move this outside of the integral but if it's not you can't) and also how the differentials work (i.e. the dx like terms) and how you treat these.

The rules for doing these kind of things are usually taught in an first year university course on calculus.

 

[HallsofIvy]

ok no algebra, can I still do algebriac rearrangment? (ie)
\int \frac{{x^2}}{x^{-2}}\;dx = \int \frac{x^2}{(\frac{1}{x^2})}\;dx = \int \frac{{x^2}{x^2}}{1}\;dx = \int x^{2+2}\;dx = \int x^4\;dx

My point was that you cannot treat the "d" and "d^2" in the derivative as if they were numbers and powers of numbers. They are not.