Question

Can someone explain this?

If

`a=(dv)/(dt)`

Then

`dv=adt`

`dv=int(a)dt`

Why does `adt` turn into an integral. I really don't get this.

Answer 1

Hence, `adt` turns into an integral, and `a` turns into an integrand.

Explanation:

`a=(dv)/dt`

This equation is most likely derived from kinematics, where `a` is acceleration and `v` is velocity.

Therefore, acceleration is the velocity differentiated.

`a=(dv)/dt`

To find velocity, just integrate velocity differentiated.

`v=int" "(dv)/dt" "dt`
`color(white)(v)=int" "a" "dt`

Hence, acceleration is velocity integrated.

For example

Let acceleration follow the formula: `4x-5`

`a=4x-5`

Hence,

`(dv)/dt=4x-5`

To find velocity,

`v=int" "4x-5" "dt`
`color(white)(v)=2x^2-5x+c`, where `c` is the constant of integration.

*As Steve M pointed out, a derivative should not be taken lightly as a fraction, despite it behaving like a fraction under Leibniz notation, as it is an operation.

Answer 2

It is the process of separation of variables, but you must do the same thing on the left and right sides.

---

Acceleration is just how the velocity changes over time,

`(Deltavecv)/(Deltat) = veca`

and if we examine such a change as `Delta->d`, i.e. the interval of time shrinks to an infinitesimally short time span,

`(dvecv)/(dt) = veca`

`dvecv = vecadt`

Integrating the right side with respect to time and the left side with respect to velocity states,

`int_(v_0)^(v_f) dvecv = int_(t_1)^(t_2)vecadt`

If the acceleration is an average acceleration, and we set `t_1 = 0` and `t_2 = t`, then `veca -= bara` is a constant with respect to time and we get the familiar average acceleration equation:

`v_f - v_0 = baraint_(0)^(t)dt`

`color(blue)(v_f = v_0 + barat)`

Suppose `veca = veca(t) = -t`, i.e. the acceleration decreases over time (because the person gets tired of pushing something?).

Instead, we would have (with `t_1 = 0` and `t_2 = t`):

`int_(v_0)^(v_f) dvecv = -int_(0)^(t)tdt`

`v_f - v_0 = -t^2/2`

`=> color(blue)(v_f = v_0 - t^2/2)`

Answer 3

A more formal explanation is to investigate some fundamentals of calculus: What the derivative is, and what it represents, how to undo the process of differentiation and its relationship to integration.

Recall the fundamentals of differential calculus, where we solve the general problem of finding the gradient of the tangent to a curve `y=f(x)` at a general coordinate. This leads us to the limit definition of the derivative of a function, which gives us the required gradient:

`f'(x) = lim_(h rarr 0) (f(x+h)-f(x))/h`

We often use the notation, `f'(x) = d/dx( f(x))`, and we write:

`f'(x) = dy/dx`

Where `d/dx` is the differential operator of a function wrt `x`

The slope of the tangent is critically important as it tells us how quickly the function is changing. ie it tells us the rate of change of the function.

Next, we examine the fundamentals of integral calculus , where we solve the problem of calculating the exact area `A(a,b)` under a curve `y=f(x)`, between two x-coordinates `x=a` and `x=b`, this leads us to a riemann sum:

`A(a,b) = 1/(b-a) lim_(Delta x_i rarr 0) sum_(i=1)^n f(x_i) Delta x_i`

And we more concisely write this sum using the notation:

`A(a,b) = int_a^b \ f(x) \ dx`

Where `int \ ... \ dx` is the integral operator of a function wrt `x`

Note that the use of the symbol `x` does not appear in the final result as the area is number, here the "variable" `x` is a dummy variable

Then, we can show using the Fundamental Theorem of Calculus, that (rather surprisingly) there is a relationship between these two process, and in fact for some function `y=f(x)`, then:

`int_a^b \ f'(t) \ dt = f(b) - f(a)`

In other words the process of integration "undoes" the process of differentiation.

So returning to the initial question we are asked to explain why

`a = (dv)/dt iff v = int \ a \ dt`

Here, acceleration `a(t)` at time `t` is defined to be the rate of change of velocity `v(t)` with respect to time, ie it tells us how velocity changes wrt time, so using our results from differential calculus we can write:

`a = (dv)/dt`, or `v'(t) = a`

And then to get the velocity function, we must "undo" the process of differentiation, so we apply the integral operator to get:

`int \ v'(t) \ dt = int \ a \ dt`

And again using our results from the Fundamental Theorem of Calculus

`v(t) = int \ a \ dt + C`

Knowing that the process works, we often become rather lax with the notation (especially in the sciences, and we often write:

`(dv)/(dt) = a`

And so:

`dv = a \ dt`

But, really this is lazy shorthand for:

`v = int \ a \ dt`