Difference of differential equations

Question

First of all, I would like to mention that I have problems understanding how to use differentials. I know how to solve some of them and I know that they represent slopes, but I lack a good understanding of the practicality of them.

For example, I remember seeing once a formulation for velocity that looked like this:

$\frac{dx_{tot}}{dt}=\frac{dx_1}{dt}-\frac{dx_2}{dt}$

where $x_{tot}$ was the total distance, $x_1$ was one distance and $x_2$ was another distance. At this point, things become blurry and I can't understand the meaning of this equation. Can anyone help me to understand how to interpret such an equation? And secondly, how should a beginner tackle such an equation, in order to understand it? What should one look at first? I would highly appreciate a step-by-step explanation.

I know this is maybe high school problem, but I really have a huge gap in my brain when trying to understand it. Any help would be highly appreciated.

Thank you!

Answer 1

Distance (more properly here, displacement) is always relative to some point. In most cases, this point of reference is fixed, but it does not have to be (search moving reference frames). In this case $x_1$ is the position of the object you are referencing in the coordinate system you are using, and $x_2$ is the position of the reference point in that same coordinate system. If the reference point $x_2$ is fixed, then its time derivative is zero. But if the coordinate system ("reference frame") is moving with respect to $x_2$, then the time derivative is not zero. Displacement is always measured as a distance between $x_1$ and $x_2$m, $d=x_1-x_2$, so the velocity is the time derivative $$v=\frac{d(x_1-x_2)}{dt}=\frac{dx_1}{dt}-\frac{dx_2}{dt}$$

Answer 2

When an object is in motion, its speed is computed as the ratio of the space it traversed over the time it took:

$$v=\frac{x(t)-x(0)}{t-0}=\frac{\Delta x}{\Delta t}.$$

This is true when the speed is constant. When the speed varies during the motion, you can still consider an average speed, computed the same way:

$$\bar v=\frac{\Delta x}{\Delta t}.$$

But if you are interested in the speed "at any time", you must use the derivative, denoted as

$$v(t)=\frac{dx(t)}{dt}.$$

Technically speaking, this instantaneous speed is defined as the average speed during a time lapse so small that the speed variation is neglectible.

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Now your equation simply says that the relative speed of two points $1$ and $2$ equals the difference of the speeds,

$$v(t)=v_1(t)-v_2(t)$$

and directly comes from the fact that their distance is the difference of their abscissas:

$$x(t)=x_1(t)-x_2(t).$$

This shows that the differentiation operator is linear

$$\frac{d(x_1-x_2)}{dt}=\frac{dx_1}{dt}-\frac{dx_2}{dt}.$$

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And what's the relation to the slope ?

If you plot the abscissa as a function of time, you will get a curve. If you draw a line tangent to this curve at some point, the line represents a body moving at constant speed. As the slopes coincide at the tangency point, the slope of the line represents the instantaneous speed.