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I would like to know whats the pratical use of derivate calculus? Or what it means?

If you can give some pratical example I'll be grateful.

Eg.:

I can use an definite integral to know area of a function knowing the expression of the function and the limits.

Know how to solve exercises make no sense if i cant imagine the real purpose and the objective meaning.

Thanks in advance.

Cold
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4 Answers4

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it was originally invented to calculate the trajectory of planets.

More generally, almost all physical laws describing how a system evolves are naturally expressed in term of derivatives. To compute the evolution of a system according to these laws, you need derivate calculus...

Denis
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  • I know the history @Denis, and the usage (well, since i was born i have seen this on formulas, and where other guys decide use it). I wonna take the same base as the same guy who express : vel = dS/st – Cold Oct 14 '14 at 14:46
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There are a variety of applications especially in physics.

Suppose you know the force that acts on a particle with mass $m$ as a function of time, i.e. $F = F(t)$. Suppose you furthermore know that at $t = 0$, the velocity of the particle is given by $v_0$. It can be shown that

\begin{align*} v(t) = v_0 + \int_0^t \frac{F(t^\prime)}{m} \mathrm{d} t^\prime \end{align*}

where $v(t)$ denotes the velocity of the particle at time $t$ for $t \geq 0$.

hickslebummbumm
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  • My question that @hick: Why use derivate on this formula? I can memorize the formula but i wonna understand why to use. – Cold Oct 14 '14 at 14:48
  • I used an integral in this formula which is the opposite of a derivative. Alternatively: suppose you know the velocity of the particle as a function of time. The acceleration is per definition $a = \frac{\mathrm{d} v}{\mathrm{d} t}$. As the force acting on the particle is given by $F(t) = m a(t)$, we have $F(t) = m \frac{\mathrm{d} v}{\mathrm{d} t}$. – hickslebummbumm Oct 14 '14 at 14:51
  • Thanks again @Hick. But a insist bro, my problem is not know the formulas, but understand why is the formula as it is. I know how to get the aceleration, but why use derivate when we have a variation on time. – Cold Oct 14 '14 at 14:54
  • The time derivative of the velocity gives you the rate of change of the velocity at a given time $t$. We call that the acceleration. – hickslebummbumm Oct 14 '14 at 14:57
  • Hum... I'm getting sarted to understand. Thnks again @hick – Cold Oct 14 '14 at 15:00
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[Copying my answer from here]

The derivative function says how fast the original function is changing at each point. If $f(t)$ is the position of a particle or a rocket ship at each time $t$, then the derivative $f'(t)$ is the speed of the particle or the rocket ship at time $t$.

Consider as an example $f(t) = -5t^2 + 20t$. Suppose this describes the height of a rocket above the ground at time $t$. This curve is a downward-facing parabola with $f(0) = f(4) = 0$ and the peak of the parabola at $f(2) = 20$:

parabola

At $t=0$ we have $f(t) = 0$ and the rocket is on the ground. The rocket goes up, quickly at first, then more slowly, until at $t=2$ it stops going up and starts to come down, slowly at first, then more quickly as time goes by, until it hits the ground again at $t=4$.

What if you want to know the speed of the rocket? That is the derivative, $$f'(t) = -10 t+20.$$

parabola with derivative

The derivative is the blue line in the picture. It represents the upward speed of the rocket at each point. When $t=0$, the derivative has the value $20$, representing a fast upward motion. When $t=1$, the upward speed has decreased to $10$. When $t=2$, the rocket has reached the peak of its flight and has stopped going up and is about to come back down. $f'(t) = 0$, meaning that the rocket has no motion at this instant. Then at $t=3$ the derivative is $-10$, which represents a downward motion, and at $t=4$ when the rocket hits the ground its downward motion is twice as fast, since $f(4) = -20$.

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MJD
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One practical use: Approximate ugly functions with nice ones. In fact, we like Lines, they are simple. The equation of a line is just

$$ Ax + By = C $$

Where $A,B,C$ are numbers. This is a nice function. Sometimes you may encounter functions like this one:

$$ f(x) = \sin ( e^{x^2} )\cdot e^{\arctan x} + x^8 - \ln x $$

$f$ is a horrible function. Calculus then help us. Derivative help us because we can approximate this $f(x)$ in a vicinity of a point $(x_0,y_0)$ where $f$ would make sense with a line, in fact explicitly given by

$$ \mathcal{L}(x) = y_0 + f'(x_0)(x-x_0)$$