Help me understand first derivative in the context of distance and velocity

Question

While studying math in high school, we learned a lot about derivatives and integrals, but we used handbooks of common derivatives, so we never got the chance to link derivatives to something from the real world. So I tried to understand them by myself using the simple example of distance, velocity, acceleration.

I have artificially created a distance profile for, let's say, a jogger/hiker.

Then I plotted this and got a trendline using Excel:

So, according to the theory, if we take the derivative of this function, one should get the slope, correct?

So if this function is:

f(x) = -0.625x^4 + 5.8796x^3 - 15.347x^2 + 19.854x + 0.0397

The slope would be:

f'(x) = -2.5x^3 + 17.6388x^2 - 30.694x + 19.854

So far so good. Using my math knowledge so far, I could derivate this but now is the challenging part. What does this mean? What kind of information can one get from this?

Can one use this to get the velocity at any time? In my mind, one would think that the derivative would help to get the velocity at any time, just by replacing x with the time (in hours) - like in the figure below. Is this a correct interpretation?

What puzzles me, even more, is the fact that at x = t = 5h, that slope is -5.146 km/h, which doesn't make any sense in my mind. According to some explanation that I found, a negative number means a downward slope, which makes sense (if one looks at the profiles above and below), but it cancels the idea of getting the velocity at any time.

As you can see, I am confused about what this means, but I would appreciate it very much if someone could help me understand the significance of the derivative in this context.

Answer 1

The problem is that your function $f$ comes from interpolating your points with a polynomial: this is, finding the lowest degree polynomial that goes by those points. As soon as you leave these points $f$ can behave very differently from your expectations. This is the shape of your function indeed: https://www.wolframalpha.com/input/?i=-0.625x%5E4+%2B+5.8796x%5E3+-+15.347x%5E2+%2B+19.854x+%2B+0.0397

As you have reasoned the derivative is the slope of the distance: this is, the speed. Negative speed means you are travelling ‘backwards’, back to the origin: this is the case since, as you can see in the link, $f$ is decreasing at t=5.

Your understanding of derivatives seem to be right though, your falt was assuming the function generated by Excel would be a reasonable trajectory profile.

Answer 2

Okay soo...when I was learning about derivatives in school (which was like a few months ago) I also asked a similar question to my teacher.

Assuming you've differentiated the function correctly -

One thing you need to understand is that term "derivative" represents a number. You find the derivative of a function at a certain point on the graph - which means its a numerical answer. The numerical answer is the value of the slope of the tangent at that point. Remember derivative at the end of the day is the $dy \over{dx}$ which is nothing but the minuscule change is y over the same in x i.e. in a way its just $\tan(\theta)$ if you were to consider a triangle that small (though you usually don't put it that way) And this is exactly what we're taught in math when our teachers teach us the concept of slope.

$f'(x)$ is a representative function of all the slopes of all tangents at every point on $f(x)$. It is one function that can give you the slope of any tangent at any point. Take for example, $f(x) = x^2 \implies f'(x) = 2x$. Now if you were to take any point on the parabola (i.e. $f(x)$), draw a tangent at that point and calculate the slope, you'll find that the slope values at every point satisfies $f'(x) = 2x$.

The significance of derivative when you're calculating velocity from displacement is that, you're essentially just asking how much has the hiker displaced in a particular time period i.e. displacement between two timestamps. When you use calculus, we consider two timestamps that are infinitesimally close to each other (i.e. $\Delta t \rightarrow 0$).

As for the negative answer that you got at $t=5h$, it can mean that the body is moving in the opposite direction with respect to the direction it was moving in the time before. Remember velocity is a vector so it can be negative.