math2001's blog

Motion and calculus

15 February 2019

If you studied physics, one of the first things you probably did was graph the “motion” of objects. That is the displacement, velocity, and acceleration at an instant t. Didn't it remind you of a mathematical concept?


Let's say we have an object whos displacement is such that:

$$ s(t) = t^2 \quad \text{where t is the time} $$

If we graphed this function, we'd get a parabola:

Now, let's graph its velocity. You can do it manually, but if you try to feel it, you'll see that since as the time goes by the displacement keeps on increasing faster and faster, the velocity could increase linearly. The velocity of that object at an instant $t$ could be given by:

$$ v(t) = t $$

Which would look something like this:

What if we graphed the acceleration? Since the velocity is increasing at a constant rate, the acceleration must be constant, something like:

$$ v(t) = 1 $$

Now, doesn't that make you think of a derivating the function $v$? Everytime, the function is one degree lower.

The meaning

The derivative

Say we graphed the displacement of a ball with time on x-axis and displacement on the y-axis. What would be the meaning of the gradient at any point?

$$ m = \frac{rise}{run} = \frac{displacement}{time} $$

The rise over the run is a displacement over some time, which is a velocity. And it makes sense! The gradient (which is the rate of change) of any point on this displacement/time graph gives us the velocity (which is the change of position per unit time, ie. how quickly it's changing position, its rate).

Now, from a velocity/time graph, the gradient gives us the change in velocity, which is called acceleration.

The integral

What if you want to go the other way? From a velocity/time graph, the displacement is given by the area under the curve.

If you're not convinced, it's easier to think of a simple example where the velocity is constant (a straight flat line). If you want to know by how much an object traveled knowing it's speed (2 m/s) and the time it took (5 s), you'd just multiply (2 * 5 = 10 m). Now, this happens to be area under the graph (width * height). And with a bit of thinking, you can convince yourself that this isn't a coincidence.

Can you see that this also works for an acceleration/time graph (it gives you the velocity).

So, the derivative of a function let's you go from displacement to velocity to acceleration, and the integral let's you go the other way.

No wonder Newton was into calculus…