Application of Derivatives

Understanding Motion Through Derivatives

Derivatives are not just an abstract concept in mathematics; they are a powerful tool for understanding change. One of their most tangible applications is in physics, especially for analyzing the motion of an object. If we know the position function of an object at any given time, we can use derivatives to find its velocity and even its acceleration at any moment.

Instantaneous Velocity from Position

Imagine you are driving a car. Your position is constantly changing over time. Velocity is the rate of change of that position. If we have a function that describes position, $s(t)$ , then the instantaneous velocity, $v(t)$ , at time $t$ is the first derivative of the position function.

v(t) = s'(t) = \frac{ds}{dt}

This means we can determine the precise velocity at any instant, not just the average velocity over a journey.

Determining the Velocity Function

For example, the motion of a particle is determined by the position function $s(t) = t^3 - 6t^2 + 9t$ , where $s$ is in meters and $t$ is in seconds. To find its velocity function, we simply differentiate the function $s(t)$ with respect to $t$ .

v(t) = \frac{d}{dt}(t^3 - 6t^2 + 9t)

v(t) = 3t^2 - 12t + 9

With this function, we can calculate the particle's velocity at any time. For instance, at $t = 1$ second, the velocity is $v(1) = 3(1)^2 - 12(1) + 9 = 0$ meters/second. This zero result indicates that the particle is momentarily at rest, perhaps pausing for a moment or even changing direction.

Acceleration from Velocity

Now, what if the velocity itself is changing? Perhaps you're pressing the gas pedal or the brakes. This change in velocity is called acceleration. Just as velocity is the derivative of position, acceleration, $a(t)$ , is the first derivative of the velocity function. This also makes it the second derivative of the position function.

a(t) = v'(t) = s''(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}

Determining the Acceleration Function

Continuing with the previous particle example, we already have the velocity function $v(t) = 3t^2 - 12t + 9$ . We can find its acceleration function by differentiating the function $v(t)$ .

a(t) = \frac{d}{dt}(3t^2 - 12t + 9)

a(t) = 6t - 12

From this, we can find the particle's acceleration at any time. For example, at $t = 3$ seconds, its acceleration is $a(3) = 6(3) - 12 = 6$ meters/second².

Peak Application in Vertical Motion

A ball is thrown straight up. Its height, $h(t)$ in meters, after $t$ seconds is given by the equation $h(t) = 50t - 5t^2$ .

First, we can find its velocity and acceleration functions.

Velocity function:

v(t) = h'(t) = 50 - 10t

Acceleration function:

a(t) = v'(t) = -10

The acceleration here is constant and negative. This represents Earth's gravitational acceleration, which always pulls objects downward. Its value is approximately $-9.8$ m/s², but it is often rounded to $-10$ in physics calculations to simplify things.

One of the most interesting questions is: when does the ball reach its maximum height?

The ball reaches its highest point exactly when it stops for a moment to change direction and fall back down. In other words, this occurs when its velocity is exactly zero.

v(t) = 0

50 - 10t = 0

10t = 50

t = 5

So, the ball reaches its peak at 5 seconds. To find out what the maximum height is, we substitute $t=5$ back into the initial height function.

h(5) = 50(5) - 5(5)^2 = 250 - 125 = 125