Application of Derivatives

Using Derivatives to Analyze Motion

Derivatives describe how a quantity changes. Motion gives a concrete example: from an object's position function, we can determine its instantaneous velocity and then its acceleration.

Instantaneous Velocity from Position

As a car moves, its position changes over time. Velocity is the rate of change of that position. If $s(t)$ describes position, then the instantaneous velocity $v(t)$ at time $t$ is the first derivative of the position function.

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

At every time where $s$ is differentiable, this derivative gives the velocity at that exact moment. An average velocity uses a time interval.

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)

At $t = 1 \text{ second}$ , the velocity is $v(1) = 3(1)^2 - 12(1) + 9 = 0 \text{ m/s}$ . A zero velocity means the particle is momentarily at rest, but it does not by itself prove a change of direction. Factor the velocity to inspect its sign on either side of $t=1$ :

v(t)=3(t-1)(t-3)

For $0<t<1$ , both factors are negative, so $v(t)>0$ . For $1<t<3$ , the factors have opposite signs, so $v(t)<0$ . The velocity therefore changes sign at $t=1$ , which proves that the particle reverses direction there.

Acceleration from Velocity

Velocity can itself change, as when a car speeds up or brakes. Its rate of change is called acceleration. The acceleration $a(t)$ is the first derivative of velocity and therefore the second derivative of position.

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)

From this, we can find the particle's acceleration at any time. For example, at $t = 3 \text{ seconds}$ , its acceleration is $a(3) = 6(3) - 12 = 6 \text{ meters/second}^2$ .

Finding Maximum Height in Vertical Motion

A ball is thrown straight up. Its height, $h(t)$ in meters, after $t \text{ seconds}$ 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

In this simplified model, upward is set as the positive direction and air resistance is ignored. Gravitational acceleration near Earth's surface is rounded from $-9.8 \text{ m/s}^2$ to $-10 \text{ m/s}^2$ . The negative sign shows that the acceleration points downward.

Because $h(0)=0$ and $h(10)=0$ , the modeled flight runs from launch at $t=0$ until the ball returns to its starting height at $t=10$ .

We now determine when the ball reaches its maximum height.

At the highest point, the ball's vertical velocity changes from positive to negative. We therefore solve $v(t)=0$ within the modeled flight interval.

v(t) = 0

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

\begin{aligned} h(5) &= 50(5) - 5(5)^2 \\ &= 250 - 125 \\ &= 125 \end{aligned}

The maximum height is $125 \text{ meters}$ . To find a maximum height, differentiate the position function, find when the velocity is zero while the object is moving, and substitute that time into the original height function.