We study a special type of differential equation.
Consider a falling object. Recall that the acceleration due to gravity is about m/s. Since the first derivative of the velocity function is the acceleration and the second derivative of a position function is the acceleration, we have the differential equations
From these simple equations, we can derive expressions for the velocity and for the position of the object using antiderivatives.
Since it is tossed up with an initial velocity of m/s, and we see that . Therefore, represents the initial velocity of the ball. Hence .
Now when , m/s, and the ball is rising, and at , m/s, and the ball is falling.
Now let’s do a similar problem, but instead of finding the velocity, we will find the position.
Here represents the initial velocity of the ball. Since it is tossed up with an initial velocity of m/s, and Since the velocity is the derivative of the position, we can write, alternatively Now let’s take the antiderivative again.
Since we know the initial height was meters, write Hence . We need to know when the ball hits the ground, this is when . Solving the equation we find two solutions and . Discarding the negative solution, we see the ball will hit the ground after approximately seconds.
The power of calculus is that it frees us from rote memorization of formulas and enables us to derive what we need.