21. Antiderivatives, Areas and the FTC

d. Differential Equations and Initial Value Problems

A differential equation is an equation involving derivatives which we solve for a function.

So the equation \(\dfrac{dx}{dt}=v(t)\) to be solved for \(x(t)\) is an example of a differential equation. So is \(\dfrac{dv}{dt}=a(t)\) to be solved for \(v(t)\). When we solve these differential equations, there is an arbitrary constant in the solution.

An initial value problem is a differential equation together with an initial condition which is used to determine the arbitrary constant.

So the equation \(\dfrac{dx}{dt}=v(t)\) together with the initial condition \(x(t_0)=x_0\) is an example of an initial value problem. So is \(\dfrac{dv}{dt}=a(t)\) together with the initial condition \(v(t_0)=v_0\).

The examples and exercises on the previous three pages were examples of solving an initial value problem. Here is another exercise with two equations and two initial conditions. This time the constants of integration are not just the initial conditions.

A rocket is fired straight up. At time \(t=0\,\text{sec}\), it starts from a platform which is \(y(0)=10\,\text{m}\) above the ground with no initial velocity, \(v(0)=0\,\dfrac{\text{m}}{\text{sec}}\). If its acceleration is \(a(t)=100 e^{-t}\,\dfrac{\text{m}}{\text{sec}^2}\), find its height at \(t=60\,\text{sec}\).

Although the initial velocity is \(0\), the constant is not.

\(y(60)=100e^{-60}+5910\,\text{m}\approx5910\,\text{m}\)

We start from the acceleration, \(a(t)=100 e^{-t}\).
Since the velocity is an antiderivative of the acceleration, \[ v(t)=-100e^{-t}+C \] To find C, we use the initial condition for the velocity: \[ v(0)=-100+C=0 \qquad \Longrightarrow \qquad C=100 \] So the velocity at time \(t\) is: \[ v(t)=-100e^{-t}+100 \] Since the position is an antiderivative of the velocity, \[ y(t)=100e^{-t}+100t+K \] To find \(K\), we use the initial condition for the position: \[ y(0)=100+K=10 \qquad \Longrightarrow \qquad K=-90 \] So the position at time \(t\) is: \[ y(t)=100e^{-t}+100t-90 \] and the height at \(t=60\,\text{sec}\) is: \[\begin{aligned} y(60)&=100e^{-60}+100\cdot60-90 \\ &=100e^{-60}+5910\,\text{m}\approx5910\,\text{m} \end{aligned}\]

We check by differentiating the position \(y(t)=100e^{-t}+100t-90\) to get the velocity and the acceleration: \[\begin{aligned} v(t)&=y'(t)=-100e^{-t}+100 \\ a(t)&=v'(t)=100e^{-t} \end{aligned}\] Further, we check the initial conditions: \[\begin{aligned} y(0)&=100e^{-0}+100\cdot0-90=10 \\ v(0)&=-100e^{-0}+100=0 \end{aligned}\]

You can also practice computing the velocity and position from the acceleration by using the following Maplet (requires Maple on the computer where this is executed):

From Acceleration to Velocity and Position Rate It

21. Antiderivatives, Areas and the FTC

d. Differential Equations and Initial Value Problems

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