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 example and exercise on the previous page were examples of solving an initial value problem. Here is another exercise with two equations and two 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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