Problem 50 - Calculus: Early Transcendental Functions

Question

Find the position function \(s(t)\) from the given velocity or acceleration function and initial value(s). Assume that units are feet and seconds. $$a(t)=16-t^{2}, v(0)=0, s(0)=30$$

Step-by-Step Solution

Answer

The position function is \(s(t) = 8t^{2} - \frac{1}{12}t^{4} + 30\).

Step 1: Find the Integral of the Acceleration

To find the velocity \(v(t)\), take the integral of the acceleration function \(a(t)=16-t^{2}\). The integral \(16-t^{2} dt\) results in \(16t - \frac{1}{3}t^{3} + C\).

Step 2: Solve for the Constant of Integration

Substitute the given initial condition \(v(0)=0\) into the velocity function to solve for the constant of integration. Plugging these values in to the equation gives \(0 = 16*0 - \frac{1}{3}*0^{3} + C\), so \(C = 0\). The velocity function is then \(v(t) = 16t - \frac{1}{3}t^{3}\).

Step 3: Find the Integral of the Velocity

To find the position function \(s(t)\), take the integral of the velocity function \(v(t) = 16t - \frac{1}{3}t^{3}\). The integral of \(16t - \frac{1}{3}t^{3} dt\) results in \(8t^{2} - \frac{1}{12}t^{4} + C\).

Step 4: Solve for the Constant of Integration Again

Substitute the given initial condition \(s(0)=30\) into the position function to solve for the constant of integration. Plugging these values in to the equation gives \(30 = 8*0^{2} - \frac{1}{12}*0^{4} + C\), so \(C = 30\). The position function is then \(s(t) = 8t^{2} - \frac{1}{12}t^{4} + 30\).

Key Concepts

Integration

Integration is an essential tool in calculus used to find quantities like area, displacement, and accumulation, derived from rates of change. In this context, when given an acceleration function, integration allows us to reverse the process of differentiation to find other related functions such as velocity and position.

• To find a velocity function from an acceleration function, we integrate the acceleration. Given the acceleration function, say, \(a(t) = 16-t^2\), we calculate the integral \(\int (16-t^2) \, dt\), which results in a velocity function of the form \(v(t) = 16t - \frac{1}{3}t^3 + C\).
• The constant \(C\) is critical, as initially determined conditions are used to solve for this constant, yielding a unique function that satisfies these conditions.

This ability to transform a differential form into an integral form bridges the gap between acceleration, velocity, and position, representing the dynamic movement of objects.

Velocity function

The velocity function represents the rate of change of position over time. Velocity can be derived directly from an acceleration function through integration. In this specific problem, starting with the acceleration function \(a(t) = 16-t^2\), we integrated to find the velocity function:

• Performing the integration yields \(v(t) = 16t - \frac{1}{3}t^3 + C\), where \(C\) is a constant determined by initial conditions.
• The integration process reverses differentiation of position to expose how quick and in what direction the position changes occur at any moment \(t\).

After substituting the given initial condition \(v(0)=0\), we find that \(C = 0\), thus simplifying our velocity function to \(v(t) = 16t - \frac{1}{3}t^3\). This is useful in understanding motion by giving instantaneous "snapshot" speeds at any time \(t\).

Position function

The position function, denoted as \(s(t)\), indicates the location of an object at any time \(t\). To obtain this from a velocity function, another round of integration is necessary. For our velocity function \(v(t) = 16t - \frac{1}{3}t^3\), integrating gives:

• The integral \(\int (16t - \frac{1}{3}t^3) \, dt\) results in \(s(t) = 8t^2 - \frac{1}{12}t^4 + C\).
• Here, \(C\) is again an integration constant that is found using initial conditions specific to the problem at hand.

Substituting the initial condition \(s(0) = 30\), we solve for \(C\) and find it to be \(30\). So, the complete position function becomes \(s(t) = 8t^2 - \frac{1}{12}t^4 + 30\). This function details the object's path over time, crucial for predictions and analyses in physics and engineering.

Initial conditions

Initial conditions are crucial in calculus for uniquely determining solutions to differential equations involving integration. These conditions help resolve the arbitrary constants that arise during integration by pinning functions to specific points in time and space. In our exercise, you can see their importance in both velocity and position determinations:

• For velocity, given \(v(0)=0\) allowed us to solve for the constant \(C\) in \(v(t) = 16t - \frac{1}{3}t^3 + C\), revealing \(C = 0\). This determined the specific velocity curve.
• Similarly, the initial position \(s(0)=30\) helped pinpoint the position function constant, effectively determining the exact trajectory path with \(C = 30\).

Understanding initial conditions is fundamental, as they ensure that the mathematical models applied reflect real-world scenarios accurately and correspond uniquely to particular states or points defined within calculations.