Question

A particle moves along a line with a velocity \((\text { in } \mathrm{m} / \mathrm{s})\) given by \(v(t)=\sec ^{4} \frac{\pi t}{12},\) for \(0 \leq t \leq 5,\) where \(t\) is measured in seconds. Determine the position function \(s(t),\) for \(0 \leq t \leq 5 .\) Assume \(s(0)=0\)

Short answer

Answer: The position function is \(s(t) = \frac{12}{\pi}\left(\frac{1}{3}\sec^3{\frac{\pi t}{12}}\tan{\frac{\pi t}{12}} - \frac{2}{9}\tan^3{\frac{\pi t}{12}}\right)\) for \(0 \leq t \leq 5\).

Step-by-step solution

Integrate the given velocity function

To find the position function, integrate the given velocity function with respect to time: $$ s(t) = \int v(t) dt = \int \sec^4 \frac{\pi t}{12} dt $$ This integral is quite complex, and it might be challenging for a high school student to solve by hand. Instead, we will use integration techniques that involve substitution and integral tables. Let: $$ u = \frac{\pi t}{12} $$ $$ du = \frac{\pi}{12} dt \\ dt = \frac{12}{\pi} du $$ Now substitute \(u\) into the integral: $$ s(t) = \frac{12}{\pi}\int \sec^4u\ du $$ To solve \(\int \sec^4u\ du\), we can use integration by parts and the reduction formula: $$ \int \sec^4u\ du = \frac{1}{3}\sec^3u \tan{u} - \frac{2}{3}\int \sec^2u\tan^2u\ du $$ To evaluate the remaining integral, we can use another substitution: $$ v = \tan{u} $$ $$ dv = \sec^2u\ du $$ Replace \(v\) in the integral: $$ \int \sec^4u\ du = \frac{1}{3}\sec^3u \tan{u} - \frac{2}{3}\int v^2dv $$ This integral is much more manageable, so we evaluate it: $$ \int \sec^4u\ du = \frac{1}{3}\sec^3u \tan{u} - \frac{2}{3}\cdot\frac{1}{3}v^3 + C_1 $$ Now replace \(u\) and \(v\) with the original expressions: $$ s(t) = \frac{12}{\pi}\left(\frac{1}{3}\sec^3\frac{\pi t}{12}\tan\frac{\pi t}{12} - \frac{2}{9}\tan^3\frac{\pi t}{12}\right) + C_1 $$

Apply the initial condition

Now, we need to find the constant of integration, \(C_1\), using the initial condition \(s(0) = 0\): $$ 0 = \frac{12}{\pi}\left(\frac{1}{3}\sec^3\frac{\pi 0}{12}\tan\frac{\pi 0}{12} - \frac{2}{9}\tan^3\frac{\pi 0}{12}\right) + C_1 \\ C_1 = 0 $$

Write the specific position function

With the constant of integration, \(C_1\), identified, we now have the specific position function, \(s(t)\): $$ s(t) = \frac{12}{\pi}\left(\frac{1}{3}\sec^3{\frac{\pi t}{12}}\tan{\frac{\pi t}{12}} - \frac{2}{9}\tan^3{\frac{\pi t}{12}}\right) $$ This function holds for \(0 \leq t \leq 5\).

Key concepts

Velocity Function

Understanding a velocity function is crucial for determining how an object's position changes over time. In physics, velocity is the rate at which an object's position changes, with direction considered. It's a vector quantity and is typically expressed as a function of time, like our example:
- \( v(t) = \sec^4\left(\frac{\pi t}{12}\right) \)
Here, this specific function gives the velocity of a particle moving along a line at time \( t \) measured in seconds. The function uses the secant trigonometric function raised to the fourth power, which can hint at periodic and complex behaviors. By integrating this velocity function, we can find the position function, \( s(t) \), which tells us the particle's position at any time \( t \). Recognizing the form of the velocity function helps in selecting the right integration technique.

Integration by Parts

The method of integration by parts is pivotal when dealing with integrals that are products of functions. It is formulated from the product rule of differentiation and is given by:
- \( \int u \, dv = uv - \int v \, du \)
In the context of finding \( s(t) \), integration by parts is one technique used to tackle the more complicated function \( \sec^4 u \). This approach can sometimes simplify the integration process by reducing it to an integral that is more straightforward or familiar.
However, remember: integration can sometimes be iterative, requiring multiple applications of this technique, or alongside other methods like substitution. Here, it's essential to express \( \sec^4u \) in a way that plays well with integration by parts, often needing some initial algebraic manipulation. This strategy helps break down complex formulas into manageable calculations.

Initial Condition

An initial condition gives us specific information about the function at a starting point, serving as a boundary that helps in determining any constants that arise during integration. When calculating \( s(t) \), the initial condition provided is:
- \( s(0) = 0 \)
This means that at time \( t = 0 \), the position of the particle is known to be 0. After performing the integration of the velocity function, you will find a constant term, commonly denoted as \( C \) or \( C_1 \) from the integration process.
Apply the initial condition by substituting \( t = 0 \) and \( s(0) = 0 \) into the integrated function to solve for this constant. This ensures the position function accurately describes the particle's motion from the start point on the defined interval. Without the initial condition, we would have an indefinite integral that is not uniquely tied to our problem's physical context.

Substitution Method

The substitution method, also known as u-substitution, is a handy technique in calculus for simplifying integrals. It involves substituting a portion of the integral with a new variable \( u \), and redefining the differentials accordingly. This simplifies the integral, turning it into a form more easily solved.

• In our problem, we set \( u = \frac{\pi t}{12} \)
• Then find \( du = \frac{\pi}{12} dt \)
• Resolve to \( dt = \frac{12}{\pi} du \)

This substitution simplifies the integral \( s(t) = \int \sec^4 \left(\frac{\pi t}{12}\right) dt \) by converting it into a function of \( u \), making further mathematical manipulation more straightforward.
The substitution method is powerful as it can transform functions into forms that are more amenable to standard integration techniques and also sometimes contributes to solving the problem using integration by parts. By correctly substituting parts of the function, complex integrals reduce to simpler calculations.