Question

In Exercises 87-89, consider a particle moving along the \(x\) -axis where \(x(t)\) is the position of the particle at time \(t, x^{\prime}(t)\) is its velocity, and \(\int_{a}^{b}\left|x^{\prime}(t)\right| d t\) is the distance the particle travels in the interval of time. The position function is given by \(x(t)=t^{3}-6 t^{2}+9 t-2\) \(0 \leq t \leq 5 .\) Find the total distance the particle travels in 5 units of time.

Short answer

The total distance the particle travels in 5 units of time can be computed by adding the integrals of the absolute value of the velocity function over the time intervals [0,1], [1,3] and [3,5].

Step-by-step solution

Differentiate Position Function

To find how fast and in which direction the particle is travelling at any given time, the first derivative of the position function \(x(t)\) must be computed. Differentiating the position function \(x(t) = t^{3} - 6t^{2} + 9t - 2\) with respect to \(t\) yields the velocity function \(x'(t) = 3t^{2} - 12t + 9\).

Find Zeroes of the Velocity Function

To evaluate the integral of the absolute value of \(x'(t)\), it is necessary to work out the time intervals in which the velocity is positive and negative. This requires finding the zeroes of the velocity function, that is finding the solutions to the equation \(x'(t) = 0\). After factoring, the equation \(3t^{2} -12t + 9 = 0\) simplifies to \(t^{2} - 4t + 3 = 0\), which in turn factors to \((t-1)(t-3)=0\). This gives the solutions \(t = 1\) and \(t=3\). Therefore, the time intervals of interest are \([0,1]\), \([1,3]\) and \([3,5]\).

Compute Integral for Each Interval

The absolute value of the velocity function, \(|x'(t)|\), must now be integrated over each of the intervals of time identified. In each interval \([a, b]\), this is given by \(\int_{a}^{b} |x'(t)| dt\). Find the sign of \(x'(t)\) in each of the intervals by evaluating \(x'(t)\) at any point within each interval. If \(x'(t)\) is positive, then \(|x'(t)| = x'(t)\) and if \(x'(t)\) is negative, then \(|x'(t)| = -x'(t)\). Then apply the Fundamental Theorem of Calculus to compute the integrals.

Find Total Distance Traveled

Add together each of the results from Step 3 to find the total distance. The total distance the particle travels in 5 units of time is given by \(\int_{0}^{1} x'(t) dt + \int_{1}^{3} -x'(t) dt + \int_{3}^{5} x'(t) dt\).

Key concepts

Differentiate Position Function

Understanding the concept of differentiating a position function is crucial when studying particle motion in calculus. It allows you to determine the velocity or speed of a particle at a particular moment in time. Consider the position function of a particle moving along the x-axis provided in our example, which is written as x(t) = t^3 - 6t^2 + 9t - 2.

To find the derivative, or the velocity function, you apply the power rule of differentiation to each term with respect to time t. This rule states that for any term of the form at^n, the derivative is an*t^(n-1). Using this rule, the derivative of the position function becomes x'(t) = 3t^2 - 12t + 9. This process effectively provides us the speed at which the particle is traveling along the x-axis at any given moment. The sign of the velocity function indicates the direction: a positive value suggests movement to the right while a negative indicates movement to the left on the x-axis.

It's important to remember that the highest power of t gives the term with the greatest influence on the behavior of the function for larger values of t. This differentiation step sets the stage for understanding the overall trajectory of the particle's motion over a given time period.

Velocity Function Zeros

Finding the zeroes of the velocity function helps us to understand when a particle is changing direction or momentarily coming to a stop. In the example provided, students are asked to look at the velocity function x'(t) = 3t^2 - 12t + 9 and find its zeroes.

The zeroes are the values of t where the velocity is zero, which corresponds to the times when the particle is changing its direction of motion. To find these crucial points, one would set the velocity function equal to zero and solve for t. For the given function, factoring leads to (t-1)(t-3)=0, giving the solutions t=1 and t=3. These zeroes divide the time interval into parts where the particle's velocity is either positive or negative. This informs us that the particle reverses direction at these points. Consequently, as an educator, it's essential to emphasize to students that understanding where and why a function reaches zero is fundamental to analyzing particle motion.

Distance Integral Calculus

The concept of distance travelled by a particle over a time interval involves taking the integral of the absolute value of the velocity function. This application of integral calculus provides a powerful tool to determine the total distance travelled, irrespective of direction. In calculus, distance is an accumulation of the absolute velocity over time, which is why integrals are suitably employed.

In the exercise, students must use the given velocity function and integrate it across different time intervals to calculate the distance. The key is to separate the time intervals based on the zeroes of the velocity function, as the particle's direction may change at these points. To find the distance the particle travels between time a and b, you evaluate the integral t_{a}^{b} |x'(t)| dt. The absolute value ensures we are always adding distance, never subtracting, because distance is a scalar that only increases.

By computing the integral for each segment and adding them together, you arrive at the total distance travelled. When instructing students, stress the importance of the absolute value and that the total distance is not the same as the net displacement, which does not consider direction and only accounts for the starting and ending positions of the particle.