Problem 8 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

The displacement of an object is given by \(d(t)=2 t^{5}-6 t^{3}+2 t^{2}+1\) miles over the time interval \(-2 \leq t \leq 2\) where \(t\) is measured in hours. (a) Approximately when does the object change direction? Please give answers accurate to within \(0.1\). When you zoom in on the graph here, what do you observe? (b) Approximately when is the object's velocity positive? Negative? (c) Approximate the object's velocity at time \(t=0\).

Step-by-Step Solution

Answer

The object changes direction at \(t = 0\). The object's velocity is positive on \(-2 \leq t < 0\) and negative on \(0 < t \leq 2\). The object's velocity at time \(t=0\) is 0 miles per hour.

Step 1: Find the derivative (velocity function)

The first step to solving this problem is to derive the displacement function. This gives us the object's velocity at any time \(t\). The derivative of \(d(t)=2 t^{5}-6 t^{3}+2 t^{2}+1\) is \(v(t)=10t^{4}-18t^{2}+4t\).

Step 2: Find the roots of the velocity

To find when the object changes direction, set \(v(t)\) equal to zero and solve for \(t\). This gives us the values of \(t\) when the velocity is zero, the points where the object may change direction. \(v(t) = 10t^{4}-18t^{2}+4t = 0\). Guessing or using an algebraic approach, we find that \(t = 0, -1, 1\) are solutions to this equation.

Step 3: Determine the direction change

We could graph the function or make a sign chart to see where the function is positive or negative. On \(-2 \leq t < 0\), \(v(t)\) is positive. On \(0 < t \leq 2\), \(v(t)\) is negative. Hence, the object changes its direction at \(t = 0\). We cannot conclude anything about \(t = -1\) and \(t = 1\) because these points are outside the given interval.

Step 4: Determine when the velocity is positive or negative

As found in the previous step, the velocity is positive on the interval \(-2 \leq t < 0\) and negative on the interval \(0 < t \leq 2\). That means the object is moving 'forward' on \(-2 \leq t < 0\) and 'backward' on \(0 < t \leq 2\). Thus, the object's velocity is positive on \(-2 \leq t < 0\) and negative on \(0 < t \leq 2\).

Step 5: Find the object's velocity at \(t = 0\)

To find the object's velocity at \(t=0\), substitute \(t=0\) into the velocity function. \(v(0) = 10(0)^{4}-18(0)^{2}+4(0) = 0\). Hence, the object's velocity at time \(t=0\) is 0 miles per hour.

Key Concepts

Understanding Derivatives in Calculus

In the realm of calculus, derivatives are fundamental. They help describe how a function changes at any given point. In simpler terms, by finding the derivative of a function, you determine its slope, or its rate of change. For this exercise, you had a displacement function, expressed as \(d(t) = 2t^5 - 6t^3 + 2t^2 + 1\). The main task was to derive this function to find the velocity.

• The first derivative of a function provides the velocity function \(v(t)\).
• Mathematically, it captures the object's instantaneous rate of change in position.
• In this exercise, taking the derivative turned the displacement function into the velocity function \(v(t) = 10t^4 - 18t^2 + 4t\).

Understanding derivatives allows you to analyze changing quantities, central to both physics and engineering.

Exploring the Concept of Velocity

Velocity is a vector quantity; it not only tells you how fast something is moving, but it also gives the direction of movement. In this context, the velocity function derived from the displacement function gives us crucial insights into the object's motion.

• The velocity function \(v(t)\) is \(10t^4 - 18t^2 + 4t\).
• To determine when the object's velocity is positive or negative, we need to examine where this function changes sign over the interval \([-2, 2]\).

Positive velocity indicates forward movement. Conversely, a negative velocity points towards backward movement. By analyzing the function, we discern when the object is moving forward or backward within the specified time frame. Hence, the determination of velocity provides a detailed understanding of the object's motion over time.

Understanding Displacement and Its Effect on Motion

Displacement offers a snapshot of how far an object is from its starting point after a certain time. It’s crucial to solving problems involving motion.

• Displacement is the function \(d(t)\) provided in the exercise.
• It calculates the net change in position from the initial to the final state.

Using the displacement function, we can determine if and when the object alters its course. By taking the derivative and solving for zero, we identify potential points where the direction changes. In our context, the object changed its direction at \(t = 0\). This is indicative of the point where its velocity became zero.

Algebraic Approach to Solving Derivatives

An algebraic approach can often greatly simplify the process of solving for derivatives and finding solutions. By employing algebraic techniques, we can isolate variables and obtain meaningful results more efficiently.

• Algebra simplifies the process of solving the equation \(v(t) = 0\) by allowing us to perform factorization.
• Deriving the displacement function required handling polynomial terms, which calculus and algebra complement each other to solve.
• Steps included solving \(10t^4 - 18t^2 + 4t = 0\) to find roots that potentially represent direction changes.

By factoring or using polynomial identities, we can find values of \(t\) where velocity equals zero, indicating potential changes in the direction of motion. Algebra forms a bridge in tackling derivative problems, offering a smooth path to finding reliable answers.