Problem 51 - Calculus of a Single Variable

Question

Applying the First Derivative Test In Exercises \(41-48\) , consider the function on In Exercises \(49-54,\) (a) use a computer algebra system to differentiate the function, (b) sketch the graphs of \(f\) and \(f^{\prime}\) on the same set of coordinate axes over the given interval, (c) find the critical numbers of \(f\) in the open interval, and (d) find the interval(s) on which \(f^{\prime}\) is positive and the interval(s) on which it is negative. Compare the behavior of \(f\) and the sign of \(f^{\prime}\) $$ f(t)=t^{2} \sin t, \quad[0,2 \pi] $$

Step-by-Step Solution

Answer

Firstly, the derivative \(f'(t) = 2t \sin(t) + t^2 \cos(t)\) is gotten, then set to zero yielding \(2t \sin(t) + t^2 \cos(t) = 0\). Afterwards, the expression is evaluated for positive and negative values to determine where the function increases or decreases. Sketches of \(f\) and \(f'(t)\) show this visually.

Step 1: Differentiate the function

To differentiate the function \( f(t) = t^2 \sin(t)\), use the product rule which states \((fg)'=f'g+fg'\). Hence, \(f'(t) = 2t \sin(t) + t^2 \cos(t)\)

Step 2: Find the critical points

Obtain the critical point by equating the derived expression to zero, and solving for \(t\). Therefore: \(2t \sin(t) + t^2 \cos(t) = 0\)

Step 3: Identify intervals where f'(t) is positive or negative

To get the intervals for which \(f'(t)\) is positive or negative, plugin values from the intervals into \(f'(t)\). Where \(f'(t)\) is greater than 0, function increases. If it is less than 0, function decreases.

Step 4: Generate sketches for f and f'(t)

A computer algebra system can be used to generate the graphs for the function and the derivative, highlighting critical points where derivative is 0 and the intervals of decrease and increase.

Key Concepts

Derivatives

Understanding derivatives is essential in calculus. A derivative represents the rate of change of a function's value with respect to a change in its input value. It provides a powerful tool to understand how a function behaves, especially when trying to determine where it increases or decreases.
When working with derivatives, you're often tasked with deriving a new function, called the derivative, from an original function. This new function, denoted as \( f'(x) \), shows how \( f(x) \) changes as \( x \) changes.
For example, given a function \( f(t) = t^2 \sin(t) \), its derivative can be calculated using the product rule (which we'll explore in detail below). The derivative in this case is \( f'(t) = 2t \sin(t) + t^2 \cos(t) \).

• The purpose of this derivative is to provide insight into the original function's behavior. In mathematical terms, it allows us to analyze things like how sharply a curve turns (curvature) and determine the function's slope at particular points.

Understanding derivatives not only paves the way for these analyses but also builds a foundation for more complex concepts.

Critical Points

Critical points are key values in calculus that help identify where a function may have a local maximum or minimum, or a change in its increasing or decreasing behavior. To find these points, you first need to find where the derivative of the function is equal to zero or does not exist.
For the function \( f(t) = t^2 \sin(t) \), the derivative is \( f'(t) = 2t \sin(t) + t^2 \cos(t) \). To locate critical points, solve the equation \( f'(t) = 0 \). This solution reveals the values of \( t \) where the rate of change of the function is zero, indicating potential critical points.

• These points are significant because they often indicate key features of a graph, such as peaks (local maxima) and valleys (local minima), or where the function changes direction.

Finding critical points is an essential skill for understanding the full behavior of a function over a given interval.

Interval Analysis

Interval analysis involves examining sections of a function's domain to determine where the function increases or decreases. In calculus, after identifying the critical points, you often divide the domain into intervals to perform this analysis.
For the function \( f(t) = t^2 \sin(t) \), once critical points are found by solving \( f'(t) = 0 \), you can assess each interval around these points by choosing test values. By evaluating \( f'(t) \) at these test values, you determine whether the function is increasing (where \( f'(t) > 0 \)) or decreasing (where \( f'(t) < 0 \)).

• This helps visualize the graph of the function and understand the nature of its peaks and troughs.
• Interval analysis provides a way to compare the behavior of a function against the sign of its derivative, which can assist in graph sketching and understanding overall curve behavior.

Through interval analysis, one gains detailed insight into a function's structure and motion within its domain.

Product Rule

The product rule is a differentiation technique used when deriving a function that is the product of two other functions. When you have a function like \( f(x) = g(x) \cdot h(x) \), the product rule states that its derivative is \((g \cdot h)' = g' \cdot h + g \cdot h'\).
Applying the product rule to \( f(t) = t^2 \sin(t) \), we treat \( g(t) = t^2 \) and \( h(t) = \sin(t) \). The derivatives of these components are \( g'(t) = 2t \) and \( h'(t) = \cos(t) \), respectively. Employing the product rule gives us:
\[ f'(t) = 2t \cdot \sin(t) + t^2 \cdot \cos(t) \]

• This rule is especially useful for functions that involve complex products, simplifying the process to find their derivatives accurately.
• Understanding and using the product rule properly is a fundamental skill for solving advanced calculus problems that involve multiple terms.

With the product rule in hand, differentiating products of functions becomes systematically manageable.