Problem 17 - Calculus

Question

Graph the function defined by \(g(r)=\left\\{\begin{array}{lll}1+\cos (\pi r / 2) & \text { for } & -2 \leq r \leq 2 \\ 0 & \text { for } & r<-2 & \text { or } \quad r>2\end{array}\right.\) (a) Is \(g\) continuous at \(r=2 ?\) Explain your answer. (b) Do you think \(g\) is differentiable at \(r=2 ?\) Explain your answer.

Step-by-Step Solution

Answer

(a) Yes, \( g \) is continuous at \( r=2 \). (b) Yes, \( g \) is differentiable at \( r=2 \).

Step 1: Understand the Function

The function \( g(r) \) is piecewise with two parts: For \( -2 \leq r \leq 2 \), \( g(r) = 1 + \cos(\pi r / 2) \). For \( r < -2 \) and \( r > 2 \), \( g(r) = 0 \). Notice that the first part uses the cosine function.

Step 2: Evaluate at \( r = 2 \)

Calculate \( g(2) \) by substituting \( r = 2 \) into the first piece: \( g(2) = 1 + \cos(\pi \times 2 / 2) = 1 + \cos(\pi) = 1 - 1 = 0 \).Check the value of \( g(r) \) for \( r > 2 \): According to the definition, \( g(r) = 0 \) for \( r > 2 \). Hence \( g(2) = 0 \) is already zero as expected from the piecewise definition.

Step 3: Determine Continuity at \( r = 2 \)

For \( g \) to be continuous at \( r = 2 \), the limit needs to exist and be equal to \( g(2) \). Calculate the limit from both sides. For the left-hand limit as \( r \to 2^- \), \( g(r) = 1 + \cos(\pi r / 2) \). As \( r \to 2 \), \( g(r) \to 0 \).The right-hand limit as \( r \to 2^+ \) is \( g(r) = 0 \). Thus, both limits match and are equal to \( g(2) \). Thus, \( g \) is continuous at \( r = 2 \).

Step 4: Evaluate Differentiability at \( r = 2 \)

Next, check if \( g \) is differentiable at \( r = 2 \), which requires that the derivative exists and is the same from both sides.For \( -2 \leq r < 2 \), the derivative \( g'(r) = -(\pi/2) \sin(\pi r / 2) \).For \( r > 2 \), the derivative \( g'(r) = 0 \), as the function is constant.The derivative from the left as \( r \to 2^- \) is \( -(\pi/2)\sin(\pi \times 2 / 2) = 0 \).From the right as \( r \to 2^+ \), the derivative is already 0.Since the derivatives match, \( g \) is, surprisingly, differentiable at \( r = 2 \).

Key Concepts

Continuity

A piecewise function is said to be continuous at a particular point if it does not "break" or "jump" at that point. This means that the function’s value and the approaching values from both sides of that point are the same. In the exercise, we analyze the function \( g(r) \) at \( r = 2 \) to determine its continuity. The function is defined in two parts, and we care most about the behavior as \( r \) approaches 2 from the negative and positive directions.

• Left-hand limit \((r \to 2^-)\): \( g(r) = 1 + \cos(\pi r / 2) \) approaches 0.
• Right-hand limit \((r \to 2^+)\): \( g(r) = 0 \).

Since both limits are equal to \( g(2) = 0 \), the function does not have any jumps or sudden changes at this point. Therefore, the function \( g(r) \) is continuous at \( r = 2 \). Continuity ensures a smooth transition which is a crucial aspect when graphing or analyzing functions in calculus.

Differentiability

For a function to be differentiable at a point, it means there’s a defined tangent or slope at that precise spot. Differentiability is a step further from continuity, involving the existence of a derivative. For \( g(r) \) at \( r = 2 \), we need to verify if the derivative from either side matches and is well-defined.

• The derivative from the left \((r \to 2^-)\): calculated as \(-\frac{\pi}{2} \sin(\pi r / 2)\), which reaches 0 at \( r = 2 \).
• The derivative from the right \((r \to 2^+)\): Since the function is constant \( (0) \), the derivative is 0.

Both derivatives being zero confirms that \( g(r) \) is differentiable at \( r = 2 \). This outcome might seem unexpected given that piecewise functions often introduce discontinuities or abrupt changes, yet this highlights an exception where careful design maintains differentiability.

Calculus

Calculus provides powerful tools such as derivatives and limits to study functions thoroughly. When evaluating \( g(r) \) in the context of this exercise, calculus allows us to dissect the function into its components, observing behavior at specific points like \( r = 2 \). This piecewise function demonstrates key calculus concepts:

• Continuity: Ensures function cohesion and smoothness, verified by matching limits and defined function values.
• Differentiability: Describes the function's readiness for slope calculation, which transitions from continuity by considering derivatives.
• Piecewise Definition: Illustrates functions that follow different rules in separate sections, common in real-world applications requiring calculus for analysis.

Such concepts are fundamental in calculus, empowering us to handle complex functions by breaking them into understandable parts to explore properties like continuity and differentiability.