Problem 66 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that the curve $$ y=\left\\{\begin{array}{rll} 0 & \text { if } & x \leq 0 \\ x^{3} & \text { if } & x>0 \end{array}\right. $$ has continuous first derivatives and curvature at all points.

Step-by-Step Solution

Answer

The curve is continuous in both derivatives and curvature at all points.

Step 1: Understand the Piecewise Function

The curve is given by a piecewise function. For \(x \leq 0\), \(y = 0\). For \(x > 0\), \(y = x^3\). The function changes its definition at \(x = 0\). To determine continuity, we need to ensure that both the function and its derivative are continuous at \(x = 0\).

Step 2: Check Continuity of the Function

To check continuity at \(x = 0\), evaluate the left-hand limit and right-hand limit for \(y\). The left-hand limit as \(x\) approaches 0 from the negative side is 0. The right-hand limit as \(x\) approaches 0 from the positive side is also 0 (since \(0^3 = 0\)). Thus, the function is continuous at \(x = 0\) as both limits are equal and \(f(0) = 0\).

Step 3: Find the Derivative of \(y\)

First, differentiate each piece of the piecewise function separately. The derivative of \(y = 0\) for \(x \leq 0\) is 0. For \(x > 0\), differentiate \(y = x^3\) to get \(y' = 3x^2\). Now check if these derivatives match at \(x = 0\).

Step 4: Check Continuity of the Derivative at \(x = 0\)

Evaluate the derivative at \(x = 0\). The left-hand limit of the derivative as \(x\) approaches 0 is 0. The right-hand limit is also 0 since \(3(0)^2 = 0\). Since both limits are equal and \(y'(0) = 0\), the derivative is continuous at \(x = 0\).

Step 5: Examine the Curvature

The curvature \(k\) is given by the formula \(k = \frac{|y''|}{(1 + (y')^2)^{3/2}}\). First, find \(y''\): for \(x \leq 0\), \(y'' = 0\). For \(x > 0\), \(y'' = 6x\). At \(x = 0\), \(y''\) from the right-hand side approaches 0. Since all derivatives are 0 at \(x = 0\), \(k = 0\). Thus, the curvature is well-defined and continuous at \(x = 0\).

Step 6: Final Verification

The first derivatives and curvature of the curve are continuous at all points because the function and its derivatives coincide at the transition point \(x = 0\) and the curvature does not present a singularity.

Key Concepts

Understanding Piecewise Functions

A piecewise function is a function composed of multiple sub-functions, each defined over a particular interval. These sub-functions are put together to create one entire function but are defined by different formulas depending on the input value. In this exercise, the curve is defined with a piecewise function:

• For any value of \( x \leq 0 \), the function is \( y = 0 \).
• For values of \( x > 0 \), the function is \( y = x^3 \).
• The transition occurs at \( x = 0 \).

At the boundary, the function must be checked for continuity, meaning that the left-hand limit, right-hand limit, and the function value at the point must all be identical to ensure there is no jump in the graph. In our case, at \( x = 0 \), this means both limits and \( f(0) \) must equal 0.

Examining Curvature

Curvature refers to how sharply a curve bends at a particular point. A straight line has zero curvature, whereas a curve with sharp twists has high curvature. For a function, the curvature \( k \) is mathematically defined as:\[k = \frac{|y''|}{(1 + (y')^2)^{3/2}}\]To compute curvature:

• We calculate the second derivative \( y'' \).
• For \( x \leq 0 \), \( y'' = 0 \), since the original function is a flat line (\( y = 0 \)).
• For \( x > 0 \), \( y'' = 6x \). Near \( x = 0 \), the second derivative, and hence curvature, bow towards zero.

Simply put, since both the first and second derivatives become zero at \( x = 0 \), the curvature at this transition point is well-defined and continuous, making it smooth across the function.

Continuous Derivatives

For a function to be nicely behaved, one key property it must have is continuous derivatives. This means its derivative, such as slope (\( y' \), the first derivative), doesn't jump at any point. A continuous derivative ensures that as you move along the graph, there are no abrupt changes.

• The first derivative of \( y = 0 \) (for \( x \leq 0 \)) is \( y' = 0 \).
• For \( x > 0 \), differentiate \( y = x^3 \) resulting in \( y' = 3x^2 \).
• At \( x = 0 \), the limits of these derivatives from both left and right approach 0, ensuring continuity.

By making sure the derivatives match at the interface point \( x = 0 \), we confirm a smooth transition without bumps or breaks in the slope, guaranteeing a continuous blend between the two pieces of the function.