Ch. 3 - Derivatives

Hass15th EditionThomas' CalculusISBN: 9780137616077Not the one you use?Change textbook

Hass 15th Edition

Ch. 3 - Derivatives

Problem 63

Chapter 3, Problem 63

Derivative of y = |x| Graph the derivative of f(x) = |x|. Then graph y = (|x| − 0)/(x − 0) = |x|/x. What can you conclude?

Verified step by step guidance

To find the derivative of y = |x|, consider the piecewise definition of the absolute value function: y = x for x >= 0 and y = -x for x < 0.

Differentiate each piece separately: For x > 0, the derivative of y = x is 1. For x < 0, the derivative of y = -x is -1. At x = 0, the derivative is undefined because the function has a sharp corner.

Graph the derivative: For x > 0, plot a horizontal line at y = 1. For x < 0, plot a horizontal line at y = -1. At x = 0, indicate that the derivative is undefined, often shown as an open circle or a gap.

Now, consider the function y = |x|/x. This is also a piecewise function: y = 1 for x > 0 and y = -1 for x < 0. At x = 0, the function is undefined.

Graph y = |x|/x: For x > 0, plot a horizontal line at y = 1. For x < 0, plot a horizontal line at y = -1. At x = 0, indicate that the function is undefined. Notice that the graph of the derivative of |x| matches the graph of |x|/x, except at x = 0 where both are undefined.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Derivative of Absolute Value Function

The derivative of the absolute value function, y = |x|, is not defined at x = 0 because the function has a sharp corner at this point. For x > 0, the derivative is 1, and for x < 0, the derivative is -1. This discontinuity at x = 0 is crucial for understanding the behavior of the derivative graph.

Piecewise Functions

Piecewise functions are defined by different expressions based on the input value. For f(x) = |x|, it can be expressed as f(x) = x for x ≥ 0 and f(x) = -x for x < 0. Understanding this helps in graphing and analyzing the derivative, as it shows how the function behaves differently on either side of x = 0.

Graphing Rational Functions

Graphing the function y = |x|/x involves understanding how the numerator and denominator affect the graph. This function is defined for all x ≠ 0 and equals 1 for x > 0 and -1 for x < 0, creating a step function. Recognizing this helps in visualizing the discontinuity and the behavior of the function around x = 0.

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Related Practice

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The total surface area S of a right circular cylinder is related to the base radius r and height h by the equation S = 2πr² + 2πrh.

a. How is dS/dt related to dr/dt if h is constant?

Textbook Question

Temperature and the period of a pendulum For oscillations of small amplitude (short swings), we may safely model the relationship between the period T and the length L of a simple pendulum with the equation:

T = 2π√(L/g),

where g is the constant acceleration of gravity at the pendulum’s location. If we measure g in centimeters per second squared, we measure L in centimeters and T in seconds. If the pendulum is made of metal, its length will vary with temperature, either increasing or decreasing at a rate that is roughly proportional to L. In symbols, with u being temperature and k the proportionality constant,

dL/du = kL.

Assuming this to be the case, show that the rate at which the period changes with respect to temperature is kT/2.

Textbook Question

[Technology Exercise]

Graph y = 1/(2√x) in a window that has 0 ≤ x ≤ 2. Then, on the same screen, graph

y = (√(x + h) − √x)/h

for h = 1, 0.5, 0.1. Then try h = −1, −0.5, −0.1. Explain what is going on.

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Find the value of dy/dt at t = 0 if y = 3 sin 2x and x = t² + π.

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If x¹/³ + y¹/³ = 4, find d²y/dx² at the point (8, 8).

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Assume that a particle’s position on the x-axis is given by

x = 3 cos t + 4 sin t,

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b. Find the particle’s velocity when t = 0, t = π/2, and t = π.