Textbook Question
Slopes and Tangent Lines
In Exercises 1–4, use the grid and a straight edge to make a rough estimate of the slope of the curve (in y-units per x-unit) at the points P₁ and P₂.
Ch. 3 - Derivatives
Hass15th EditionThomas' CalculusISBN: 9780137616077
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Table of contents
- Ch. 1 - Functions155
- Ch. 2 - Limits and Continuity240
- Ch. 3 - Derivatives337
- Ch. 4 - Applications of Derivatives293
- Ch. 5 - Integrals41
- Ch. 6 - Applications of Definite Integrals25
- Ch. 7 - Transcendental Functions489
- Ch. 8 - Techniques of Integration398
- Ch. 9 - First-Order Differential Equations60
- Ch. 10 - Infinite Sequences and Series119
- Ch. 11 - Parametric Equations and Polar Coordinates58
Chapter 3, Problem 3.3.65
Cylinder pressure If gas in a cylinder is maintained at a constant temperature T, the pressure P is related to the volume V by a formula of the form
P = (nRT / (V − nb)) − (an² / V²),
in which a, b, n, and R are constants. Find dP/dV. (See accompanying figure.)
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Verified step by step guidance1
Identify the given formula for pressure P in terms of volume V: P = (nRT / (V − nb)) − (an² / V²).
Recognize that you need to find the derivative of P with respect to V, denoted as dP/dV.
Apply the derivative rules: For the first term (nRT / (V − nb)), use the quotient rule, which states that if you have a function f(x) = u(x)/v(x), then its derivative f'(x) = (u'(x)v(x) - u(x)v'(x)) / (v(x))².
For the second term (an² / V²), apply the power rule for derivatives, which states that if you have a function f(x) = x^n, then its derivative f'(x) = n*x^(n-1).
Combine the derivatives of both terms to express dP/dV as a single expression, ensuring to simplify where possible.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Differentiation
Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes with respect to one of its variables. In this context, dP/dV represents the derivative of pressure P with respect to volume V, indicating how pressure changes as volume changes. Understanding differentiation is crucial for solving problems involving rates of change.
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Product Rule
The product rule is a technique used in calculus to differentiate functions that are products of two or more functions. It states that the derivative of a product of two functions u and v is given by (du/dx)v + u(dv/dx). In the given formula, the pressure P is expressed as a product of terms involving V, requiring the application of the product rule to find dP/dV.
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05:18The Product Rule
Chain Rule
The chain rule is a method for differentiating composite functions, where one function is nested within another. It is essential when dealing with expressions where variables are interdependent. In the formula for P, terms like (V - nb) and V² involve nested functions, necessitating the use of the chain rule to correctly compute the derivative dP/dV.
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05:02Intro to the Chain Rule
Related Practice
Textbook Question
Slopes, Tangent Lines, and Normal Lines
In Exercises 31–40, verify that the given point is on the curve and find the lines that are (a) tangent and (b) normal to the curve at the given point.
y = 2 sin(πx – y), (1,0)
Textbook Question
If L = √(x² + y²), dx/dt = –1, and dy/dt = 3, find dL/dt when x = 5 and y = 12.
Textbook Question
Find the derivatives of the functions in Exercises 19–40.
g(t) = (1 + sin(3t) / (3 − 2t))⁻¹
Textbook Question
Find the derivatives of the functions in Exercises 19–40.
y = (5 − 2x)⁻³ + (1 / 8)(2 / x + 1)⁴
Textbook Question
Finding Linearizations
In Exercises 1–5, find the linearization L(x) of f(x) at x = a.
f(x) = ∛x, a = −8