Textbook Question
In Exercises 53 and 54, find both dy/dx (treating y as a differentiable function of x) and dx/dy (treating x as a differentiable function of y). How do dy/dx and dx/dy seem to be related?
54. x³ + y² = sin²y
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 3.7.30
In Exercises 29 and 30, find the slope of the curve at the given points.
(x² + y²)² = (x – y)² at (1,0) and (1,–1)
Verified step by step guidance1
First, recognize that the given equation \((x^2 + y^2)^2 = (x - y)^2\) is an implicit function of \(x\) and \(y\). To find the slope of the curve at a given point, we need to find \(\frac{dy}{dx}\) using implicit differentiation.
Differentiate both sides of the equation with respect to \(x\). For the left side, use the chain rule: \(\frac{d}{dx}((x^2 + y^2)^2) = 2(x^2 + y^2) \cdot (2x + 2y \frac{dy}{dx})\). For the right side, use the chain rule: \(\frac{d}{dx}((x - y)^2) = 2(x - y)(1 - \frac{dy}{dx})\).
Set the derivatives equal to each other: \(2(x^2 + y^2)(2x + 2y \frac{dy}{dx}) = 2(x - y)(1 - \frac{dy}{dx})\).
Simplify the equation and solve for \(\frac{dy}{dx}\). This involves expanding both sides, collecting like terms, and isolating \(\frac{dy}{dx}\) on one side of the equation.
Substitute the given points \((1, 0)\) and \((1, -1)\) into the expression for \(\frac{dy}{dx}\) to find the slope of the curve at these points. Evaluate the expression to find the numerical value of the slope at each point.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Implicit Differentiation
Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly solved for one variable in terms of another. In this problem, the equation (x² + y²)² = (x – y)² involves both x and y, requiring implicit differentiation to find dy/dx, the slope of the curve at given points.
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Finding The Implicit Derivative
Chain Rule
The chain rule is a fundamental principle in calculus used to differentiate composite functions. When applying implicit differentiation to the given equation, the chain rule helps differentiate terms like (x² + y²)², where the outer function is raised to a power and the inner function involves both x and y.
Recommended video:
05:02Intro to the Chain Rule
Evaluating Derivatives at Specific Points
After finding the derivative using implicit differentiation, it is crucial to evaluate it at specific points to determine the slope of the curve at those points. For this problem, once dy/dx is obtained, substitute the coordinates (1,0) and (1,–1) into the derivative to find the respective slopes at these points.
Recommended video:
04:50Critical Points
Related Practice
Textbook Question
Derivative Calculations
In Exercises 1–12, find the first and second derivatives.
y = x² + x + 8
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.
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
Falling meteorite The velocity of a heavy meteorite entering Earth’s atmosphere is inversely proportional to √s when it is s km from Earth’s center. Show that the meteorite’s acceleration is inversely proportional to s².