Problem 21
Question
A slope field of the form \(y^{\prime}=f(y)\) is said to be \(autonomous.\) (a) Explain why the tangent segments along any horizontal line will be parallel for an autonomous slope field. (b) The word autonomous means "independent." In what sense is an autonomous slope field independent? (c) Suppose that \(G(y)\) is an antiderivative of \(1 /[f(y)]\) and that \(C\) is a constant. Explain why any differentiable function defined implicitly by \(G(y)-x=C\) will be a solution to the equation \(y^{\prime}=f(y).\)
Step-by-Step Solution
Verified Answer
(a) Slopes depend only on \(y\), causing parallel tangents on horizontal lines. (b) The slope is independent of \(x\). (c) \(G(y) - x = C\) matches the differential equation solution form.
1Step 1: Understand Autonomous Slope Fields
A slope field is a graphical representation of differential equations of the form \(y' = f(y)\). An autonomous slope field implies that the derivative \(y'\) only depends on \(y\), not on \(x\). This means the slope equation is independent of \(x\).
2Step 2: Explain Horizontal Line Parallelism
In an autonomous slope field, since \(y' = f(y)\) depends only on \(y\), each horizontal line (constant \(y\)) will have the same slope value. This implies that all tangent segments along any horizontal line are parallel because they have the same slope at every point where \(y\) is constant.
3Step 3: Define Independence in Autonomous Slope Fields
Autonomous slope fields are independent in the sense that the slope \(y'\) at any point is determined solely by the value of \(y\), regardless of \(x\). There is no dependency on the horizontal position, making these slope fields 'autonomous' or 'self-governing'.
4Step 4: Discuss Implicit Differentiable Function Solution
An antiderivative \(G(y)\) of \(1/[f(y)]\) implies that integrating \(1/[f(y)]\) gives \(G(y)\). The expression \(G(y) - x = C\) is an implicit function form of the solution which means \(d(G(y))/dy = 1/f(y)\). Differentiating \(G(y) - x = C\) with respect to \(x\) yields \(G'(y)y' - 1 = 0\) or \(y' = 1/G'(y) = f(y)\). This shows that \(G(y) - x = C\) solves \(y' = f(y)\).
Key Concepts
Understanding Differential EquationsTangent Segments in Autonomous Slope FieldsAntiderivatives in Solving Differential Equations
Understanding Differential Equations
Differential equations are fundamental in mathematics, describing a relationship involving an unknown function and its derivatives. They allow us to model various natural phenomena, from physics to economics. At the core, a differential equation has the form \( y' = f(x, y) \), where \( y' \) suggests the rate of change of \( y \) with respect to \( x \).
In an autonomous slope field, the differential equation simplifies to \( y' = f(y) \), indicating the derivative depends solely on \( y \), not on \( x \). This simplifies the analysis since the focus is on how \( y \) changes, independently of the "location" \( x \). Recognizing whether a differential equation is autonomous or not is essential as it implies specific properties in the slope field, such as parallel tangent segments along horizontal lines.
In an autonomous slope field, the differential equation simplifies to \( y' = f(y) \), indicating the derivative depends solely on \( y \), not on \( x \). This simplifies the analysis since the focus is on how \( y \) changes, independently of the "location" \( x \). Recognizing whether a differential equation is autonomous or not is essential as it implies specific properties in the slope field, such as parallel tangent segments along horizontal lines.
- Autonomous differential equations: Depend only on \( y \).
- Non-autonomous differential equations: Depend on both \( x \) and \( y \).
Tangent Segments in Autonomous Slope Fields
Tangent segments are small line segments that illustrate the slope or direction of the solution to a differential equation at various points. Each segment provides a snapshot of how the curve behaves at that point.
In autonomous slope fields, tangent segments have a unique property: they are parallel along any horizontal line. Since the equation \( y' = f(y) \) dictates that the slope only depends on \( y \), not on \( x \), the slope value for a particular \( y \) remains constant across all horizontal lines. This results in parallel segments because:
In autonomous slope fields, tangent segments have a unique property: they are parallel along any horizontal line. Since the equation \( y' = f(y) \) dictates that the slope only depends on \( y \), not on \( x \), the slope value for a particular \( y \) remains constant across all horizontal lines. This results in parallel segments because:
- The slope function \( f(y) \) creates uniform slope values across each horizontal line.
- This uniformity implies that any changes in the function are solely due to changes in \( y \), not \( x \).
Antiderivatives in Solving Differential Equations
The concept of an antiderivative is central in solving differential equations, particularly when it comes to finding solution curves. An antiderivative of a function \( f(y) \) is a function \( G(y) \) such that \( G'(y) = f(y) \). Finding an antiderivative helps us reverse the process of differentiation to understand the original function's behavior.
In relation to autonomous slope fields, suppose \( G(y) \) is an antiderivative of \( 1/[f(y)] \). The implicit form \( G(y) - x = C \) represents a family of differentiable functions where \( C \) is a constant. This form effectively captures the solution to \( y' = f(y) \) because:
In relation to autonomous slope fields, suppose \( G(y) \) is an antiderivative of \( 1/[f(y)] \). The implicit form \( G(y) - x = C \) represents a family of differentiable functions where \( C \) is a constant. This form effectively captures the solution to \( y' = f(y) \) because:
- Integrating \( 1/[f(y)] \) involves determining \( G(y) \), allowing us to reinterpret the differential relationship.
- Given \( G(y) - x = C \), differentiating both sides concerning \( x \) leads back to \( y' = f(y) \), confirming it as a valid solution.
Other exercises in this chapter
Problem 21
Find a solution to the initial-value problem. $$ y^{\prime}+4 x=2, y(0)=3 $$
View solution Problem 21
At time \(t=0,\) a tank contains 25 oz of salt dissolved in 50 gal of water. Then brine containing 4 oz of salt per gallon of brine is allowed to enter the tank
View solution Problem 22
Find a solution to the initial-value problem. $$ y^{\prime \prime}+6 x=0, y(0)=1, y^{\prime}(0)=2 $$
View solution Problem 22
A tank initially contains 200 gal of pure water. Then at time \(t=0\) brine containing 5 lb of salt per gallon of brine is allowed to enter the tank at a rate o
View solution