/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 1 Sketch the slope field for \(y^{... [FREE SOLUTION] | 91Ó°ÊÓ

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Sketch the slope field for \(y^{\prime}=x y / 4\) at the 25 gridpoints \((x, y)\), where \(x=-2,-1, \ldots, 2\) and \(y=-2,-1, \ldots, 2\).

Short Answer

Expert verified
Sketch slopes \( y' = \frac{xy}{4} \) at all grid points with given \( x \) and \( y \) values.

Step by step solution

01

Understand the Differential Equation

The given differential equation is \( y' = \frac{xy}{4} \). This equation describes how the slope \( y' \) changes at different points \((x, y)\) on the coordinate plane. Our task is to sketch the slope field to visualize these slopes at specific grid points.
02

List the Involved Grid Points

We need to find the slope \( y' \) at 25 grid points. These points are generated using \( x = -2, -1, 0, 1, 2 \) and \( y = -2, -1, 0, 1, 2 \). This covers all combinations of these \( x \) and \( y \) values.
03

Calculate Slopes at Grid Points

Calculate \( y' = \frac{xy}{4} \) for each combination of \( x \) and \( y \) grid points. For example, at \((x,y) = (0,0)\), the slope \( y' \) is \( 0 \). At \((x,y) = (1, 2)\), \( y' = \frac{1 \cdot 2}{4} = 0.5 \). Repeat this for all 25 grid points.
04

Sketch Slope Vectors on a Grid

Draw a coordinate grid covering the points from \( x = -2 \) to \( x = 2 \) and \( y = -2 \) to \( y = 2 \). At each grid point \((x, y)\), draw a short line segment with the slope calculated in Step 3. The line's orientation represents the direction of the solution curve at that point.
05

Review the Slope Field

Observe the pattern formed by the line segments. Notice any trends as you move through the grid, such as changes in direction or magnitude. This visual representation helps understand how solutions to the differential equation behave.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Differential Equations
Differential equations are fundamental tools in calculus, used to describe how quantities change. Often, they involve a function and its derivatives, which show rates of change. In our exercise, the differential equation given is \( y' = \frac{xy}{4} \). Here, \( y' \) represents the rate of change of the function \( y \), and \( x \) is an independent variable. The equation describes precisely how the slope, \( y' \), is determined at various points \((x, y)\) on a coordinate plane.
This allows us to examine relationships between the variables: as \( x \) or \( y \) changes, so does the slope. Such equations help model real-world phenomena, predicting everything from population growth to motion of particles. Understanding these relationships is crucial to solving and interpreting differential equations.
Visualization with Slope Fields
Visualization is key to understanding differential equations. Slope fields, also known as direction fields, provide a graphical way of looking at solutions to differential equations, like the one given. They consist of small line segments or arrows drawn at points on a grid. Each segment corresponds to the slope calculated from the differential equation at that point.
For instance, at each grid point in the exercise, we calculate the slope \( y' = \frac{xy}{4} \). A line segment with this slope is then sketched at that point. The overall pattern formed by these segments allows us to visualize how solutions might behave across the entire plane.
Slope fields make it easier to see the broader picture without explicitly solving the equation. They provide intuition about the behavior of solutions, helping us understand whether the system will, for example, stabilize or oscillate.
Calculus and Its Role
Calculus plays a fundamental role in the study and application of differential equations. It provides the mathematical framework for understanding change, which is vital when working with differential equations. Derivatives, a core concept of calculus, denote rates of change and are central in differential equations. In our exercise, the derivative \( y' \) represents the change in \( y \) with respect to \( x \).
With calculus, we learn to compute these derivatives, solve differential equations, and apply them to various fields. This includes physics, engineering, economics, and biology, where they model dynamic systems. For example, the differential equation \( y' = \frac{xy}{4} \) can be approached with calculus techniques to understand solutions' behavior without explicitly solving it.
By studying slope fields within calculus, we gain insights into the 'direction' of solutions. This holistic view is crucial in theoretical and applied aspects of calculus and mathematics.

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Most popular questions from this chapter

Solve the differential equation. If you have a CAS with implicit plotting capability, use the CAS to generate five integral curves for the equation. $$ y^{\prime}=\frac{x^{2}}{1-y^{2}} $$

Solve the differential equation by the method of integrating factors. \(\frac{d y}{d x}+2 x y=x\)

Solve the differential equation by separation of variables. Where reasonable, express the family of solutions as explicit functions of \(x\) $$ \frac{d y}{d x}-\frac{y^{2}-y}{\sin x}=0 $$

Suppose that \(P\) dollars is invested at an annual interest rate of \(r \times 100 \%\). If the accumulated interest is credited to the account at the end of the year, then the interest is said to be compounded annually; if it is credited at the end of each 6-month period, then it is said to be compounded semiannually; and if it is credited at the end of each 3 -month period, then it is said to be compounded quarterly. The more frequently the interest is compounded, the better it is for the investor since more of the interest is itself earning interest. (a) Show that if interest is compounded \(n\) times a year at equally spaced intervals, then the value \(A\) of the investment after \(t\) years is $$ A=P\left(1+\frac{r}{n}\right)^{n t} $$ (b) One can imagine interest to be compounded each day, each hour, each minute, and so forth. Carried to the limit one can conceive of interest compounded at each instant of time; this is called continuous compounding. Thus, from part (a), the value \(A\) of \(P\) dollars after \(t\) years when invested at an annual rate of \(r \times 100 \%\) compounded continuously, is $$ A=\lim _{n \rightarrow+\infty} P\left(1+\frac{r}{n}\right)^{n t} $$ Use the fact that \(\lim _{x \rightarrow 0}(1+x)^{1 / x}=e\) to prove that \(A=P e^{r t}\) (c) Use the result in part (b) to show that money invested at continuous compound interest increases at a rate proportional to the amount present.

Find a solution to the initial-value problem. \(x y^{\prime}+y=e^{x}, y(1)=1+e \quad\) (See Exercise 25.)

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