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Chapter 8: Problem 16

Use the window \([-2,2] \times[-2,2]\) to sketch a direction field for the following equations. Then sketch the solution curve that corresponds to the given initial condition. \(A\) detailed direction field is not needed. $$y^{\prime}(x)=\sin y, y(-2)=\frac{1}{2}$$

Short Answer

Expert verified
Answer: The solution curve that passes through the initial condition \(y(-2) = \frac{1}{2}\) bends slightly away from regions with negative slopes (i.e., \(y<-2\)), but remains in the same general direction since the slopes are positive for \(y>\frac{1}{2}\).

Step by step solution

01

Plot the direction field of the given differential equation

For each point \((x,y)\) in the given window, we only need to determine the slope of the tangent line to the solution curve at that point. The slope is equal to \(y'(x) = \sin y\). We then sketch short line segments at various points in the window, using the slopes determined from the equation to indicate the direction. Most importantly, we focus on the points nearest the initial condition \(y(-2) = \frac{1}{2}\).
02

Determine the behavior of the solution curve around the initial condition$

Now we need to find the solution curve that matches the initial condition \(y(-2) = \frac{1}{2}\). We will evaluate the slope \(y'= \sin y\) at these points: a) \(y=-2\): \(y'=\sin (-2)<0\) b) \(y=\frac{1}{2}\): \(y'=\sin (\frac{1}{2})>0\) c) \(y=2\): \(y'=\sin (2)>0\) We start at the initial condition \((-2, \frac{1}{2})\), where the slope is positive as per point b. Looking at how the slope changes in positive and negative direction of x-axis: - As x increases and \(y> \frac{1}{2}\), it remains positive (per point c). - As x increases and \(y< \frac{1}{2}\), it becomes negative (per point a), indicating that the solution curve bends away from the region below the line \(y=\frac{1}{2}\), and go back toward it. - As x decreases, the slope changes in similar way, i.e., it remains positive for \(y> \frac{1}{2}\) and becomes negative for \(y< \frac{1}{2}\). With this information, we can now sketch the solution curve starting at the initial condition \(y(-2)=\frac{1}{2}\). The solution curve should bend slightly away from regions with negative slopes (i.e., \(y<-2\)), but remain in the same general direction since the slopes are positive for \(y>\frac{1}{2}\).

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Key Concepts

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

Differential Equations
A differential equation is a mathematical equation that involves functions and their derivatives. It represents how a particular quantity changes as it depends on one or more variables, usually represented by derivatives such as \( y' \) which is the derivative of \( y \) with respect to \( x \). Differential equations are fundamental in modeling scenarios where change is continuous such as in physics, engineering, and biological systems.
In our exercise, the differential equation given is \( y'(x) = \sin(y) \). This means that the rate of change of \( y \) with respect to \( x \) is determined by the function \( \sin(y) \). Understanding how these equations work helps us predict how a system evolves over time or space. The solution to a differential equation is a function or a set of functions that satisfies the equation within a certain domain.
Initial Condition
An initial condition is a value or a set of values, given at a certain point, that specifies the state of a system at that point. It helps to determine a unique solution of a differential equation from the family of possible solutions. Without an initial condition, a differential equation can have multiple solutions.
In this problem, the initial condition is given as \( y(-2) = \frac{1}{2} \). This condition tells us that at \( x = -2 \), the value of \( y \) is \( \frac{1}{2} \). It is crucial to know this initial starting point because it provides the information necessary to sketch the specific solution curve that corresponds to this particular setup. Initial conditions are used to solve initial value problems in differential equations.
Slope Field
A slope field, also known as a direction field, is a graphical representation of a differential equation in the plane. It visualizes the slopes or directions of the solution curves for different values at various points across a grid.
To create a slope field for the equation \( y'(x) = \sin(y) \), you evaluate the derivative \( y' \) at several grid points within the window \([-2,2] \times [-2,2]\). Each point on the grid gets a small line segment with a slope equal to \( \sin(y) \) at that point. This helps you visualize how the solution curves might flow across the plane.
For instance, at \( (x, y) = (0, \frac{\pi}{2}) \), \( y' = \sin(\frac{\pi}{2}) = 1 \), indicating steep positive slopes. Meanwhile, at \( (x, y) = (0, 0) \), \( y' = \sin(0) = 0 \), where the slope is horizontal. Slope fields are crucial in understanding the overall behavior and tendencies of solutions to differential equations.
Solution Curve
A solution curve is the curve in the plane that represents the solution to a differential equation that also satisfies a given initial condition. Using the initial condition as a starting point, you trace along the slopes indicated by the slope field to visualize how the function behaves.
In our exercise, we were instructed to sketch the solution curve starting from \( (-2, \frac{1}{2}) \). Here, we first look at the slope at this initial point, which determines the initial angle of the tangent to the curve. As per our slope field, the slope \( y' = \sin(\frac{1}{2}) \) is positive, meaning the curve starts by moving upwards.
A solution curve will weave through the slope field, adjusting its path in response to the positive or negative slope values, showing us visually how the equation's solutions behave. The curve helps in understanding not just the correct direction of solutions but also how those solutions evolve over the window examined, such as \([-2, 2] \times [-2, 2]\) in this scenario.

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Most popular questions from this chapter

Consider a loan repayment plan described by the initial value problem $$B^{\prime}(t)=0.03 B-600, \quad B(0)=40,000$$ where the amount borrowed is \(B(0)=\$ 40,000,\) the monthly payments are \(\$ 600,\) and \(B(t)\) is the unpaid balance in the loan. a. Find the solution of the initial value problem and explain why \(B\) is an increasing function. b. What is the most that you can borrow under the terms of this loan without going further into debt each month? c. Now consider the more general loan repayment plan described by the initial value problem $$B^{\prime}(t)=r B-m, \quad B(0)=B_{0}$$ where \(r>0\) reflects the interest rate, \(m>0\) is the monthly payment, and \(B_{0}>0\) is the amount borrowed. In terms of \(m\) and \(r,\) what is the maximum amount \(B_{0}\) that can be borrowed without going further into debt each month?

For the following separable equations, carry out the indicated analysis. a. Find the general solution of the equation. b. Find the value of the arbitrary constant associated with each initial condition. (Each initial condition requires a different constant.) c. Use the graph of the general solution that is provided to sketch the solution curve for each initial condition. $$\begin{array}{l} e^{-y / 2} y^{\prime}(x)=4 x \sin x^{2}-x ; y(0)=0 \\ y(0)=\ln \left(\frac{1}{4}\right), y(\sqrt{\frac{\pi}{2}})=0 \end{array}$$

Two curves are orthogonal to each other if their tangent lines are perpendicular at each point of intersection. A family of curves forms orthogonal trajectories with another family of curves if each curve in one family is orthogonal to each curve in the other family. Use the following steps to find the orthogonal trajectories of the family of ellipses \(2 x^{2}+y^{2}=a^{2}\) a. Apply implicit differentiation to \(2 x^{2}+y^{2}=a^{2}\) to show that $$ \frac{d y}{d x}=\frac{-2 x}{y} $$ b. The family of trajectories orthogonal to \(2 x^{2}+y^{2}=a^{2}\) satisfies the differential equation \(\frac{d y}{d x}=\frac{y}{2 x} .\) Why? c. Solve the differential equation in part (b) to verify that \(y^{2}=e^{C}|x|\) and then explain why it follows that \(y^{2}=k x\) Therefore, the family of parabolas \(y^{2}=k x\) forms the orthogonal trajectories of the family of ellipses \(2 x^{2}+y^{2}=a^{2}\)

Solve the following initial value problems and leave the solution in implicit form. Use graphing software to plot the solution. If the implicit solution describes more than one curve, be sure to indicate which curve corresponds to the solution of the initial value problem. $$z^{\prime}(x)=\frac{z^{2}+4}{x^{2}+16}, z(4)=2$$

For each of the following stirred tank reactions, carry out the following analysis. a. Write an initial value problem for the mass of the substance. b. Solve the initial value problem and graph the solution to be sure that \(m(0)\) and \(\lim _{t \rightarrow \infty} m(t)\) are correct. A one-million-liter pond is contaminated and has a concentration of \(20 \mathrm{g} / \mathrm{L}\) of a chemical pollutant. The source of the pollutant is removed and pure water is allowed to flow into the pond at a rate of \(1200 \mathrm{L} / \mathrm{hr} .\) Assuming that the pond is thoroughly mixed and drained at a rate of \(1200 \mathrm{L} / \mathrm{hr}\), how long does it take to reduce the concentration of the solution in the pond to \(10 \%\) of the initial value?
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