/*! 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 67 The following exercises require ... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 67

The following exercises require the use of a slope field program. For each differential equation and initial condition: a. Use a graphing calculator slope field program to graph the slope field for the differential equation on the window [-5,5] by [-5,5] b. Sketch the slope field on a piece of paper and draw a solution curve that follows the slopes and that passes through the point (0,2) c. Solve the differential equation and initial condition. d. Use your slope field program to graph the slope field and the solution that you found in part (c). How good was the sketch that you made in part (b) compared with the solution graphed in part (d)? $$ \left\\{\begin{array}{l} \frac{d y}{d x}=\frac{4 x}{y^{3}} \\ y(0)=2 \end{array}\right. $$

Short Answer

Expert verified
The solution to the differential equation is \( y = \sqrt[4]{8x^2 + 16} \).

Step by step solution

01

Graph Slope Field using a Graphing Calculator

Input the differential equation \( \frac{d y}{d x} = \frac{4x}{y^3} \) into the graphing calculator that can perform slope fields. Set the window to be from -5 to 5 on both axes. Observe the slope field that is generated by the calculator.
02

Sketch and Draw a Solution Curve

On a piece of paper, sketch the slope field observed from the calculator. Draw a curve that follows the slope field and passes through the initial condition point (0,2). Pay attention to the direction and curvature suggested by the slopes.
03

Solve the Differential Equation

To solve \( \frac{d y}{d x} = \frac{4x}{y^3} \) with initial condition \( y(0) = 2 \), separate variables: \( y^3 \, dy = 4x \, dx \). Integrate both sides: \( \int y^3 \, dy = \int 4x \, dx \). This results in \( \frac{y^4}{4} = 2x^2 + C \). Solve for \( y \) to determine the function. Use the initial condition \( y(0) = 2 \) to find \( C \). Substitute \( x = 0 \) and \( y = 2 \): \( \frac{2^4}{4} = 2(0)^2 + C \Rightarrow C = 4 \). Then, solve for \( y \): \( y^4 = 8x^2 + 16 \Rightarrow y = \sqrt[4]{8x^2 + 16} \).
04

Graph the Solution on Slope Field Program

Input the solution \( y = \sqrt[4]{8x^2 + 16} \) into the slope field program. Compare this graph with the sketch made in Step 2. Reflect on the accuracy of your sketch based on the computer-generated solution graph.

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Key Concepts

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

Slope Fields
Slope fields are a great way to visually grasp differential equations without solving them explicitly. Essentially, a slope field is a graph that shows tiny line segments or slopes at various points, which represent the slope of the solution curve of a differential equation at those points. This visual representation provides insight into how a solution might behave over an area.

When plotting a slope field for a differential equation like \( \frac{d y}{d x} = \frac{4x}{y^3} \), we take each point \((x, y)\) in our chosen window \([-5,5]\) by \([-5,5]\) and calculate the slope at that point using the equation. The slope dictates the direction and angle of the small line that represents the differential equation there.

The biggest benefit of slope fields is their ability to show at a glance how different initial conditions might lead to different types of behavior in solutions. This is particularly useful in understanding how solutions may change when initial conditions are varied.
Graphing Calculators
Graphing calculators are an essential tool for visualizing mathematical concepts, especially in calculus and differential equations. They help students and professionals by allowing them to input complex equations and instantly receive a graphical output.

To graph a slope field using a graphing calculator, the differential equation needs to be entered in a specific format that the calculator can interpret. For the equation \( \frac{d y}{d x} = \frac{4x}{y^3} \), you would input it using the slope field function of your calculator. It's essential to set the viewing window appropriately, such as from \(-5\) to \(5\) on both axes, to ensure that the slope field captures the necessary range.Graphing calculators are also invaluable when comparing your hand-drawn sketches to computer-generated graphs. They provide an exact graphical representation, allowing you to see how well your manual drawing matches the calculated result. This process helps in improving manual estimation skills and understanding of differential equations.
Initial Conditions
Initial conditions play a crucial role in differentiating solutions of differential equations. Given a differential equation, its solution isn't unique unless you specify initial conditions. These conditions determine a specific path or solution curve from the infinite possibilities provided by the slope field.

For example, consider the equation \( \frac{d y}{d x} = \frac{4x}{y^3} \) with initial condition \( y(0) = 2 \). This condition means that at \( x = 0 \), the solution must pass through the point \( (0,2) \). Without this point, we couldn't precisely define which solution curve to follow among the many represented in the slope field.Detailed initial conditions make it possible to solve differential equations analytically, as they allow us to find any constant terms during integration. In this instance, using the initial condition, we calculated \( C \), which resulted in the specific solution \( y = \sqrt[4]{8x^2 + 16} \). Initial conditions ensure that our solution fits perfectly on the slope field.

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Most popular questions from this chapter

BIOMEDICAL: Dieting A person's weight \(w(t)\) after days of eating \(c\) calories per day can be modeled by the following differential equation $$ w^{\prime}+0.005 w=\frac{c}{3500} $$ where the 0.005 represents the proportional weight loss per day when eating nothing, and 3500 is the conversion rate for calories into pounds. a. If a person initially weighing 170 pounds goes on a diet of 2100 calories per day, find a formula for the person's weight after \(t\) days. b. Use your solution to find when the person will have lost 15 pounds. c. Find the "limiting weight" that will be a approached if the person continues on this diet indefinitely.

An algae bloom in a lake is a sudden growth of algae that consumes nutrients and blocks sunlight, killing other life in the water. Suppose that the number of tons \(y(t)\) of algae after \(t\) weeks satisfies $$ \begin{array}{l} y^{\prime}=t y+t \\ y(0)=2 \end{array} $$ a. Solve this differential equation and initial condition. b. Use your solution to find the amount of algae after 2 weeks. c. Graph the solution on a graphing calculator and find when the algae bloom will reach 40 tons.

For each initial value problem, use an Euler's method graphing calculator program to find the approximate solution at the stated \(x\) -value, using 50 segments. [Hint: Use an interval that begins at the initial \(x\) -value and ends at the stated \(x\) -value. \(y^{\prime}=x e^{-y}\) \(y(1)=0.5\) Approximate the solution at \(x=3\)

Solve each by the appropriate technique. a. \(y y^{\prime}=x\) b. \(y^{\prime}+y=e^{-x}\)

A medical examiner called to the scene of a murder will usually take the temperature of the body. A corpse cools at a rate proportional to the difference between its temperature and the temperature of the room. If \(y(t)\) is the temperature (in degrees Fahrenheit) of the body \(t\) hours after the murder, and if the room temperature is \(70^{\circ},\) then \(y\) satisfies $$ \begin{aligned} y^{\prime} &=-0.32(y-70) \\ y(0) &=98.6 \text { (body temperature initially } \left.98.6^{\circ}\right) \end{aligned}$$ a. Solve this differential equation and initial condition. b. Use your answer to part (a) to estimate how long ago the murder took place if the temperature of the \text { body when it was discovered was } 80^{\circ} .
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