dy Let dt Draw a direction field with 9 points to justify your answer. a 2t² + 3y + 1. Which of the following shows a trajectory through (0, 1)? a
dy Let dt Draw a direction field with 9 points to justify your answer. a 2t² + 3y + 1. Which of the following shows a trajectory through (0, 1)? a
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 91E
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Question
![### Differential Equations and Trajectories
#### Problem Statement:
Given the differential equation:
\[ \frac{dy}{dt} = -2t^2 + 3y + 1 \]
which of the following graphs shows a trajectory through the initial point (0, 1)?
#### Task:
Draw a direction field with 9 points to justify your answer.
#### Explanation:
To determine which of the graphs represents the correct trajectory, we start by analyzing the given differential equation:
\[ \frac{dy}{dt} = -2t^2 + 3y + 1 \]
Here, both \(t\) (independent variable) and \(y\) (dependent variable) affect the rate of change of \(y\).
#### Graph Analysis:
There are two graphs provided.
**Graph on the Left:**
- This graph shows a trajectory that slopes downward at first and then becomes more vertical as it progresses downward.
**Graph on the Right:**
- This graph depicts a curve that initially dips down and then sharply rises upward.
#### Direction Field:
To further support our answer, a direction field can be plotted. A direction field will show the slope of the solutions to the differential equation at a grid of points. Each mini-segment at a point represents the slope \(\frac{dy}{dt}\) at that point.
#### Steps to Draw the Direction Field:
1. Choose a set of grid points in the \(t-y\) plane.
2. Calculate \(\frac{dy}{dt}\) at each of these points using the differential equation.
3. Draw a small line segment with the slope \(\frac{dy}{dt}\) at each point.
#### Justification:
By drawing the direction field and plotting the points, we can match the slope directions to the graphs. We specifically look for the behavior around the initial point (0, 1):
- At \(t = 0\), \(y = 1\):
\[
\frac{dy}{dt} = -2(0)^2 + 3(1) + 1 = 3 + 1 = 4
\]
So, at (0,1), the slope is positive.
Compare the directional changes in the trajectory passing through (0,1) with the given graphs.
---
Upon comparison, the right graph, after conducting this analysis, should typically reflect the correct coupled behavior of \(t\) and](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3d3733a5-5e9e-433b-b6d1-2cfec636672d%2Fa4478939-5c5e-4494-bbc8-a27300f66f50%2F821guqw_processed.png&w=3840&q=75)
Transcribed Image Text:### Differential Equations and Trajectories
#### Problem Statement:
Given the differential equation:
\[ \frac{dy}{dt} = -2t^2 + 3y + 1 \]
which of the following graphs shows a trajectory through the initial point (0, 1)?
#### Task:
Draw a direction field with 9 points to justify your answer.
#### Explanation:
To determine which of the graphs represents the correct trajectory, we start by analyzing the given differential equation:
\[ \frac{dy}{dt} = -2t^2 + 3y + 1 \]
Here, both \(t\) (independent variable) and \(y\) (dependent variable) affect the rate of change of \(y\).
#### Graph Analysis:
There are two graphs provided.
**Graph on the Left:**
- This graph shows a trajectory that slopes downward at first and then becomes more vertical as it progresses downward.
**Graph on the Right:**
- This graph depicts a curve that initially dips down and then sharply rises upward.
#### Direction Field:
To further support our answer, a direction field can be plotted. A direction field will show the slope of the solutions to the differential equation at a grid of points. Each mini-segment at a point represents the slope \(\frac{dy}{dt}\) at that point.
#### Steps to Draw the Direction Field:
1. Choose a set of grid points in the \(t-y\) plane.
2. Calculate \(\frac{dy}{dt}\) at each of these points using the differential equation.
3. Draw a small line segment with the slope \(\frac{dy}{dt}\) at each point.
#### Justification:
By drawing the direction field and plotting the points, we can match the slope directions to the graphs. We specifically look for the behavior around the initial point (0, 1):
- At \(t = 0\), \(y = 1\):
\[
\frac{dy}{dt} = -2(0)^2 + 3(1) + 1 = 3 + 1 = 4
\]
So, at (0,1), the slope is positive.
Compare the directional changes in the trajectory passing through (0,1) with the given graphs.
---
Upon comparison, the right graph, after conducting this analysis, should typically reflect the correct coupled behavior of \(t\) and
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