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Chapter 5: Problem 53

Find an equation of the curve that satisfies the given conditions. At each point \((x, y)\) on the curve the slope is \(2 x+1 ;\) the curve passes through the point \((-3,0) .\)

Short Answer

Expert verified
The equation of the curve is \(y = x^2 + x - 6\).

Step by step solution

01

Understand the Problem

We need to find the equation of the curve where the slope at any point \((x, y)\) is given by the expression \(2x + 1\), and the curve passes through the point \((-3, 0)\). This indicates that we must find \(y= f(x)\) such that \(\frac{dy}{dx} = 2x + 1\) and \(f(-3) = 0\).
02

Set Up the Differential Equation

Since the slope of the curve is given as \(2x + 1\), we can set up the differential equation as \(\frac{dy}{dx} = 2x + 1\). To find \(y\), we'll integrate this expression with respect to \(x\).
03

Integrate the Differential Equation

Integrate both sides of the equation \(\frac{dy}{dx} = 2x + 1\). The integral of the right side is \(\int (2x + 1)\,dx = x^2 + x + C\) where \(C\) is the constant of integration. Thus, \(y = x^2 + x + C\).
04

Apply the Initial Condition

Use the condition that the curve passes through the point \((-3, 0)\) to find \(C\). Substitute \(x = -3\) and \(y = 0\) in the equation \(y = x^2 + x + C\): \[0 = (-3)^2 + (-3) + C\]\[0 = 9 - 3 + C\]\[C = -6\].
05

Write the Final Equation

Substitute \(C = -6\) back into the equation for \(y\). The equation of the curve is:\[y = x^2 + x - 6\].

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Key Concepts

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

integration
Integration is an essential concept in calculus, often referred to as the reverse process of differentiation. When given a slope or rate of change, integration helps find the original function or its antiderivative. In our problem, the differential equation \( \frac{dy}{dx} = 2x + 1 \) represents the rate of change of \( y \) with respect to \( x \).

Here are some important points about integration:
  • The integration of a function \( f(x) \) with respect to \( x \) is denoted as \( \int f(x) \, dx \).
  • The result of an indefinite integral is a family of functions, represented as \( F(x) + C \), where \( C \) is the constant of integration arising from the general solution of the differential equation.
  • For the differential equation \( \frac{dy}{dx} = 2x + 1 \), integrating both sides with respect to \( x \) involves finding \( \int (2x + 1) \, dx \), which equals \( x^2 + x + C \).
To connect it back to the problem, once we integrate, we obtain \( y = x^2 + x + C \), a general solution where \( C \) is determined by initial conditions.
initial value problem
An initial value problem involves solving a differential equation with a given starting point, or initial condition, which in this case is the point \((-3, 0)\). Initial conditions help to find the specific solution from the family of solutions provided by integration.

Let's break it down:
  • The initial condition places an extra requirement on the function \( y = f(x) \). It is a way to select one particular solution curve from infinitely many that satisfy the differential equation.
  • To use the initial condition \((-3, 0)\), substitute \( x = -3 \) and \( y = 0 \) into \( y = x^2 + x + C \).
  • The goal is to solve for the constant \( C \):
    \[ 0 = (-3)^2 + (-3) + C \]
  • Solving this, we get \( C = -6 \).
Thus, the specific solution to the problem, respecting the initial condition, is \( y = x^2 + x - 6 \).
slope field
A slope field, also known as a direction field, is a visual representation of a differential equation of the form \( \frac{dy}{dx} = f(x, y) \). It's like a map of small slopes drawn at grid points in the plane that show how the slope of a solution curve behaves at different points.

Important features of slope fields include:
  • Each line segment in a slope field represents the slope of the solution curve at that point.
  • They help visualize the family of solutions generated from a differential equation.
  • For our differential equation \( \frac{dy}{dx} = 2x + 1 \), the slope at any point \((x, y)\) depends only on \( x \), showing uniform change across horizontal lines.
Using a slope field, we could visually predict how solutions evolve across the plane. In practical terms, observing the slope field of \( 2x + 1 \) would provide insight into how the curve, evaluated from \( y = x^2 + x + C \), behaves globally.

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Most popular questions from this chapter

Sketch the curve and find the total area between the curve and the given interval on the \(x\) -axis. $$ y=\frac{x^{2}-1}{x^{2}} ;\left[\frac{1}{2}, 2\right] $$

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CASprograms have commands for working with most of the important nonelementary functions. Check your CAS documentantion for information about the error function erf(x) [see Formula (12)], and then complete the following. (a) Generate the graph of erf(x). (d) Use the graph to make a conjecture about the existence and location of any inflection points of erf \((x) .\) (e) Check your conjecture in part (d) using the second derivative of erf(x). (f) Use the graph to make a conjecture about the existence of horizontal asymptotes of erf(x). (g) Check your conjecture in part (f) by using the CAS to find the limits of erf(x) as \(x \rightarrow \pm \infty\). (b) Use the graph to make a conjecture about the existence and location of any relative maxima and minima of erf(x). (c) Check your conjecture in part (b) using the derivative of erf(x).

Let \(F(x)=\int_{\sqrt{3}}^{x} \tan ^{-1} t d t .\) Find $$ \begin{array}{llll}{\text { (a) } F(\sqrt{3})} & {\text { (b) } F^{\prime}(\sqrt{3})} & {\text { (c) } F^{\prime \prime}(\sqrt{3})} & {\text { . }}\end{array} $$
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