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Chapter 28: Problem 37

The general expression for the slope of a curve is \(\frac{\sin x}{3+\cos x} .\) If the curve passes through the point \((\pi / 3,2),\) find its equation.

Short Answer

Expert verified
The equation of the curve is \( y(x) = -\ln|3 + \cos x| + 2 + \ln|3.5| \).

Step by step solution

01

Understand the Derivative as Slope

The expression \( \frac{\sin x}{3 + \cos x} \) is given as the derivative \( \frac{dy}{dx} \) of the curve. This means it describes the slope of the tangent to the curve at any point \( x \). To find the original curve (\( y \)), we need to integrate this expression.
02

Set Up the Integral

To find the function \( y(x) \), we perform indefinite integration of \( \frac{\sin x}{3 + \cos x} \) with respect to \( x \). This gives us the potential function \( y(x) \) plus a constant of integration \( C \, (y = \int \frac{\sin x}{3 + \cos x} dx + C) \).
03

Integrate the Expression

Using the substitution method, let \( u = 3 + \cos x \), so \( du = -\sin x \, dx \), or \( \sin x \, dx = -du \). The integral becomes: \[ \int \frac{\sin x}{3 + \cos x} dx = -\int \frac{1}{u} du = -\ln|u| + C = -\ln|3 + \cos x| + C \].
04

Use Initial Condition

We know the curve passes through the point \( (\pi/3, 2) \). Substitute \( x = \pi/3 \) and \( y = 2 \) into the equation: \[ 2 = -\ln|3 + \cos(\pi/3)| + C \]. \( \cos(\pi/3) = 1/2 \), so \[ 2 = -\ln|3 + 1/2| + C = -\ln|3.5| + C \].
05

Solve for the Constant of Integration

To solve for \( C \), rearrange the equation:\[ 2 + \ln|3.5| = C \].
06

Write the Final Equation

Substitute \( C = 2 + \ln|3.5| \) back into the integrated function: \[ y(x) = -\ln|3 + \cos x| + 2 + \ln|3.5| \]. This is the equation of the curve passing through the given point.

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Key Concepts

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

Integration
Integration is a fundamental concept in calculus that involves finding the function whose derivative is given. In this exercise, we are given a derivative \( \frac{\sin x}{3+\cos x} \) which represents the slope of a curve. To find the original equation of the curve, we need to integrate this expression with respect to \( x \). By calculating the integral, we retrieve the potential function or curve, commonly represented as \( y(x) \), along with an arbitrary constant of integration \( C \), which reflects all possible vertical shifts of the curve.
  • Integration often involves techniques such as substitution to simplify the process. In this problem, we use \( u = 3 + \cos x \) simplifying the process significantly.
  • The result of the integral provides the general form of the function, which is adjusted using additional information, like initial conditions, to find the specific curve.
Derivative
A derivative is a primary concept in calculus, representing the rate of change of a function with respect to a variable. It is usually noted as \( \frac{dy}{dx} \), meaning how much \( y \) changes for a small change in \( x \). In this exercise, the given derivative \( \frac{\sin x}{3+\cos x} \) tells us how steep the curve is at any specific point \( x \).
  • The derivative function is crucial in finding the slope of a tangent to the curve at any point, offering insights into the pattern or behavior of the function across its domain.
  • Understanding derivatives allows us to move backwards through integration to find the original function or curve.
Slope of a curve
The slope of a curve at a point is a measure of how steep the curve is at that specific point. A positive slope indicates an upward curve as \( x \) increases, while a negative slope suggests a downward curve. In the context of calculus, the derivative of a function at a point \( x \) also provides the slope at that point.
  • In this exercise, the slope given by the derivative \( \frac{\sin x}{3+\cos x} \), varies with \( x \), indicating the curve's varying steepness depending on the location.
  • The slope is important for understanding how quickly or slowly the function values are rising or falling as \( x \) changes.
Initial condition
An initial condition is a crucial piece of information that helps determine the specific solution to a differential equation. It provides a known point on the graph of a function, allowing us to solve for the constant of integration \( C \), which appears after performing indefinite integration.
  • In this problem, the initial condition is given by the point \((\pi/3, 2)\).
  • We substitute \( x = \pi/3 \) and \( y = 2 \) into the derived function to find the value of the constant \( C \). This process ensures the function passes through the specified point, uniquely identifying the correct curve.
  • Solving for \( C \) is essential as it completes the function, tailoring it to meet the conditions provided by the problem.

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Most popular questions from this chapter

Solve the given problems by integration. Under certain conditions, the velocity \(v\) (in \(\mathrm{m} / \mathrm{s}\) ) of an object moving along a straight line as a function of the time \(t\) (in s) is given by \(v=\frac{t^{2}+14 t+27}{(2 t+1)(t+5)^{2}} .\) Find the distance traveled by the object during the first \(2.00 \mathrm{s}\).

Solve the given problems by integration. To find the electric field \(E\) from an electric charge distributed uniformly over the entire \(x y\) -plane at a distance \(d\) from the plane, it is necessary to evaluate the integral \(k d \int \frac{d x}{d^{2}+x^{2}} .\) Here, \(x\) is the distance from the origin to the element of charge. Perform the indicated integration.

Solve the given problems by integration. The vertical cross section of a highway culvert is defined by the region within the ellipse \(1.00 x^{2}+9.00 y^{2}=9.00,\) where dimensions are in meters. Find the area of the cross section of the culvert.

Solve the given problems by integration. The time \(t\) and electric current \(i\) for a certain circuit with a voltage \(E\), a resistance \(R\), and an inductance \(L\) is given by \(t=L \int \frac{d i}{E-i R} \cdot\) If \(t=0\) for \(i=0,\) integrate and express \(i\) as a function of \(t\)

Solve the given problems by integration. The current \(i\) (in \(A\) ) as a function of the time \(t\) (in s) in a certain electric circuit is given by \(i=(4 t+3) /\left(2 t^{2}+3 t+1\right) .\) Find the total charge that passes through a point during the first second.
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