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Chapter 11: Problem 22

Solve the equation \(r /(d-r \cos \theta)=e\) for \(r .[\) As stated in the text, you should obtain \(r=e d /(1+e \cos \theta) .]\)

Short Answer

Expert verified
The solution is \(r=\frac{ed}{1+e\cos\theta}\).

Step by step solution

01

Initial Equation

We start with the given equation: \[\frac{r}{d - r \cos \theta} = e\]Our goal is to solve this equation for \(r\).
02

Cross-Multiply

To eliminate the fraction, we cross-multiply:\[r = e (d - r \cos \theta)\]
03

Distribute the Factor

Distribute \(e\) across the terms in parentheses:\[r = ed - er \cos \theta\]
04

Rearrange Terms

Move the term involving \(r\) on the right side to the left side:\[r + er \cos \theta = ed\]This helps us isolate terms containing \(r\).
05

Factor Out r

Factor \(r\) out from the left side:\[r (1 + e \cos \theta) = ed\]
06

Solve for r

Divide both sides by \(1 + e \cos \theta\) to solve for \(r\):\[r = \frac{ed}{1 + e \cos \theta}\]This is the required expression for \(r\).

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Key Concepts

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

Equation Solving
Equation solving is a fundamental skill in precalculus. It involves finding the value(s) of a variable that satisfy the given equation. In this exercise, the goal is to solve the equation \(\frac{r}{d - r \cos \theta} = e\) for \(r\).
The solution process involves several steps, and it's important to approach each step logically. To solve an equation, you need to understand how to manipulate the terms and use different algebraic techniques effectively.
By isolating the desired variable, in this case, \(r\), you systematically work through rearranging and simplifying the equation.
Cross-Multiplication
Cross-multiplication is a useful algebraic technique for solving equations involving fractions. It allows you to eliminate the fraction by multiplying the numerator of one fraction by the denominator of the other. In this problem, we apply cross-multiplication to the equation \(\frac{r}{d - r \cos \theta} = e\).
When you cross-multiply, you have:
  • Multiply \(r\) by 1 (the denominator of \(e\))
  • Multiply \(e\) by \(d - r \cos \theta\)
By doing this, you transform the equation into \(r = e(d - r \cos \theta)\), effectively removing the fraction and simplifying the equation.
Algebraic Manipulation
Algebraic manipulation involves using arithmetic operations and algebraic rules to re-arrange equations and expressions. This concept comes into play after cross-multiplication. It's crucial to methodically distribute, sort, and combine terms to solve for the desired variable.
When you have the equation \(r = e(d - r \cos \theta)\), you apply algebraic manipulation by distributing \(e\) across the terms in the parentheses to get \(r = ed - er \cos \theta\).
Next, by moving terms involving \(r\) to one side, we arrange the equation as \(r + er \cos \theta = ed\), which sets the stage for the next step of factoring.
Factoring in Algebra
Factoring in algebra is a technique used to simplify equations and find solutions. It involves expressing an equation as a product of its factors. In this context, the goal is to isolate \(r\) by factoring it from the terms in the equation \(r + er \cos \theta = ed\).
By factoring out \(r\), we get \(r(1 + e \cos \theta) = ed\), which makes it easier to solve for \(r\).
Lastly, by dividing both sides by \(1 + e \cos \theta\), you obtain the final solution: \(r = \frac{ed}{1 + e \cos \theta}\).
This process not only provides the solution but also strengthens understanding of how to manipulate and solve complex algebraic expressions.

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Most popular questions from this chapter

Simplify the equation: $$\begin{array}{r}\left(\frac{2 x^{\prime}-y^{\prime}}{\sqrt{5}}\right)^{2}+4\left(\frac{2 x^{\prime}-y^{\prime}}{\sqrt{5}}\right)\left(\frac{x^{\prime}+2 y^{\prime}}{\sqrt{5}}\right)-2\left(\frac{x^{\prime}+2 y^{\prime}}{\sqrt{5}}\right)^{2}=6 \\\\\text { Answer: } 2\left(x^{\prime}\right)^{2}-3\left(y^{\prime}\right)^{2}=6\end{array}$$

The small asteroid Icarus was first observed in 1949 at Palomar Observatory in California. Given that the distance from Icarus to the Sun at aphelion is \(1.969 \mathrm{AU}\) and the eccentricity of the orbit is \(0.8269,\) compute the semimajor axis of the orbit and the distance from Icarus to the Sun at perihelion. Round the answers to two decimal places. Remark: One reason for interest in the orbit of Icarus is that it crosses Earth's orbit. On June 14,1968 there was a "close" approach in which Icarus came within approximately 4 million miles of Earth. According to the Jet Propulsion Laboratory, the next close approach will be June \(16,2015,\) at which time Icarus will come within approximately 5 million miles of the Earth.

As is the case for most asteroids in our solar system, the orbit of the asteroid Gaspra is located between the orbits of Mars and Jupiter. Given that its distances from the Sun at perihelion and aphelion are \(2.132 \mathrm{AU}\) and \(2.288 \mathrm{AU}\) respectively, compute the length of the semimajor axis of the orbit and the eccentricity. Round both answers to two decimal places. Remark: Within the past decade the Galileo spacecraft (on a mission to Jupiter) took images of Gaspra, which can be viewed on the World Wide Web. For instance, at the time of this writing, one NASA website containing these images is httpi//www.jpl.nasa.gov/galileo. From this web page, use the search feature there to find "Gaspra."

Let \(\left(x_{1}, y_{1}\right)\) be any point on the ellipse \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1(a>b)\) other than one of the endpoints of the major or minor axis. Show that the normal at \(\left(x_{1}, y_{1}\right)\) does not pass through the origin. Hint: Find the \(y\)-intercept of the normal.

Let \(F_{1}\) and \(F_{2}\) denote the foci of the ellipse \(\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1(a>b) .\) Suppose that \(P\) is one of the endpoints of the minor axis and angle \(F_{1} P F_{2}\) is a right angle. Compute the eccentricity of the ellipse.
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