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Chapter 9: Problem 78

In calculus, when finding the derivative of equations in two variables, we typically use implicit differentiation. A more direct approach is used when an equation can be solved for one variable in terms of the other variable. In Exercises \(77-80\), solve each equation for \(y\) in terms of \(x\). $$y^{2}+2 x y+4=0, y>0$$

Short Answer

Expert verified
\( y = -x + \sqrt{x^2 - 4} \).

Step by step solution

01

Identify the Equation

The equation given is \( y^2 + 2xy + 4 = 0 \). We need to solve this for \( y \) in terms of \( x \).
02

Rearrange to a Quadratic Equation

Recognize that this is a quadratic equation in the form of \( y^2 + 2xy + 4 = 0 \), where \( a = 1 \), \( b = 2x \), and \( c = 4 \).
03

Use the Quadratic Formula

Use the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to solve for \( y \). Substitute \( a = 1 \), \( b = 2x \), and \( c = 4 \) into the formula.
04

Simplify the Expression

Substitute the values into the formula:\[ y = \frac{-(2x) \pm \sqrt{(2x)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{-2x \pm \sqrt{4x^2 - 16}}{2}.\]This simplifies to:\[ y = \frac{-2x \pm \sqrt{4(x^2 - 4)}}{2} = \frac{-2x \pm 2\sqrt{x^2 - 4}}{2}.\]
05

Further Simplification and Apply Condition

Further simplify by factoring out a 2 in the numerator: \[y = -x \pm \sqrt{x^2 - 4}.\]Since \( y > 0 \), we choose the positive root to ensure that \( y \) is positive. Thus, \[y = -x + \sqrt{x^2 - 4}.\]

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Key Concepts

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

Quadratic Equation
Quadratic equations are a fundamental part of algebra and calculus and take the form \(ax^2 + bx + c = 0\). These equations have variable \(x\) raised to the power of 2, making them rather unique compared to linear equations. In the exercise you are working on, the equation provided can be rearranged to fit this form as \(y^2 + 2xy + 4 = 0\), with respect to the variable \(y\). The coefficients in a quadratic equation are crucial for solving it. They determine the nature of the equation's roots. This is where we see the role of \(a\), \(b\), and \(c\):
  • \(a = 1\) is the coefficient of \(y^2\).
  • \(b = 2x\) is the coefficient of \(y\).
  • \(c = 4\) represents the constant term.
Recognizing the structure here helps you understand how a quadratic needs to be solved using appropriate formulas.
Quadratic Formula
The quadratic formula is an efficient tool to find the values of the variable in quadratic equations. When an equation is set to the correct form, this formula helps in solving for the variable easily. For any quadratic equation \(ax^2 + bx + c = 0\), the quadratic formula is given by:\[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]This formula allows us to find the roots of the equation, and it relies on the coefficients from the equation:
  • \(-b\) helps in reversing the sign of the linear coefficient.
  • The \(\pm\) indicates that quadratic equations typically have two solutions.
  • \(\sqrt{b^2 - 4ac}\) is known as the discriminant, and it provides information about the nature of the roots.
  • \(2a\) in the denominator helps in balancing the solution.
The quadratic formula is versatile and is used not only in algebra, but also in calculus when dealing with implicit equations.
Solving for a Variable
Solving for a variable means isolating that variable on one side of the equation, expressing it in terms of other variables if necessary. In our context, you are solving for \(y\) in terms of \(x\) in the quadratic equation \(y^2 + 2xy + 4 = 0\). This is done by applying the quadratic formula to achieve a solution structure like:\[y = -x \pm \sqrt{x^2 - 4}.\]Through this process, consider the conditions that might affect which solutions are viable. For example, since \(y > 0\), only the solution that maintains this condition is selected:
  • This condition ensures that the solution is valid in its mathematical context.
  • Simplifying the equation step by step helps to clearly see which parts are necessary to solve for \(y\).
Manipulating equations to solve for a particular variable is a key skill within algebra and calculus.
Calculus Concepts
Calculus introduces the idea of solving equations not just algebraically, but also using derivatives and other advanced mathematical tools. Implicit differentiation, a concept in calculus, helps to solve equations with more than one variable. In your problem, instead of using differentiation, a simpler algebraic method is used since the equation could be rearranged to a familiar quadratic form.
  • Implicit differentiation typically deals with derivatives with respect to one variable while treating other variables as functions themselves.
  • Understanding when to apply these calculus techniques versus when to use direct algebraic approaches is crucial.
  • In calculus, knowing how to solve equations both symbolically and numerically is very important for applications in science and engineering.
This exercise shows a blend of understanding calculus principles while relying on quadratics for efficient problem-solving.

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Most popular questions from this chapter

Consider the parametric equations: \(x=a \cos t-b \cos (a t)\) and \(y=a \sin t+\sin (a t) .\) With a graphing utility, explore the graphs for \(a=3\) and \(b=1, a=4\) and \(b=2,\) and \(a=6\) and \(b=2 .\) Find the \(t\) -interval that gives one cycle of the curve.

Graph the second-degree equation. (Hint: Transform the equation into an equation that contains no \(x y\) -term.) $$8 x^{2}-20 x y+8 y^{2}+18=0$$

Consider the parametric curve \(x=a \sin ^{2} t-b \cos ^{2} t\) \(y=b \cos ^{2} t+a \sin ^{2} t, 0 \leq t \leq \frac{\pi}{2} .\) Assume that \(a\) and \(b\) are nonzero constants. Find the Cartesian equation for this curve.

In calculus, some operations can be simplified by using parametric equations. Finding the points of intersection (if they exist) of two curves given by parametric equations is a standard procedure. In Exercises \(71-74\), find the points of intersection of the given curves given \(s\) and \(t\) are any real numbers. Curve \(1: x=100 t, y=80 t-16 t^{2}\) Curve II: \(x=100-200 t, y=-16 t^{2}+144 t-224\)

Determine whether each statement is true or false. A system of equations representing a line and a parabola can intersect in at most three points.
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