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Chapter 3: Problem 15

In each exercise, find the orthogonal trajectories of the given family of curves. Draw a few representative curves of each family whenever a figure is requested. The trisectrices of Maclaurin, \((a+x) y^{2}=x^{2}(3 a-x)\).

Short Answer

Expert verified
Differentiate, simplify, find negative reciprocal, solve differential equation.

Step by step solution

01

Differentiate the Given Equation

We begin with the equation representing the family of curves: \[(a + x) y^2 = x^2 (3a - x)\]To find the orthogonal trajectories, we first need to differentiate this equation implicitly with respect to \(x\). This gives:\[(a + x)\cdot 2y \frac{dy}{dx} + y^2 = 2x(3a-x) - x^2\cdot 1 \cdot (\frac{d}{dx}(3a-x))\].Simplifying this will help us find the derivative \(\frac{dy}{dx}\).
02

Simplify the Derivative

Now we simplify the differentiated equation:\[(a + x) 2y \frac{dy}{dx} + y^2 = 2x(3a - x) - x^2 \].Rearrange this to solve for \(\frac{dy}{dx}\): \[2(a + x)y\frac{dy}{dx} = 2x(3a - x) - x^2 - y^2\].Thus, the expression for \(\frac{dy}{dx}\) becomes:\[\frac{dy}{dx} = \frac{2x(3a - x) - x^2 - y^2}{2(a + x)y}\].
03

Find the Orthogonal Trajectories

The orthogonal trajectories are found by replacing \(\frac{dy}{dx}\) with its negative reciprocal:\[\frac{dy_\perp}{dx} = -\frac{2(a + x)y}{2x(3a - x) - x^2 - y^2}\].This gives the differential equation for the orthogonal trajectories.
04

Solve the Differential Equation for the Orthogonal Trajectories

The orthogonal trajectories are found by solving the differential equation:\[\frac{dy}{dx} = -\frac{2(a + x)y}{2x(3a - x) - x^2 - y^2}\].This equation is separable, meaning we can integrate each side separately:\[\int 2(a + x)\, dy = \int -(2x^2 - x^3)\, dx\].Solving these integrals will give the equations that describe the orthogonal trajectories.
05

Graph the Curve

To visualize the family of curves and their orthogonal trajectories, plot the original equation \((a + x) y^{2}=x^{2}(3 a-x)\) for different values of \(a\), and also plot the solutions obtained from the integration in the previous step.These plots will help in understanding how the original and orthogonal trajectories intersect at right angles.

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Key Concepts

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

Implicit Differentiation
Implicit differentiation is a powerful calculus technique used when dealing with equations that cannot be solved explicitly for one variable in terms of the others. Unlike explicit differentiation, where each variable is isolated, implicit differentiation handles situations where variables are intertwined.
For instance, in the exercise provided:
  • You start with a complex equation involving two variables: \( (a + x) y^2 = x^2 (3a - x) \)
  • Since it's not possible to solve explicitly for \( y \), we differentiate both sides with respect to \( x \), treating \( y \) and \( rac{dy}{dx} \) as functions of \( x \)
To find the slope of the given family of curves, you use implicit differentiation to obtain the derivative, allowing you to determine how changes in \( x \) affect changes in \( y \). Remember, this technique helps in situations where curves are given implicitly rather than as one variable explicitly in terms of another.
Differential Equations
Differential equations are equations that involve derivatives of a function. They describe the rate of change and are fundamental in fields such as physics, engineering, and economics.
In the context of our exercise:
  • The implicit differentiation results in a differential equation: \( 2(a + x)y\frac{dy}{dx} = 2x(3a - x) - x^2 - y^2 \)
  • This equation broadly describes how one quantity changes with respect to another
The solution to this differential equation provides the change in trajectory needed for orthogonality. In our case, we aim to find a new family of curves that intersects the original family at right angles. Solving such equations is critical in modeling real-world problems and understanding dynamic systems.
Separable Equations
Separable equations are a category of differential equations where the variables can be separated on opposite sides of the equation, allowing for straightforward integration.
Our differential equation for the orthogonal trajectories \( \frac{dy}{dx} = -\frac{2(a + x)y}{2x(3a - x) - x^2 - y^2} \) can be transformed into a separable form, simplifying the integration process.
  • Separate variables by moving all the terms involving \( y \) to one side and terms involving \( x \) to the other
  • This transforms the equation into integrable parts, making it easier to solve
The separable nature allows us to integrate each side independently to find a solution, which represents the orthogonal trajectories. By using properties of separable equations, we make complex differential problems far more manageable.

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Most popular questions from this chapter

For a substance \(C,\) the time rate of conversion is proportional to the square of the amount \(x\) of unconverted substance. Let \(k\) be the numerical value of the constant of proportionality and let the amount of unconverted substance be \(x_{0}\) at time \(t=0 .\) Determine \(x\) for all \(t \geqq 0\).

In the motion of an object through a certain medium (air at certain pressures is an example), the medium furnishes a resisting force proportional to the square of the velocity of the moving object. Suppose a body falls, due to the action of gravity, through such a medium. Let \(t\) represent time, and \(v\) represent velocity, positive downward. Let \(g\) be the usual constant acceleration of gravity and let \(w\) be the weight of the body. Use Newton's law, force equals mass times acceleration, to conclude that the differential equation of the motion is $$\frac{w}{g} \frac{d v}{d t}=w-k v^{2}$$ where \(k v^{2}\) is the magnitude of the resisting force furnished by the medium.

Two substances, \(A\) and \(B\), are being converted into a single compound \(C .\) In the laboratory it has been shown that, for these substances, the following law of conversion holds: the time rate of change of the amount \(x\) of compound \(C\) is proportional to the product of the amounts of unconverted substances \(A\) and \(B\). Assume the units of measure so chosen that one unit of compound \(C\) is formed from the combination of one unit of \(A\) with one unit of \(B\). If at time \(t=0\) there are \(a\) units of substance \(A, b\) units of substance \(B,\) and none of compound \(C\) present, show that the law of conversion may be expressed by the equation $$\frac{d x}{d t}=k(a-x)(b-x)$$ Solve this equation with the given initial condition.

The radius of the moon is roughly 1080 miles. The acceleration of gravity at the surface of the moon is about \(0.165 g,\) where \(g\) is the acceleration of gravity at the surface of the earth. Determine the velocity of escape for the moon.

A college dormitory houses 100 students, each of whom is susceptible to a certain virus infection. A simple model of epidemics assumes that during the course of an epidemic the rate of change with respect to time of the number of infected students \(I\) is proportional to the number of infected students and also proportional to the number of uninfected students, \(100-I .\) (a) If at time \(t=0\) a single student becomes infected, show that the number of infected students at time \(t\) is given by $$I=\frac{100 e^{100 k t}}{99+e^{100 k t}}$$ (b) If the constant of proportionality \(k\) has value 0.01 when \(t\) is measured in days, find the value of the rate of new cases \(I^{\prime}(t)\) at the end of each day for the first 9 days.
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