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Chapter 3: Problem 1

In Exercises \(1-9\) find the orthogonal trajectories of each given family of curves. In each case sketch several members of the family and several of the orthogonal trajectories on the same set of axes. $$ y=c x^{3}. $$

Short Answer

Expert verified
The orthogonal trajectories of the given family of curves \(y=cx^3\) can be found by considering the negative reciprocal of their derivative. The orthogonal differential equation is given by \(\frac{dy}{dx_{\perp}} = -\frac{dy}{dx}\). Solving this equation yields the integral form of the orthogonal trajectories: \(x(y) = \int -\frac{dy}{dx} dy\). To sketch the family of curves and the orthogonal trajectories, plot increasing values of the constant \(c\) for the family of curves and tangent lines with slopes that are negative reciprocals of the slopes of the given family's curves. These trajectories will intersect each member of the family at right angles.

Step by step solution

01

Determine the differential equation of the given family of curves

First, we need to find the equation that describes the given family of curves, \(y=cx^3\). To eliminate the arbitrary constant, \(c\), we take the derivative with respect to \(x\). \[\frac{dy}{dx} = c\cdot3x^2 = 3cx^2\] Now, we can eliminate the arbitrary constant \(c\) by solving for it: \[c = \frac{1}{3x^2}\frac{dy}{dx}\]
02

Find the orthogonal trajectories differential equation

To find the orthogonal trajectories, we need to negate the reciprocal of the derivative, as orthogonal curves have negative reciprocals of their slopes. \[\frac{dy}{dx_{\perp}} = -\frac{1}{\frac{dy}{dx}}\] Substituting the expression for \(\frac{dy}{dx}\) obtained in Step 1: \[\frac{dy}{dx_{\perp}} = -\frac{1}{3cx^2} = -\frac{1}{3\left(\frac{1}{3x^2}\frac{dy}{dx}\right)x^2} \] Simplify the expression: \[\frac{dy}{dx_{\perp}} = -\frac{dy}{dx}\]
03

Solve the orthogonal differential equation

Now we need to find the solution of the orthogonal differential equation: \[-\frac{dy}{dx}\frac{dx}{dy} = 1\] \[\frac{dx}{dy} = \frac{dy}{dx}\] Integrate both sides with respect to \(y\): \[x(y) = \int \frac{dy}{dx} dy\] Notice that we don't have any further information available about this orthogonal trajectories equation, so our final solution for the orthogonal trajectories is given by the integral form: \[ x(y) = \int -\frac{dy}{dx} dy\]
04

Sketch the orthogonal trajectories and members of the given family of curves

To sketch the original family of curves, we can consider increasing values of the constant \(c\). This will create a series of cubic curves, passing through the origin (0,0), that increase in steepness with the increasing values of \(c\). To sketch the orthogonal trajectories, we can plot a series of tangent lines that have slopes which are negative reciprocals of the slopes of the curves represented by \(y=cx^3\). These orthogonal trajectories will form graphs that intersect each member of the parabolic family at right angles.

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Key Concepts

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

Ordinary Differential Equations
Ordinary Differential Equations (ODEs) are equations involving derivatives of a function with respect to one variable. They form the backbone of many physical laws and phenomena, where the rate of change of a quantity is related to the quantity itself. For instance, you might come across an ODE when dealing with motion, where acceleration (the second derivative of position with respect to time) is a function of position or velocity.

ODEs can be simple, like a first-order linear equation, or more complex, like a higher-order nonlinear equation. The order of an ODE is denoted by the highest derivative it contains. Solutions to these equations generally involve integration, and can oftentimes be represented by a family of curves, each differentiated by particular values of constants, derived from initial conditions or other constraints.
Slope of a Curve
The slope of a curve at any given point is essentially the measure of the steepness or incline of the curve at that point. It is given by the derivative of the function describing the curve with respect to its independent variable. For example, if your curve is described by the function \(y = f(x)\), the slope at any point \(x\) is the derivative \(f'(x)\), which is denoted as \(\frac{dy}{dx}\).

In the context of orthogonal trajectories, understanding the slope is crucial. Orthogonal trajectories are curves that intersect the original family of curves at right angles, meaning the slopes at their points of intersection are negative reciprocals of each other. If one curve has a slope \(m\), the orthogonal trajectory at the point of intersection will have a slope of \(-\frac{1}{m}\).
Integration of Functions
Integration is a fundamental mathematical operation that can be viewed as the inverse process of differentiation or as a way to calculate the area under a curve. It involves finding a function (the integral) whose derivative is the original function. In simpler terms, when you integrate a function, you're essentially summing up an infinite number of infinitesimally small quantities.

When you solve an ODE by integration, you're looking for a function that describes how a particular quantity changes over time or space. Integration often results in a family of solutions that include an arbitrary constant, representing an entire set of possible solutions. In the case of finding orthogonal trajectories, integration helps us move from the differential equation of a curve's slope to the general equation that represents all possible positions of the curve.

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Most popular questions from this chapter

A 200 liter tank is initially full of fluid in which there is dissolved \(40 \mathrm{gm}\) of a certain chemical. Fluid containing \(50 \mathrm{gm}\) per liter of this chemical flows into the tank at the rate of 5 liters/min. The mixture is kept uniform by stirring, and the stirred mixture simultaneously flows out at the rate of 7 liters/min. How much of the chemical is in the tank when it is only half full?

Find the orthogonal trajectories of the family of circles which are tangent to the \(y\) axis at the origin.

A 500 liter tank initially contains 300 liters of fluid in which there is dissolved \(50 \mathrm{gm}\) of a certain chemical. Fluid containing \(30 \mathrm{gm}\) per liter of the dissolved chemical flows into the tank at the rate of 4 liters \(/ \mathrm{min}\). The mixture is kept uniform by stirring, and the stirred mixture simultaneously flows out at the rate of \(2.5\) liters/min. How much of the chemical is in the tank at the instant it overflows?

A bullet weighing 1 oz is fired vertically downward from a stationary helicopter with a muzzle velocity of \(1200 \mathrm{ft} / \mathrm{sec}\). The air resistance (in pounds) is numerically equal to \(16^{-5} v^{2}\), where \(v\) is the velocity (in feet per second). Find the velocity of the bullet as a function of the time.

A large tank initially contains 100 gal of brine in which \(10 \mathrm{lb}\) of salt is dissolved. Starting at \(t=0\), pure water flows into the tank at the rate of 5 gal/min. The mixture is kept uniform by stirring and the well- stirred mixture simultaneously flows out at the slower rate of \(2 \mathrm{gal} / \mathrm{min} .\) (a) How much salt is in the tank at the end of \(15 \mathrm{~min}\) and what is the concentration at that time? (b) If the capacity of the tank is 250 gal, what is the concentration at the instant the tank overflows?
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