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Find the orthogonal trajectories of the family of curves. Sketch several members of each family. $$k x^{2}+y^{2}=1$$

Short Answer

Expert verified
The orthogonal trajectories are hyperbolas: \( kx^2 - y^2 = C \).

Step by step solution

01

Understand the Problem

We are given the family of curves: \( kx^2 + y^2 = 1 \), where \( k \) is a parameter. Our task is to find the orthogonal trajectories for this family. Orthogonal trajectories are curves that intersect each member of the given family of curves at right angles.
02

Differentiate the Family of Curves

Start by differentiating the given family of curves with respect to \( x \). Differentiate both sides of \( kx^2 + y^2 = 1 \). The derivative is \( 2kx + 2y \frac{dy}{dx} = 0 \).
03

Solve for dy/dx

Isolate \( \frac{dy}{dx} \) in the differentiated equation \( 2kx + 2y \frac{dy}{dx} = 0 \). This gives us \( \frac{dy}{dx} = -\frac{kx}{y} \). This is the slope of the tangent to the family of curves.
04

Find the Slopes of the Orthogonal Trajectories

The slope of the orthogonal trajectories is the negative reciprocal of the tangent slope. Therefore, the slope of the orthogonal trajectories is \( \frac{dx}{dy} = \frac{y}{kx} \).
05

Solve the Differential Equation for Orthogonal Trajectories

Follow the equation \( \frac{dx}{dy} = \frac{y}{kx} \). Rearrange as \( kx \, dx = y \ dy \). Integrate both sides: \( \int kx \, dx = \int y \, dy \), resulting in \( \frac{kx^2}{2} = \frac{y^2}{2} + C\). Simplifying, we have \( kx^2 = y^2 + C \).
06

Simplify and Represent the Orthogonal Trajectories

The orthogonal trajectories of the family are represented by the equation \( kx^2 - y^2 = C \). These are hyperbolas, where \( C \) is a constant representing different curves.
07

Sketch the Curves

Draw several members of both the family \( kx^2 + y^2 = 1 \) and its orthogonal trajectories \( kx^2 - y^2 = C \). The given family are ellipses, and the orthogonal trajectories are hyperbolas centered on the same line, meeting the ellipses at right angles.

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations involving derivatives, which represent rates of change. They play a critical role in understanding how quantities vary over time or space. In the context of finding orthogonal trajectories, differential equations help us express how the slope of the tangent to a curve changes along the curve itself.

In the original problem, by differentiating the family of curves \( kx^2 + y^2 = 1 \), we obtain \( 2kx + 2y \frac{dy}{dx} = 0 \). Solving for \( \frac{dy}{dx} \) gives us the slope of the tangent to the curve, represented by the differential equation \( \frac{dy}{dx} = -\frac{kx}{y} \). To find orthogonal trajectories, we also need the slope's negative reciprocal, which leads us to a new differential equation for the orthogonal curves.

Working with differential equations involves:
  • Identifying the relation between the curves and their derivatives.
  • Solving the equation to find related trajectories or solutions.
  • Understanding how variations in parameters like \( k \) affect the curves.
Family of Curves
A family of curves is a set of curves that can be described by a common equation with a parameter. Think of it like a big group of related shapes that change slightly with different values of this parameter. Here, \( kx^2 + y^2 = 1 \) is the equation of the given family of curves.

This family represents ellipses where changing the parameter \( k \) adjusts the ellipse's dimensions. Even though the equation looks similar for all these curves, the parameter \( k \) provides flexibility, creating a vast array of individual curves within the family.

Exploring a family of curves involves:
  • Identifying common characteristics shared by all member curves.
  • Understanding the effect of varying the parameter \( k \) on each curve.
  • Using the family to study geometric properties, like tangents and orthogonal trajectories.
Slope of Tangent
The slope of a tangent to a curve at a particular point tells us the steepness or incline of the curve at that point. It is crucial in understanding the direction in which the curve progresses.

In calculus, the slope of the tangent to the curve \( kx^2 + y^2 = 1 \), is calculated by differentiating the equation concerning \( x \), leading to \( \frac{dy}{dx} = -\frac{kx}{y} \). This formula helps determine how steep the curve is at any given point.

Key points when considering the slope of tangents:
  • The derivative \( \frac{dy}{dx} \) provides the tangent's slope for curves described by functions of \( x \) and \( y \).
  • For orthogonal trajectories, we focus on the negative reciprocal of the tangent slope to find curves intersecting perpendicularly.
  • Visualizing slopes can help in sketching curves with accurate directionality and intersections.
Hyperbolas
Hyperbolas are a type of conic section characterized by their open, curves extending to infinity on both ends. They appear when a plane intersects both pairs of opposite cones in a double cone geometry.

In the context of orthogonal trajectories, when manipulating the equation \( kx^2 - y^2 = C \), the curves formed are hyperbolas. These hyperbolas are significant because they intersect the given family of ellipses \( kx^2 + y^2 = 1 \) at right angles.

Key features of hyperbolas:
  • They have two separate branches, each extending towards infinity in opposite directions.
  • They open along a principal axis, defined by the equation's structure.
  • In this problem, hyperbolas are orthogonally related to accompanying ellipses, illustrating the contrast between curves with opposite curvatures.
Understanding hyperbolas adds depth to analyzing intersections and relationships between different families of curves. It showcases the beautiful symmetry within geometric frameworks.

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Most popular questions from this chapter

Obtain a slope field and add to it graphs of the solution curves passing through the given points.\(y^{\prime}=(y-1)(x+2)\) with a. \((0,-1)\) b.\((0,1)\) c. \((0,3)\) d. \((1,-1)\)

Use Euler's method to calculate the first three approximations to the given initial value problem for the specified increment size. Calculate the exact solution and investigate the accuracy of your approximations. Round your results to four decimal places. $$y^{\prime}=2 x e^{x^{2}}, \quad y(0)=2, \quad d x=0.1$$

Suppose that a healthy population of some species is growing in a limited environment and that the current population \(P_{0}\) is fairly close to the carrying capacity \(M_{0} .\) You might imagine a population of fish living in a freshwater lake in a wilderness area. Suddenly a catastrophe such as the Mount St. Helens volcanic eruption contaminates the lake and destroys a significant part of the food and oxygen on which the fish depend. The result is a new environment with a carrying capacity \(M_{1}\) considerably less than \(M_{0}\) and, in fact, less than the current population \(P_{0} .\) Starting at some time before the catastrophe, sketch a "before-and-after" curve that shows how the fish population responds to the change in environment.

If a body of mass \(m\) falling from rest under the action of gravity encounters an air resistance proportional to the square of velocity, then the body's velocity \(t\) seconds into the fall satisfies the equation $$m \frac{d v}{d t}=m g-k v^{2}, \quad k > 0$$ where \(k\) is a constant that depends on the body's aerodynamic properties and the density of the air. (We assume that the fall is too short to be affected by changes in the air's density.) a. Draw a phase line for the equation. b. Sketch a typical velocity curve. c. For a \(45-\mathrm{kg}\) skydiver \((m g=441)\) and with time in seconds and distance in meters, a typical value of \(k\) is \(0.15 .\) What is the diver's terminal velocity? Repeat for an \(80-\mathrm{kg}\) skydiver.

Is either of the following equations correct? Give reasons for your answers. a. \(\quad \frac{1}{\cos x} \int \cos x d x=\tan x+C\) b. \(\frac{1}{\cos x} \int \cos x d x=\tan x+\frac{C}{\cos x}\)

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