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Chapter 3: Problem 12

Find the value of \(K\) such that the parabolas \(y=c_{1} x^{2}+K\) are the orthogonal trajectories of the family of ellipses \(x^{2}+2 y^{2}-y=c_{2}\)

Short Answer

Expert verified
The value of \(K\) that satisfies the orthogonal trajectory condition is \(K = -\frac{1}{2} + c_2\), where \(c_2\) is a constant related to the ellipse family.

Step by step solution

01

Find the derivatives of each family of curves

Differentiate the given equations with respect to \(x\) using the Chain Rule. For the parabola, let \(y = y_1(x) = c_1 x^2 + K\) The derivative of \(y_1(x)\) is $$ \frac{dy_1}{dx} = 2c_1 x $$ For the ellipse, let: $$ y_2(x) = 2y^2 - y = \frac{1}{2}x^2 + c_2 $$ The derivative of \(y_2(x)\) is $$ \frac{dy_2}{dx} = x $$
02

Utilize the orthogonality condition

The product of the slopes at their intersection is equal to -1. Then we have: $$ \frac{dy_1}{dx} \cdot \frac{dy_2}{dx} = -1 $$ Plugging the derivatives from step 1: $$ (2c_1 x) \cdot (x) = -1 $$
03

Set up the equations at the intersection points

Since the intersection points belong to both families, the \(y\) values must be equal. Equate the equations for \(y_1(x)\) and \(y_2(x)\): $$ c_1 x^2 + K = \frac{1}{2}x^2 + c_2 $$
04

Solve for \(K\)

To find the value of \(K\), we will solve the system of equations formed from step 2 and step 3. From the orthogonality condition: $$ 2c_1 x^2 = -1 $$ From the equal \(y\)-values condition, substitute \(c_1 x^2\): $$ - \frac{1}{2} + K = \frac{1}{2}x^2 + c_2 $$ Since the equal y values hold for all intersection points, we can conclude that \(K = -\frac{1}{2} + c_2\). Thus, the value of \(K\) that satisfies the orthogonal trajectory condition is \(-\frac{1}{2} + c_2\), where \(c_2\) is a constant related to the ellipse family.

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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 tools used to describe various physical phenomena where the rate of change of one variable is related to the variable itself. Essentially, it's an equation that relates a function to its derivatives.

For example, when looking at the growth of a population or the decay of a radioactive substance, we use differential equations to predict future behavior based on current states. In the context of the exercise given, differential equations help us to find the slope of a curve, which is the derivative of the equation defining that curve. Understanding how to differentiate these equations enables us to explore complex relationships, such as finding orthogonal trajectories between families of curves like parabolas and ellipses.
Parabolas
Parabolas are U-shaped curves commonly encountered in algebra and geometry, and they have a variety of applications in physics, engineering, and architecture. They are defined by a quadratic function of the form y = ax^2 + bx + c, where a, b, and c are constants.

The exercise under consideration involves a set of parabolas with an added constant, K, which affects the vertical position of the parabola without changing its shape. Understanding the properties of parabolas, such as their symmetry and focus points, can yield insights into their behavior and relationship to other geometric figures like ellipses.
Ellipses
Ellipses are closed, oval-shaped curves that can be seen as stretched circles. Defined by the equation ax^2 + by^2 = c, their shape depends on the coefficients a and b.

When studying orthogonal trajectories, it鈥檚 crucial to recognize that unlike parabolas, ellipses have two axes of symmetry. This property can influence how they intersect with other curves. The elliptical equations given in the exercise open the path to exploring relationships with other curves鈥攖he key to solving the orthogonal trajectories problem.
Slope of a Curve
The slope of a curve at any given point is the rate at which y changes with respect to x; it is the tangential line's incline at that point. To find the slope, we use derivatives, which measure how a function's output changes as its input changes.

In the problem, understanding the slope helps us figure out how the parabolas and ellipses intersect under right angles, i.e., orthogonal intersections. Insights into this concept are not only foundational for calculus students but also for anyone studying the dynamics of intersecting curves in various fields.
Chain Rule
The Chain Rule is a powerful derivative technique in calculus used to calculate the derivative of a composite function. Simply put, if a variable u depends on x and another variable y depends on u, the Chain Rule allows us to find the derivative of y with respect to x easily.

In the exercise, we apply the Chain Rule to find the derivative (slope) of the ellipse as part of identifying the orthogonal trajectories. Without the Chain Rule, we would struggle to find derivatives of such composite functions, making it difficult to solve complex calculus problems like the one posed.

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Most popular questions from this chapter

A tank initially contains 100 gal of brine in which there is dissolved \(20 \mathrm{lb}\) of salt. Starting at time \(t=0\), brine containing \(3 \mathrm{Ib}\) of dissolved salt per gallon flows into the tank at the rate of \(4 \mathrm{gal} / \mathrm{min}\). The mixture is kept uniform by stirring and the well-stirred mixture simultaneously flows out of the tank at the same rate. (a) How much salt is in the tank at the end of \(10 \mathrm{~min} ?\) (b) When is there \(160 \mathrm{lb}\) of salt in the tank?

Assume that the population of a certain city increases at a rate proportional to the number of inhabitants at any time. If the population doubles in 40 years, in how many years will it triple?

A body of mass \(m\) is in rectilinear motion along a horizontal axis. The resultant force acting on the body is given by \(-k x\), where \(k>0\) is a constant of proportionality and \(x\) is the distance along the axis from a fixed point \(\mathrm{O}\). The body has initial velocity \(v=v_{0}\) when \(x=x_{0}\). Apply Newton's second law in the form (3.23) and thus write the differential equation of motion in the form $$ m v \frac{d v}{d x}=-k x $$ Solve the differential equation, apply the initial condition, and thus express the square of the velocity \(v\) as a function of the distance \(x\). Recalling that \(v=d x / d t\), show that the relation between \(v\) and \(x\) thus obtained is satisfied for all time \(t\) by $$ x=\sqrt{x_{0}^{2}+\frac{m v_{0}^{2}}{k}} \sin \left(\sqrt{\frac{k}{m}} t+\phi\right) $$ where \(\phi\) is a constant.

A certain chemical is converted into another chemical by a chemical reaction. The rate at which the first chemical is converted is proportional to the amount of this chemical present at any instant. Ten percent of the original amount of the first chemical has been converted in 5 min. (a) What percent of the first chemical will have been converted in 20 min? (b) In how many minutes will \(60 \%\) of the first chemical have been converted?

A body cools from \(60^{\circ} \mathrm{C}\) to \(50^{\circ} \mathrm{C}\) in \(15 \mathrm{~min}\) in air which is maintained to \(30^{\circ} \mathrm{C}\). How long will it take this body to cool from \(100^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\) in air that is maintained at \(50^{\circ} \mathrm{C} ?\) Assume Newton's law of cooling (Exercise 23).
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