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Find the orthogonal trajectories of each of the following families of curves. In each case sketch several of the given curves and several of their orthogonal trajectories. Be careful to eliminate the constant from \(d y / d x\) for the original curves; this constant takes different values for different curves of the original family and you want an expression for \(d y / d x\) which is valid for all curves of the family crossed by the orthogonal trajectory you are trying to find. \(y=k x^{n}\). (Assume that \(n\) is a given number; the different curves of the family have, different values of \(k_{2}\) )

Short Answer

Expert verified
The orthogonal trajectories are given by \( n y^2 + x^2 = D \).

Step by step solution

01

Differentiate the Original Curve

Given the curve equation is: \( y = k x^n \) Differentiate this equation with respect to \(x\) to find \(\frac{dy}{dx}\). Using the chain rule of differentiation, \( \frac{dy}{dx} = k n x^{n-1} \)
02

Eliminate the Constant

Since \(k\) is constant and varies for the given family of curves, eliminate \(k\) from the derivative. Solve the original equation for \(k\): \( k = \frac{y}{x^n} \). Substitute this result into \( \frac{dy}{dx} \):\( \frac{dy}{dx} = \left(\frac{y}{x^n}\right) n x^{n-1} = \frac{n y}{x} \)
03

Find the Slope of the Orthogonal Trajectories

The orthogonal trajectory must satisfy the condition that its slope \( \frac{dy'}{dx}\) is the negative reciprocal of the slope of the original curve. So,\( \frac{dy'}{dx} = -\frac{x}{n y} \)
04

Separate the Variables

Separate the variables in \(\frac{dy'}{dx}\) to solve for the orthogonal trajectories: \( n y dy' = -x dx \)
05

Integrate Both Sides

Integrate both sides to find the general solution: \( \int n y \, dy' = -\int x \, dx \)This gives: \( \frac{n y^2}{2} = -\frac{x^2}{2} + C \) where \(C\) is the constant of integration.
06

Simplify the Equation

Rearrange and simplify the equation: \( n y^2 + x^2 = D \) where \( D \) is a new constant.
07

Conclusion: Orthogonal Trajectories

The orthogonal trajectories for the family of curves \( y = k x^n \) are given by the equation: \( n y^2 + x^2 = D \)

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Key Concepts

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

calculus
Calculus is a branch of mathematics that deals with continuous change. It's divided into two main parts: differential calculus and integral calculus. In this exercise, we're focusing on differential calculus, where we use differentiation to find the rate at which a function is changing at any given point. This is crucial for understanding the slopes of curves, especially when finding orthogonal trajectories. By differentiating the function, we obtain the slope of the given curves which is essential in solving our problem. Differentiation involves applying rules like the chain rule to break down complex expressions into their simplest form.
differential equations
Curve sketching allows us to visualize the behavior of functions and their interactions on a graph. In our problem, we’re dealing with the family of curves given by \( y = k x^n \) and the orthogonal trajectories that intersect them at right angles. Sketching multiple curves from the family will show different values of \( k \). By also sketching their orthogonal trajectories, we can see the perpendicular intersections. This graphical representation helps in comprehending how the curves and their trajectories relate to each other.

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Most popular questions from this chapter

Find the general solution of each of the following differential equations. \(d y+\left(2 x y-x e^{-x^{2}}\right) d x=0\)

By separation of variables, solve the differential equation \(d y / d x=\sqrt{1-y^{2}}\) to obtain solution containing one arbitrary constant. Although this solution may be referred to as the " general solution," show that \(y=1\) is a solution of the differential equation not obtainablc from the "general solution" by any choice of the arbitrary constant. The solution \(y=1\) is called a singular selution; \(y=-1\) is another singular solution. Sketch a number of graphs of the "general solution" for different values of the arbitrary constant and observe that \(y=1\) is tangent to all of them. This is characteristic of a singular solution -its graph is tangent at each point to one of the graphs of the "general solution." Note that the given differential equation is not linear; for linear equations, all solutions are contained in the general solution, but nonlinear equations may have singular solutions which cannot be obtained from the "general solution" by specializing the arbitrary constant (or constants). Thus a nonlinear first-order equation in \(x\) and \(y\) may have two (or more) solutions passing through a given point in the \((x, y)\) plane, whereas a linear first-order equation always has just one such solution. Show that any continuous curve made up of pieces of \(y=1, y=-1\), and the sinc curves of the "general solution," gives a solution of the above differential equation. Sketch such a solution curve on your graphs.

Water with a small salt content (5 \(\mathrm{lb}\) in \(1000 \mathrm{gal}\) ) is flowing into a very salty lake at the rate of \(4 \cdot 10^{5}\) gal per hr. The salty water is flowing out at the rate of \(10^{5}\) gal per hr. If at some time (say \(t=0\) ) the volume of the lake is \(10^{9} \mathrm{gal}\), and its salt content is \(10^{7} \mathrm{Ib}\), find the salt content at time \(t\). Assume that the sah is mixed uniformly with the water in the lake at all times.

Find the "general solution" (that is, a solution containing an arbitrary constant) of each of the following differential equations, by separation of variables. Then find a particular solution of each equation satisfying the given boundary conditions. \(y^{\prime}+2 x y^{2}=0\) \(y=1\) when \(x=2\)

(a) Consider a light beam traveling downward into the ocean. As the beam progresses, it is partially absorbed and its intensity decreases. The rate at whieh the intensity is decreasing with depth at any point is proportional to the intensity at that depth. The proportionality constant \(\mu\) is called the linear absorption coefficient. Show that if the intensity at the surface is \(I_{0}\), the intensity at a distance \(s\) below the surface is \(l=\) \(I_{0} e^{-\mu s} .\) The linear absorption coefficient for water is of the order of \(10^{-2} \mathrm{ft}^{-1}\) (the exact value depending on the wavelength of the light and the impurities in the water). For this value of \(\mu\), find the intensity as a fraction of the surface intensity at a depth of \(1 \mathrm{ft}, 50 \mathrm{ft}, 500 \mathrm{ft}, 1\) mile. When the intensity of a light beam has been reduced to half. its surface intensity \(\left(I=\frac{1}{2} I_{0}\right)\), the distance the light has penetrated into the absorbing substance is called the half-value thickness of the substance. Find the half-value thickness in terms of \(\mu\). Find the half-value thickness for water for the value of \(\mu\) given above. (b) Note that the differential equation and its solution in this problem are mathematically the same as those in Example I, although the physical problem and the terminology are different. In discussing radioactive decay, we call \(\lambda\) the decay constant, and we define the half-life \(T\) of a radioactive substance as the time when \(N=\frac{1}{2} N_{0}\) (compare half-value thickness). Find the relation between \(\lambda\) and \(T\).

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