/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 18 Find a family of oblique traject... [FREE SOLUTION] | 91Ó°ÊÓ

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Find a family of oblique trajectories that intersect the family of curves \(x+\) \(y=c x^{2}\) at angle \(a\) such that \(\tan a=2\)

Short Answer

Expert verified
The family of oblique trajectories that intersect the family of curves \(x + y = cx^2\) at an angle \(a\) such that \(\tan a =2\) is given by \(y_2(x) = \int \frac{dx}{3(2cx - 1)} + C\), where \(C\) is the constant of integration.

Step by step solution

01

Find the derivative of the given family of curves

Differentiate the given equation \(x + y = cx^2\) with respect to \(x\): \(\frac{dy}{dx} = y' = -1 + 2cx\).
02

Substitute the relationship of \(\tan a = 2\)

We know \(\tan a = 2\), which implies \(\tan(\pi - a) = -2\) (since \(\tan(\pi - x) = -\tan x\)). So, replacing \(\tan(\pi - a)\) value in the tangent angle relationship formula, we get \(\frac{y' - dy_2/dx}{1 + y'(dy_2/dx)} = -2\).
03

Solve for the second derivative, and find the general solution

Let \(y_2=f(x)\), then let's differentiate \(f(x)\): \(\frac{dy_2}{dx} = f'(x)\). Now substitute the first derivative \(y'\) from step 1 into the equation obtained from step 2: \(\frac{-1 + 2cx - f'(x)}{1 + (-1 + 2cx)f'(x)} = -2\). To solve for \(f'(x)\), we first cross multiply and then group the terms containing \(f'(x)\), and finally solve it to find the derivative: \(-1 + 2cx - f'(x) = -2(1 + (-1 + 2cx)f'(x))\), \(2cx-1-f'(x)=-2-2(2cx-1)f'(x)\), \(1+2f'(x)=2(2cx-1)f'(x)\). Now we can solve for \(f'(x)\): \(f'(x) = \frac{1}{3(2cx - 1)}\).
04

Express the resulting equation in the form of the family of curves

Now, integrating the equation that we found for \(f'(x)\) with respect to \(x\) gives us the family of oblique trajectories: \(y_2(x) =\int f'(x) dx = \int \frac{dx}{3(2cx - 1)} + C\), where \(C\) is the constant of integration. This is the family of oblique trajectories that intersect the family of curves \(x + y = cx^2\) at an angle \(a\) such that \(\tan a = 2\).

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Key Concepts

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

Family of Curves
A family of curves refers to a set of curves that share a common mathematical relationship or equation but differ when a certain parameter within that equation is varied. In our problem, the family of curves is given by the equation \(x + y = cx^2\), where \(c\) is a parameter that can take on different values.
This means each value of \(c\) produces a different curve, but all curves will have the same basic form. You can imagine a family of curves as contours on a topography map, where each contour (curve) represents a different level (value of \(c\)).
Finding a family of curves is useful because it allows us to study how shifting one aspect of a system (in this case, the parameter \(c\)) affects the entire set of possible outcomes, thus offering a deeper understanding of the relationship between \(x\) and \(y\) for varying \(c\).
This shared structural form is how we begin to explore their intersecting trajectories, especially when these intersections happen at a specific angle.
Derivative
The derivative is a fundamental concept in calculus that represents the rate of change of a function. Derivatives indicate how a function's output value changes as its input changes.
For our exercise, we start by finding the derivative of the curve equation \(x + y = cx^2\). This derivative \(\frac{dy}{dx}\) will help us understand how steep the curve is at any point, which is crucial for figuring out how to intersect it at an angle.
Calculating the derivative, we differentiate with respect to \(x\), resulting in \(\frac{dy}{dx} = y' = -1 + 2cx\). This means that at any point \((x, y)\) on the curve, the slope is \(-1 + 2cx\).
Knowing how to calculate and interpret derivatives is key to exploring more complex properties of functions, like finding where two functions might intersect at a given angle.
Tangent Angle Formula
The tangent angle formula is used to determine the angle between two curves at the point where they intersect. It's a powerful tool in this problem as we're looking for trajectories intersecting at a specific angle.
When two curves intersect, their corresponding derivatives at the intersection point indicate their steepness. The tangent angle formula is
\[\frac{y' - f'(x)}{1 + y' f'(x)} = m\]
where \(y'\) and \(f'(x)\) are the derivatives of the intersecting curves, and \(m\) is the tangent of the angle between them.
In this scenario, we have \(\tan a = 2\) for the original intersection angle, and hence we use the negative reciprocal for the desired intersection which becomes \(-2\). Plugging this value into the tangent angle formula allows us to find the derivative of the second curve that intersects our given family of curves at this specific angle.
Integration
Integration is essentially the reverse process of differentiation. While derivatives concern themselves with rates of change, integration is about accumulation and finding areas under curves.
After finding \(f'(x) = \frac{1}{3(2cx - 1)}\), the next step is to integrate this derivative to obtain the family of trajectories that intersect our initial family of curves.
Integrating \(\frac{1}{3(2cx - 1)}\) with respect to \(x\) results in the antiderivative \(y_2(x) = \int \frac{dx}{3(2cx - 1)} + C\), where \(C\) is an arbitrary constant representing vertical shifts of the trajectories. This integral represents the new family of curves—our "oblique trajectories."
Integration often involves finding these antiderivatives and setting up initial conditions or boundaries described by additional parameters or constants like \(C\) to fully define a particular solution within the family.

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Most popular questions from this chapter

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 \(5 \mathrm{~min}\). (a) What percent of the first chemical will have been converted in \(20 \mathrm{~min} ?\) (b) In how many minutes will \(60 \%\) of the first chemical have been converted?

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