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Chapter 3: Problem 57

Find the normals to the curve \(x y+2 x-y=0\) that are parallel to the line $2 x+y=0 .

Short Answer

Expert verified
The 'x' coordinates at which the normals to the given curve are parallel to the specified line are the roots of the equation \(x^3 - x + 1 = 0\). The corresponding 'y' coordinates can be obtained by substituting these 'x' values into the equation \(y = 2 - 2x^2\).

Step by step solution

01

Find the derivative of the given curve

Differentiate the given curve equation with respect to \(x\). Applying the product rule on \(xy\) and differentiating \(2x\) and \(-y\) separately, we get:\(\frac{dy}{dx} = \frac{1-y-2}{x}\)
02

Find the slope of normals

The slope of the curve at a point is given by \(\frac{dy}{dx}\), so the slope of the normals is \(-\frac{1}{\frac{dy}{dx}}\). This follows because the usual rule applies that the product of the slopes of two perpendicular lines equals -1.
03

Equate the slope of normals to the slope of given line

The slope of the given line \(2x + y = 0\) is -2. Since the normals are parallel to this line, their slope also will be -2. So, \(-\frac{1}{\frac{dy}{dx}} = -2\). Substituting the value of \(\frac{dy}{dx}\) from Step 1, we obtain the equation \(2x^2 + y - 2 = 0\).
04

Incorporate this into the original equation of the curve

The existing equation of the curve is \(x y+2 x-y=0\). When we substitute \(y = 2 - 2x^2\) into this, we get, after simplifying, \(x^3 - x + 1 = 0\). The roots of this equation will give possible 'x' coordinates of the points where the normals exist.
05

Find the corresponding y-coordinates

When 'x' coordinates have been found, solve for 'y' by substituting the obtained 'x' values into the simplified equation from step 4 to find the corresponding y coordinates.

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Key Concepts

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

Differentiation
Differentiation is a key concept in calculus. It involves finding the derivative of a function, which indicates how the function's output changes as the input changes. Essentially, the derivative tells us the rate of change or the slope of the tangent line to a curve at any given point.
For the given problem, we start by differentiating the equation of the curve, which utilizes the product rule. The product rule states that for two functions, say \(u\) and \(v\), the derivative of their product is \(u'v + uv'\).
In the equation \(x y + 2x - y = 0\), applying differentiation helps us identify how changes in \(x\) influence \(y\) and, hence, the slope at any point on the curve. This process leads us to find \(\frac{dy}{dx}\), indicating the slope of the tangent at any given point on the curve.
Slopes of Lines
The slope of a line is a measure of its steepness and is calculated as the ratio of the vertical change to the horizontal change between two points on the line. In mathematical terms, it is represented as \(m = \frac{\Delta y}{\Delta x}\).
For a straight line in the form \(ax + by + c = 0\), the slope can be determined by rearranging the equation to \(y = mx + b\). Here, \(m\) represents the slope.
In this exercise, the line equation \(2x + y = 0\) has a slope of -2, as evident when rewritten as \(y = -2x\). This slope is crucial for comparing with other line slopes to determine parallelism or perpendicularity between lines.
Normals to Curves
Normals to curves are lines perpendicular to the tangent of the curve at a given point, providing important geometric insights.
To find a normal to a curve, it's essential first to understand the slope of the tangent line, given by the derivative \(\frac{dy}{dx}\). Once the tangent's slope is known, the slope of the normal is the negative reciprocal of the tangent slope.
In this problem, we have the normal lines parallel to a line with slope -2. Therefore, our normal's slope should also be -2. By setting the normal slope to -2, we can solve for points on the curve where the normal line has this slope, thus finding where these specific normals exist.

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Most popular questions from this chapter

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