/*! 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 57 Find an equation of the normal l... [FREE SOLUTION] | 91Ó°ÊÓ

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Find an equation of the normal line to the parabola \(y=x^{2}-5 x+4\) that is parallel to the line \(x-3 y=5\)

Short Answer

Expert verified
The equation of the normal line is \(y = \frac{1}{3}x - \frac{1}{3}\).

Step by step solution

01

Identifying the Slope of the Given Line

The given line is in the form \(x - 3y = 5\). We convert it to the slope-intercept form \(y = mx + b\) by solving for \(y\). This gives us \(3y = x - 5\), or \(y = \frac{1}{3}x - \frac{5}{3}\). The slope, \(m\), is \(\frac{1}{3}\).
02

Finding the Slope of the Normal Line

The normal line is parallel to the given line, therefore its slope is the same. Thus, the slope of the normal line is also \(\frac{1}{3}\).
03

Differentiating the Parabola

The equation of the parabola is \(y = x^2 - 5x + 4\). We find its derivative to get the slope of the tangent line: \(\frac{dy}{dx} = 2x - 5\). This gives us the slope of the tangent at any point \((x, y)\) on the parabola.
04

Finding the Point of Tangency

Since the normal line is parallel to the given line, its slope is \(\frac{1}{3}\). The slope of the tangent line, \(2x - 5\), must satisfy \((2x - 5) \cdot \frac{1}{3} = -1\) because the product of the slopes of two perpendicular lines is \(-1\). Solving \((2x - 5) \times \frac{1}{3} = -1\) gives us \(2x - 5 = -3\), hence \(2x = 2\) and \(x = 1\).
05

Finding the Y-Coordinate of the Point

Substitute \(x = 1\) into the parabola \(y = x^2 - 5x + 4\). We get \(y = 1^2 - 5 \cdot 1 + 4 = 0\). Thus, the point of tangency is \((1, 0)\).
06

Writing the Equation of the Normal Line

Using the point-slope form of the line equation \(y - y_1 = m(x - x_1)\), where \(m = \frac{1}{3}\) and the point is \((1, 0)\), the equation of the normal line becomes \(y - 0 = \frac{1}{3}(x - 1)\). Simplifying gives \(y = \frac{1}{3}x - \frac{1}{3}\).

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Key Concepts

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

Normal line
A normal line is a line perpendicular to another line at a given point. In this problem, we want to find the normal line to a parabola at a specific point. Understanding the relationship between normal lines and tangents is important, as they are interconnected through their slopes.

When a line is normal to a curve at a point, it is perpendicular to the tangent line at that point. Therefore, if we know the slope of the tangent line, we can find the slope of the normal line using the following property:
  • The slopes of two perpendicular lines multiply to be -1.
In the exercise, the slope of the normal line is given as \(\frac{1}{3}\). It must match the slope of the parallel line provided in the problem, since both lines share the same direction but are positioned differently.
Parabola
A parabola is a U-shaped curve described by a quadratic function. In this exercise, the parabola is defined by the function \(y = x^2 - 5x + 4\). The distinctive feature of a parabola is that it can open upwards or downwards depending on the sign of the quadratic term's coefficient.

Some characteristics of parabolas include:
  • The vertex, which is the highest or lowest point of the curve.
  • An axis of symmetry that passes vertically through the vertex.
  • The y-intercept, where the curve crosses the y-axis.
In the exercise, the vertex isn't directly needed, but understanding its form helps in visualizing how the parabola behaves. Specifically, parabolas are essential in various physical applications, like projectile motion or satellite dishes, where paths or signals follow parabolic trajectories.
Slope
Slope is a measure of how steep a line is. The slope of a line between two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is calculated as the change in y divided by the change in x:$$m = \frac{y_2 - y_1}{x_2 - x_1}$$In this exercise, slopes allow us to find both the tangent and normal lines to the parabola. The problem specifies a line with a slope of \(\frac{1}{3}\), which must be matched by the normal line to make them parallel.

Important things to remember about slope include:
  • A positive slope means the line ascends as x increases.
  • A negative slope means the line descends as x increases.
  • A zero slope means a horizontal line.
  • An undefined slope indicates a vertical line.
Using these principles helps in analyzing the behavior of lines and curves within coordinate spaces effectively.
Derivative
The derivative represents the rate of change of a function with respect to its variable, often giving us the slope of a curve at any given point. For the parabola in this exercise, finding the derivative of \(y = x^2 - 5x + 4\) results in the expression \(\frac{dy}{dx} = 2x - 5\). This derivative function provides the slope of the tangent line for each point on the curve.

Here's a straightforward explanation of derivatives:
  • They measure how a function value (y) changes as the input (x) changes.
  • The derivative function is sometimes called the "instantaneous" rate of change.
  • In graphical terms, it describes the slope of the tangent to a curve at any given point.
Understanding and using derivatives is crucial in calculus, as they connect algebraic representations to geometrical interpretations, providing insight into how functions behave and change over their domains.

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