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Chapter 3: Problem 33

Show that the line \(y=m x+b\) is own tangent line at any point \(\left(x_{0}, m x_{0}+b\right)\)

Short Answer

Expert verified
The line is tangent to itself since it always has the same slope as its tangent at any point.

Step by step solution

01

Understand the Exercise

The exercise asks us to prove that the line defined by the equation \(y = mx + b\) is its own tangent at any given point \((x_0, mx_0 + b)\). This means that the slope of the tangent at the point on the line is the same as the slope of the line itself.
02

Recall the Slope-Intercept Form

The equation \(y = mx + b\) is in slope-intercept form, where \(m\) represents the slope of the line, and \(b\) is the y-intercept. The line has a constant slope \(m\) throughout.
03

Identify the Point on the Line

Consider the point \((x_0, mx_0 + b)\) which is a generic point on the line. At this point on the line, with \(x_0\) as the x-coordinate, the y-coordinate will be \(mx_0 + b\).
04

Find the Derivative

Find the derivative of the line's equation to determine the slope of the tangent. Since the line equation is linear, its derivative is simply the constant slope \(m\).
05

Show Tangency at the Given Point

At the point \((x_0, mx_0 + b)\), the tangent slope, given by the derivative, is \(m\). Since the slope of the line \(y = mx + b\) is also \(m\), the line is tangent to itself.
06

Conclusion

Since at any point on the line, the slope of the tangent (derived from the derivative of the line function) equals the slope of the line itself, the line \(y = mx + b\) is always tangent to itself at any given point on the line.

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Key Concepts

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

Slope-Intercept Form
The slope-intercept form of a linear equation is a straightforward way to represent a line on a graph. It is written as \(y = mx + b\).
In this equation:
  • \(m\) is the slope of the line, indicating how steep the line is.
  • \(b\) is the y-intercept, the point where the line crosses the y-axis.
The slope \(m\) shows how much \(y\) increases when \(x\) increases by one unit. When you visualize this on a graph, starting from the y-intercept \((0, b)\), you can move up or down by \(m\) and over one unit to find another point on the line.
The consistency of \(m\) across the entire line means each point on the line has the same steepness, which is key for understanding tangency.
Derivative of Linear Functions
The derivative tells us the rate at which a function's output changes with respect to changes in input. For a linear function like \(y = mx + b\), the derivative is very simple: it is the constant \(m\).
This comes from understanding that a linear function has a constant rate of change. As a result,
  • The derivative, or slope, is the same at every point on the line.
To find the derivative:
  • Differentiate \(y = mx + b\) with respect to \(x\).
  • The derivative, \(\frac{dy}{dx} = m\), remains constant.
The fact that the derivative is a constant \(m\) here reflects the line鈥檚 consistent slope throughout, reinforcing why the line can be its own tangent.
Self-Tangency
A line being tangent to itself might seem a bit puzzling at first, but with linear functions, it makes perfect sense.
Self-tangency means that at any spot on the line, the slope of the tangent (derived from the function's derivative) is exactly equal to the slope of the line itself.
Consider the line \(y = mx + b\) and a point \((x_0, mx_0 + b)\) on it:
  • The slope of the line is \(m\).
  • The derivative of the line is also \(m\).
  • Thus, the slope of the tangent at this point is consistent with the line's slope.
The concept of self-tangency highlights the idea that every point on the line possesses the same directional tendency as the line as a whole. Thus, it is self-tangent everywhere on its domain. This property underscores the uniformity of linear equations, where the slope never changes.

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Most popular questions from this chapter

A balloon and a bicycle \(A\) balloon is rising vertically above a level, straight road at a constant rate of 1 \(\mathrm{ft} / \mathrm{sec} .\) Just when the balloon is 65 \(\mathrm{ft} /\) above the ground, a bicycle moving at a constant rate of 17 \(\mathrm{ft} / \mathrm{sec}\) passes under it. How fast is the distance \(s(t)\) between the bicycle and balloon increasing 3 sec later?

$$If r=\sin (f(t)), f(0)=\pi / 3, and f^{\prime}(0)=4, then what is d r / d t at t=0 ?$$

Surface area Suppose that the radius \(r\) and surface area \(S=4 \pi r^{2}\) of a sphere are differentiable functions of \(t\) . Write an equation that relates \(d S / d t\) to \(d r / d t\) .

Use a CAS to perform the following steps. \begin{equation} \begin{array}{l}{\text { a. Plot the equation with the implicit plotter of a CAS. Check to }} \\ {\text { see that the given point } P \text { satisfies the equation. }} \\ {\text { b. Using implicit differentiation, find a formula for the deriva- }} \\ {\text { tive } d y / d x \text { and evaluate it at the given point } P .}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. Use the slope found in part (b) to find an equation for the tan- }} \\ {\text { gent line to the curve at } P \text { . Then plot the implicit curve and }} \\ {\text { tangent line together on a single graph. }}\end{array} \end{equation} \begin{equation} y^{2}+y=\frac{2+x}{1-x}, \quad P(0,1) \end{equation}

In Exercises \(57-60,\) use a CAS to estimate the magnitude of the error in using the linearization in place of the function over a specified interval I. Perform the following steps: $$ \begin{array}{l}{\text { a. Plot the function } f \text { over } I} \\ {\text { b. Find the linearization } L \text { of the function at the point } a \text { . }} \\ {\text { c. Plot } f \text { and } L \text { together on a single graph. }} \\ {\text { d. Plot the absolute error }|f(x)-L(x)| \text { over } I \text { and find its max- }} \\ {\text { imum value. }}\end{array} $$ $$ \begin{array}{l}{\text { e. From your graph in part (d), estimate as large a } \delta>0 \text { as you }} \\ {\text { can, satisfing }}\end{array} $$ $$ \begin{array}{c}{|x-a|<\delta \quad \Rightarrow \quad|f(x)-L(x)|<\epsilon} \\\ {\text { for } \epsilon=0.5,0.1, \text { and } 0.01 . \text { Then check graphically to see if }} \\ {\text { your } \delta \text { -estimate holds true. }}\end{array} $$ $$ f(x)=x^{3}+x^{2}-2 x, \quad[-1,2], \quad a=1 $$
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