/*! 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 59 Show that the triangle that is f... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 59

Show that the triangle that is formed by any tangent line to the graph of \(y=1 / x, x>0,\) and the coordinate axes has an area of 2 square units.

Short Answer

Expert verified
The area is always 2 square units, regardless of the point of tangency.

Step by step solution

01

Find the Point of Tangency

Let the point of tangency be \((a, \frac{1}{a})\). We know this point lies on the curve \(y = \frac{1}{x}\).
02

Determine the Slope of the Tangent Line

Differentiate \(y = \frac{1}{x}\) to obtain\[\frac{dy}{dx} = -\frac{1}{x^2}\].At \(x = a\), the slope (\(m\)) is \(-\frac{1}{a^2}\).
03

Write the Equation of the Tangent Line

The equation of the tangent line at \((a, \frac{1}{a})\) can be expressed as:\[y - \frac{1}{a} = -\frac{1}{a^2}(x - a)\].Simplifying, we obtain:\[y = -\frac{1}{a^2}x + \frac{2}{a}\].
04

Find the Intercepts

The x-intercept occurs when \(y = 0\):\[0 = -\frac{1}{a^2}x + \frac{2}{a}\].Solving, \(x = 2a\).The y-intercept occurs when \(x = 0\):\[y = \frac{2}{a}\].
05

Calculate the Area of the Triangle

The base of the triangle is \(2a\) (x-intercept) and the height is \(\frac{2}{a}\) (y-intercept).The area \(A\) is given by:\[A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2a \times \frac{2}{a} = 2\] square units.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus that allows us to determine the rate at which a function changes at any given point.
  • In simpler terms, differentiation helps us find the slope or steepness of the tangent line to a curve at a particular point.
  • To differentiate a function means to compute its derivative, which can give us valuable information, like the slope of a tangent line.
In the exercise, we have a curve defined by the function \(y = \frac{1}{x}\), where \(x > 0\). To find the derivative, we apply differentiation rules to obtain:\[\frac{dy}{dx} = -\frac{1}{x^2}\]Here, \(\frac{dy}{dx}\) represents the derivative of \(y\) with respect to \(x\), which tells us the slope of the tangent line at any point on the curve. When evaluated at a specific \(x=a\), it gives \(-\frac{1}{a^2}\). This is the slope of the tangent line at the point \((a, \frac{1}{a})\) on the graph.
Area of a Triangle
The area of a triangle can be easily calculated if you know the base and the height.
  • These dimensions are typically perpendicular to each other, making it straightforward to use the area formula for triangles.
  • The formula to find the area \(A\) is: \[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
In the context of this exercise, the triangle's base and height are derived from the x-intercept and y-intercept of the tangent line.
  • The base of the triangle is the segment along the x-axis from the origin to the x-intercept \(2a\).
  • The height is the y-intercept, \(\frac{2}{a}\), which is the segment along the y-axis from the origin to where the tangent line intersects.
Therefore, the area of the triangle can be calculated as:\[A = \frac{1}{2} \times 2a \times \frac{2}{a} = 2 \text{ square units}\]
Equation of a Line
The equation of a line in the slope-intercept form is crucial for understanding how lines behave on a graph.
  • The general form of this equation is \(y = mx + c\), where \(m\) represents the slope and \(c\) the y-intercept.
  • The slope \(m\) tells us how steep the line is, and in which direction it is slanting.
For the tangent line in this problem, starting from the point \((a, \frac{1}{a})\), and with a slope \(-\frac{1}{a^2}\), the equation is:\[y - \frac{1}{a} = -\frac{1}{a^2}(x - a)\]Simplifying this gives us:\[y = -\frac{1}{a^2}x + \frac{2}{a}\]
  • Here, \(-\frac{1}{a^2}\) is the slope, which indicates the tangent's direction and steepness.
  • The y-intercept \(\frac{2}{a}\) is the point where the tangent crosses the y-axis.
This is how the equation helps in forming and visualizing the line on a graph.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let \(f(x)=5 \sqrt{x}\) and \(g(x)=4+\cos x\) (a) Find \((f \circ g)(x)\) and \((f \circ g)^{\prime}(x)\) (b) Find \((g \circ f)(x)\) and \((g \circ f)^{\prime}(x)\)

Find the indicated derivative. $$ y=\cot ^{3}(\pi-\theta) ; \text { find } \frac{d y}{d \theta} $$

Find \(f^{\prime}(x)\) $$ f(x)=\cos ^{3}\left(\frac{x}{x+1}\right) $$

Find \(f^{\prime}(x)\) $$ f(x)=\sqrt{\cos (5 x)} $$

Find \(d^{2} y / d x^{2}\) $$ y=\frac{1+x}{1-x} $$
See all solutions

Recommended explanations on Math Textbooks

Geometry

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Decision Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.