   Chapter 4.7, Problem 58E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# What is the smallest possible area of the triangle that is cut off by the first quadrant and whose hypotenuse is tangent to the parabola y = 4 − x2 at some point?

To determine

To find: The smallest possible area of the triangle.

Explanation

Given:

The curve is y=4x2

Formula used:

Let the curve be y=f(x).

Differentiate y with respect to x to get dydx

Then the equation of the tangent line of the curve y=f(x) at (x1,y1) is,

yy1=dydx(xx1)

Calculation:

The parabola is y=4x2.

Hence, any point on the parabola is of the form (a,4a2).

Differentiate y=4x2 with respect to x,

y=2x

Hence, the equation of the tangent line on the parabola at (a,4a2) is,

y(4a2)=2a(xa)y=4a22ax+2a2y=a22ax+4

From Figure 1, it is noticed that AB is the tangent line of y=4x2 at (a,4a2).

To find y-intercept of AB,

Substitute x=0 in y=a22ax+4,

y=a2+4

Hence, the height of ΔAOB is a2+4

To find x-intercept of AB,

Substitute y=0 in y=a22ax+4,

0=a22ax+42ax=a2+4x=a2+42a

The base of ΔAOB is a2+42a

### Still sussing out bartleby?

Check out a sample textbook solution.

See a sample solution

#### The Solution to Your Study Problems

Bartleby provides explanations to thousands of textbook problems written by our experts, many with advanced degrees!

Get Started

#### Find more solutions based on key concepts 