   Chapter 4.7, Problem 57E ### 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 shortest possible length of the line segment that is cut off by the first quadrant and is tangent to the curve y = 3/x at some point?

To determine

To find: The shortest possible length of the line segment that is cut off by the first quadrant.

Explanation

Given:

The curve is y=3x.

Formula used:

If a straight line goes through a point (x1,y1) and has slope m, then the equation of the straight line is yy1=m(xx1).

Calculation:

Use the online graphing calculator and draw the graph of y=3x as shown below in Figure 1.

Differentiate y with respect to x,

dydx=3x2

An equation of the tangent at the point (a,3a) is,

y3a=3a2(xa)y=3a2x+6a

Substitute x=0 in y=3a2x+6a then,

y=3a2×0+6a=6a

The y intercept is 6a

Substitute y=0 in y=3a2x+6a then,

y=3a2x+6a0=3a2x+6ax=2a

Thus, the x intercept is x=2a.

The distance of the line segment that has endpoints at the intercepts is

d=(2a0)2+(06a)2

Let s=d2

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