Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Question

Chapter 5.4, Problem 42E

To determine

To find the minimum distance of semicircle.

Expert Solution & Answer

Answer to Problem 42E

The minimum distance is d=2 .

Explanation of Solution

Given information: Given equation y=16−x2 ,the point is (1,3) .

Calculation:

The equation of the semicircle is given by y=16−x2.

Distance of the point (1,3) from the point (x,16−x2) is d=(x−1)2+(16−x2−3)2.

Square of the distance is ,

D=(x−1)2+(16−x2−3)2 =x2+1−2x+(16−x2)+3−23(16−x2) =−2x+20−248−3x2

For any function f(x) the condition of minimum and maximum is obtained by find f′(x)=0 ,

Critical points and end points

D=−2x+20−248−3x2dDdx=−2−2(−6x)248−3x2 =−2+(6x)48−3x2

Critical points occur at

D′(x)=0,−4<x<4−2+(6x)48−3x2=0(6x)48−3x2=2 36x2=4(48−3x2) 12x2=48 x=±2

Consider x=2

Here dDdx<0 for−4<x<2anddDdx>0 for 2<x<4. this corresponds to minimum distance.

For x=2,

D=−2x+20−248−3x2 =−4+20−248−12 =4d=4=2

Thus, the minimum distance is d=2.

Chapter 5.4, Problem 41E

Chapter 5.4, Problem 43E

Chapter 5 Solutions

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

Ch. 5.1 - Prob. 1QR Ch. 5.1 - Prob. 2QR Ch. 5.1 - Prob. 3QR Ch. 5.1 - Prob. 4QR Ch. 5.1 - Prob. 5QR Ch. 5.1 - Prob. 6QR Ch. 5.1 - Prob. 7QR Ch. 5.1 - Prob. 8QR Ch. 5.1 - Prob. 9QR Ch. 5.1 - Prob. 10QR

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