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Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 4, Problem 17P
To determine

To find:the point P on the arc AOB of the parabola that maximizes the area of the triangle PAB.

Expert Solution

Answer to Problem 17P

The pointPonthearcAOBoftheparabola is [P(12m,14m2)] .

Explanation of Solution

Given information:

The givenline is,

  y=mx+b

Intersects theparabola y=x2 in points A and B

Concept used:

The quadraticequationis

  y2=mx+b

Bythequadraticequation (y2=mx+b) thepointis,

  A=(x1,x12) , B=(x2,x22)

Thepoint P=(x,x2)

Set A1=(x1,0),B1=(x2,0) and p1=(x1,0)

Theareasofthreetrapezoidsare

  f(x)=area(A1ABB1)area(A1APP1)area(B1BPP1)=12(x12+x22)(x2x1)12(x12+x22)(xx1)12(x2+x22)(x2x)

Bycalculated,

  f(x)=12(x2x12x1x22+x1x2x2x2+xx22)=12[x12(x2x)+x22(xx1)+x2(x1x2)]

Todifferentiatetheequationwithrespectto x

  f'(x)=12(x12+x22+2x(x1x2))

To differentiate the equation again with respect to x

  f''(x)=12[2(x1x2)]=x1x2=x1x2<0

Since x2>x1

  f'(x)=02x(x1x2)=x12x22xp=12(x1+x2)

Simplified as,

  f(xp)=12(x12[12(x2x1)]+x22[12(x2x1)]+14(x1+x2)2(x1x2))=12(12[(x2x1)(x12+x22)]14[(x2x1)x(x1+x2)2])=18(x2x1)(2(x12+x22)(x12+2x1x2+x22))=18(x2x1)(x1x2)2

Simplify further,

  f(xp)=18(x2x1)(x1x2)2=18(x2x1)3...(1)

By the quadratic equation (y2=mx+b) with respect x1

  y=mx1+bx12mx1b=0x1=12(mm2+4b)

By the quadratic equation (y2=mx+b) with respect x2

  x2=12(m+m2+4b)

TheareaofthetrianglePABis

  18(x2x1)3=18(m2+4b)3(xp,xp2)=P(12m,14m2)

Therefore, thepoint P on the arc AOB of the parabola is [P(12m,14m2)] .

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