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Question Number 108099 by bemath last updated on 14/Aug/20

    ((★BeMath⊚)/⊓)   (1) ∫ x tan^(−1) (x) dx ?  (2) Find the distance of the point   (3,3,1) from the plane Π with equation  (r^→ −i^→ −j^→ )•(i^→ −j^→ +k^→ ) = 0 , also   find the point on the plane that is  nearest to (3,3,1).

BeMath(1)xtan1(x)dx?(2)Findthedistanceofthepoint(3,3,1)fromtheplaneΠwithequation(rij)(ij+k)=0,alsofindthepointontheplanethatisnearestto(3,3,1).

Commented by bemath last updated on 14/Aug/20

thank you both

thankyouboth

Answered by Dwaipayan Shikari last updated on 14/Aug/20

∫xtan^(−1) x  (x^2 /2)tan^(−1) x−(1/2)∫(x^2 /(1+x^2 ))  (x^2 /2)tan^(−1) x−(x/2)+(1/2)tan^(−1) x+C

xtan1xx22tan1x12x21+x2x22tan1xx2+12tan1x+C

Answered by mathmax by abdo last updated on 14/Aug/20

1) A =∫ x arctan(x)dx  by parts  A =(x^2 /2) arctanx −∫ (x^2 /2) (dx/(1+x^2 ))   =(x^2 /2)arctan(x)−(1/2)∫  ((1+x^2 −1)/(1+x^2 ))dx  =(x^2 /2) arctan(x)−(x/(2 )) +(1/2) arctan(x) +C  =(1/2)(x^2  +1)arctanx −(x/2) +C

1)A=xarctan(x)dxbypartsA=x22arctanxx22dx1+x2=x22arctan(x)121+x211+x2dx=x22arctan(x)x2+12arctan(x)+C=12(x2+1)arctanxx2+C

Answered by 1549442205PVT last updated on 15/Aug/20

i^→ =(1,0,0),j^(→) =(0,1,0),k^(→) =(0,0,1)  ⇒i ^(→) −j ^(→) +k ^(→) =(1,−1,1),r ^(→) =(x,y,z)  ⇒r ^(→) −i ^(→)  −j ^(→) =(x−1,y−1,z)  (r ^(→) −i  ^(→) −j ^(→) )•(i ^(→) −j ^(→)  +k ^(→) )  =(x−1,y−1,z)•(1,−1,1)=0⇔x−1−y+1+z=0  ⇔x−y+z=0  d[(3,3,1),Π]=((∣3−3+1∣)/( (√(1^2 +(−1)^2 +1^2 ))))=(1/( (√3)))  ii)Let M(x,y,y−x)∈Πbe any point of  the plane Π which has the equation:  x−y+z=0 and A(3,3,1).Then  d(A,M)=(√((x−3)^2 +(y−3)^2 +(y−x−1)^2 ))  we need to find smallest of the expression  P=(x−3)^2 +(y−3)^2 +(y−x−1)^2   =2x^2 +2y^2 −4x−8y−2xy+19  =2x^2 −(2y+4)x+2y^2 −8y+19  =2[(x−((y+2)/2))^2 +2y^2 −8y+19−(((y+2)^2 )/2)  =2[(x−((y+2)/2))^2 +((3y^2 −20y+34)/2)  =2[(x−((y+2)/2))^2 +((3(y−((10)/3))^2 +(2/3))/2)  2[(x−((y+2)/2))^2 +((3(y−2)^2 )/2)+(1/3)≥(1/3)  ⇒d(A,M)≥(1/( (√3)))  The equality ocurrs if and only if   { ((y−((10)/3)=0)),((x−((y+2)/2)=0)) :} ⇔ { ((x=8/3)),((y=10/3)) :}⇒z=2/3  Thus the nearest distance between the  point A(3,3,1)and Π be (1/( (√3))) and  the point  M lie on the plane Πwhich  is nearest to   A having  coordinate be M((8/3),((10)/3),(2/3))  other way   Let H(x,y,y−x)∈Π.Then A(3,3,1)is nearest  to the plane Π if and only if AH^(→) ⊥Π  ⇔AH^(→) //n_Π ^(→) =(1,−1,1),since AH^(→) =(x−3,y−3,y−x−1),  so AH^(→) //n_Π ^(→) ⇔((x−3)/1)=((y−3)/(−1))=((y−x−1)/1)  ⇔ { ((x+y=6)),((2x−y=2)) :}⇔ { ((x=8/3)),((y=10/3)),((z=2/3)) :}⇒H((8/3),((10)/3),(2/3))  The distance between A and the plane  Π equal to d(A,H)=(√((3−(8/3))^2 +(3−((10)/3))^2 +(1−(2/3))^2 ))  =(√(3((1/3))^2 ))=(1/( (√3)))

i=(1,0,0),j=(0,1,0),k=(0,0,1)ij+k=(1,1,1),r=(x,y,z)rij=(x1,y1,z)(rij)(ij+k)=(x1,y1,z)(1,1,1)=0x1y+1+z=0xy+z=0d[(3,3,1),Π]=33+112+(1)2+12=13ii)LetM(x,y,yx)ΠbeanypointoftheplaneΠwhichhastheequation:xy+z=0andA(3,3,1).Thend(A,M)=(x3)2+(y3)2+(yx1)2weneedtofindsmallestoftheexpressionP=(x3)2+(y3)2+(yx1)2=2x2+2y24x8y2xy+19=2x2(2y+4)x+2y28y+19=2[(xy+22)2+2y28y+19(y+2)22=2[(xy+22)2+3y220y+342=2[(xy+22)2+3(y103)2+2322[(xy+22)2+3(y2)22+1313d(A,M)13Theequalityocurrsifandonlyif{y103=0xy+22=0{x=8/3y=10/3z=2/3ThusthenearestdistancebetweenthepointA(3,3,1)andΠbe13andthepointMlieontheplaneΠwhichisnearesttoAhavingcoordinatebeM(83,103,23)otherwayLetH(x,y,yx)Π.ThenA(3,3,1)isnearesttotheplaneΠifandonlyifAHΠAH//nΠ=(1,1,1),sinceAH=(x3,y3,yx1),soAH//nΠx31=y31=yx11{x+y=62xy=2{x=8/3y=10/3z=2/3H(83,103,23)ThedistancebetweenAandtheplaneΠequaltod(A,H)=(383)2+(3103)2+(123)2=3(13)2=13

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