Tiêu đề Maths1_ Winter-2019.pdf ✅

9/6/22, 3:23 PM Find the equations of the tenagent plane and normal line to the surface x2+y2+z2=3 at the point (1,1,1). | Winter-2019 gtu-paper-solution.com/Paper-Solution/Mathematics-I-3110014/Winter-2019/Question-1a 1/18 GTU APY Material Mathematics-I (3110014) (Winter-2019) Find the equations of the tenagent plane and normal line to the surface at the point . . Q1) (a) 3 Marks x 2 + y 2 + z 2 = 3 (1, 1, 1) Here,f(x, y, z) = x 2 + y 2 + z 2 − 3 and p = (x1, y1, z1) = (1, 1, 1) Now, ∂f ∂x = ∂ ∂x (x 2 + y 2 + z 2 − 3) = 2x ⟹ ∂f ∂x = (2x) (1,1,1) [Replace, p = (1, 1, 1)] ⟹ ( ∂f ∂x ) p = 2(1) = 2 ∂f ∂y = ∂ ∂y [x 2 + y 2 + z 2 − 3] = 2y ⟹ ∂f ∂y = (2y) (1,1,1) [Replace, p = (1, 1, 1)] ⟹ ( ∂f ∂y ) p = 2(1) = 2 ∂f ∂z = ∂ ∂z [x 2 + y 2 + z 2 − 3] = 2z ⟹ ∂f ∂z = (2z) (1,1,1) [Replace, p = (1, 1, 1)] ⟹ ( ∂f ∂z ) p = 2(1) = 2 Equation of TANGENT PLANE is, (x − x1) ( ∂f ∂x ) p + (y − y1) ( ∂f ∂x ) p + (z − z1) ( ∂f ∂x ) p = 0 ⟹ (x − 1) (2) + (y − 1) (2) + (z − 1) (2) = 0 ⟹ 2x − 2 + 2y − 2 + 2z − 2 = 0 ⟹ 2x + 2y + 2z − 6 = 0 ⟹ 2x + 2y + 2z = 6 ⟹ x + y + z = 3 [Cancelling by "2" from eqch term] Equation of NORMAL LINE is, x−x1 ( ∂f ∂x ) p = y−y1 ( ∂f ∂y ) p = z−z1 ( ∂f ∂z ) p ⟹ x−1 2 = y−1 2 = z−1 2 [Cancelling ′′2′′ from denomenator] ⟹ x − 1 = y − 1 = z − 1 [Adding ′′1′′ in all terms] ⟹ x = y = z
9/6/22, 3:23 PM Find the equations of the tenagent plane and normal line to the surface x2+y2+z2=3 at the point (1,1,1). | Winter-2019 gtu-paper-solution.com/Paper-Solution/Mathematics-I-3110014/Winter-2019/Question-1a 2/18 Find the equations of the tenagent plane and normal line to the surface x2+y2+z2=3 at the point (1,1,1). Here,fx,y,z=x2+y2+z2-3 and p=(x1,y1,z1)=(1, 1, 1). Now, ∂f∂x=∂∂x(x2+y2+z2-3)=2x ⟹∂f∂x=(2x)(1,1,1) [Replace,p=(1,1,1)] ⟹∂f∂xp=2(1)=2 ∂f∂y=∂∂y[x2+y2+z2-3]=2y ⟹∂f∂y=(2y)(1,1,1) [Replace,p=(1,1,1)] ⟹∂f∂yp=2(1)=2 ∂f∂z=∂∂z[x2+y2+z2-3]=2z ⟹∂f∂z=(2z)(1,1,1) [Replace,p=(1,1,1)] ⟹∂f∂zp=2(1)=2 Equation of TANGENT PLANE is, (x-x1) ∂f∂xp +(y-y1) ∂f∂xp +(z-z1) ∂f∂xp=0 ⟹(x-1) 2+(y-1) 2+(z-1) 2=0 ⟹2x-2+2y-2+2z-2=0 ⟹2x+2y+2z-6=0 ⟹2x+2y+2z=6 ⟹x+y+z=3 [Cancelling by "2" from eqch term] Equation of NORMAL LINE is, x-x1∂f∂xp = y-y1∂f∂yp = z-z1∂f∂zp ⟹x-12 = y-12 = z-12 Cancelling "2" from denomenator ⟹x-1=y-1=z-1 Adding "1" in all terms ⟹x=y=z Evaluate limx→0xex-log1+xx2 Now, limx→0xex-log1+xx2 00 type and applying L'Ho ̈spital's Rule Note that, ddxxex=xex+ex and ddxlog1+x=11+x =limx→0xex+ex-11+x2x 00 type and applying L'Ho ̈spital's Rule Note that, ddx11+x=-11+x2 Q1) (a) 3 Marks Q1) (b) 4 Marks
9/6/22, 3:23 PM Find the equations of the tenagent plane and normal line to the surface x2+y2+z2=3 at the point (1,1,1). | Winter-2019 gtu-paper-solution.com/Paper-Solution/Mathematics-I-3110014/Winter-2019/Question-1a 3/18 =limx→0xex+ex+ex+11+x22 =0e0+e0+e0+11+022 Substituting limit of x =0+1+1+12 =32 Using Gauss Elimination method solve the following system -x+3y+4z =30 3x+2y-z =9 2x-y+2z =10 Here, -x+3y+4z =30 3x+2y-z =9 2x-y+2z =10 co-efficient matrix A=-1 3 4 3 2-1 2-1 2 Variable matrix X=xyz Constant matrix B=30910 Now, Augmented Matrix is, A B=-1 3 4 3 2-1 2-1 2 30910 Taking R1(-1) A B~1-3 -43 2 -12-1 2 -30910 Taking R12(-3) , R13(-2) A B~1-3 -43-3(1) 2-3(-3) -1-3(-4)2-2(1)-1-2(-3) 2-2(-4) -309-3(-30)10-2(-30) A B~1 -3-40 11 110 5 10 -30 99 70 Taking R2111 , R315 A B~1 -3-40 1 10 1 2 -30 9 14 Taking R32-1 A B~1-3 -40 1 10-1(0)1-1(1) 2-1(1) -30914-1(9) A B~1-3 -40 1 10 0 1 -30 9 5 By back substitution, • z=5 • y+z=9 ⟹y=9-z ⟹y=9-5 ⟹y=4 • x-3y-4z=-30 ⟹x=-30+3y+4z Q1) (c) 7 Marks