1) let f(X,y)=x^2+xy+y^2 Diff. W.r.to X we get fx(x,y)=2x+y+0=2x+y fy(X,y)=0+X+2y=X+2y Then fx=0, 2x+y=0 .....(1) fy=0, x+2y=0 .....(2) Solving (1) & (2) we get Therefore 2x+y=0 2x+4y=0 - - ---------- 0-3y=0 y=0 Therefore from (1), X=0 The point is (0,0). Then the extreme values at (0,0): A=fxx(x,y)=2, B=fxy(x,y)=1, C=fyy(x,y)=2 Then at point (x,y)=(0,0), A=2,B=1,C=2 Therefore AC-B^2=4 greater than 0 and A=2 greater than 0. Hence the given function is maximum at point (0,0) andit's maximum value at point (0,0) is f(0,0)=0.
a) let y=sin6xcos2x then use the formula 2sinAcosB=sin(A+B)+sin(A-B), Therefore y=[sin(6x+2x)+sin(6x-2x)]/2. y=[sin8x+sin4x]/2 then taking n th derivatives to the both sides, we get yn=[8^nsin(8x+nπ/2)+4^nsin(4x+nπ/2)]/2 This is the answer.
b) let u(x,y)=[x+y]/[x^1/2+y^1/2] By Euler's theorem, we have xdu/dx+ydu/dy=nu ....(1) First find n: By definition of homogeneous function, we have u(kx,ky)=[kx+ky]/[(kx)^1/2+(ky)^1/2] =k[x+y]/k^1/2[x^1/2+y^1/2] =k^1/2 u(x,y) Hence n=1/2. Now to find du/dx and du/dy: du/dx=[(x^1/2+y^1/2)(1+0)-(x+y)(1/2x^-1/2+0)]/[x^1/2+y^1/2]^2 xdu/dx=1/2[(x+y)/(x^1/2+y^1/2)] .....(2) Similarly, we find ydu/dy=1/2[(x+y)/(x^1/2+y^1/2)].....(3) Adding (2) and (3) we get xdu/dx+ydu/dy=1/2 u(x,y) Hence verify it.
y=sin-1(x), diff. w.r.to x , we get y1=1/√(1-x^2) Therefore √(1-x^2) y1=1 then again diff. w.r.to x, we get √(1-x^2) y2-x/√(1-x^2) y1=0 Therefore multiplying by √(1-x^2), we get (1-x^2)y2-xy1=0 Taking n th derivatives and by using Leibnitz's theorem we get nc0yn+2(1-x^2)+nc1yn+1(-2x)+nc2yn(-2)- nc0yn+1x-nc1yn(1)=0 Since nc0=1,nc1=n,nc2=n(n-1)/2, we have (1-x^2)yn+2-(2n+1)xyn+1-n^2yn=0
Find the extreme value of (1) x^2 + xy + y^2 (2) x^3 + y^3 - 3axy
ReplyDelete1) let f(X,y)=x^2+xy+y^2
DeleteDiff. W.r.to X we get
fx(x,y)=2x+y+0=2x+y
fy(X,y)=0+X+2y=X+2y
Then fx=0, 2x+y=0 .....(1)
fy=0, x+2y=0 .....(2)
Solving (1) & (2) we get
Therefore 2x+y=0
2x+4y=0
- -
----------
0-3y=0
y=0
Therefore from (1), X=0
The point is (0,0).
Then the extreme values at (0,0):
A=fxx(x,y)=2, B=fxy(x,y)=1,
C=fyy(x,y)=2
Then at point (x,y)=(0,0),
A=2,B=1,C=2
Therefore AC-B^2=4 greater than 0 and A=2 greater than 0.
Hence the given function is maximum at point (0,0) andit's maximum value at point (0,0) is f(0,0)=0.
2) Similarly you will solve this.
DeleteLet f(x,y) =x^3 + y^3 - 3axy . Now by sufficient condition for extreme values, fx(x,y)=3x^2 -3ay...(1) fy(x,y)=3y^2 -3ax...(2) now how it can be solve
Delete(a) Find nth derivative of sin6x.cos2x (b) verify euler's theorem for x+y /x^1/2 + y^1/2
ReplyDeletea) let y=sin6xcos2x then use the formula 2sinAcosB=sin(A+B)+sin(A-B),
DeleteTherefore y=[sin(6x+2x)+sin(6x-2x)]/2.
y=[sin8x+sin4x]/2 then taking n th derivatives to the both sides, we get
yn=[8^nsin(8x+nπ/2)+4^nsin(4x+nπ/2)]/2
This is the answer.
b) let u(x,y)=[x+y]/[x^1/2+y^1/2]
DeleteBy Euler's theorem, we have
xdu/dx+ydu/dy=nu ....(1)
First find n:
By definition of homogeneous function, we have
u(kx,ky)=[kx+ky]/[(kx)^1/2+(ky)^1/2]
=k[x+y]/k^1/2[x^1/2+y^1/2]
=k^1/2 u(x,y)
Hence n=1/2.
Now to find du/dx and du/dy:
du/dx=[(x^1/2+y^1/2)(1+0)-(x+y)(1/2x^-1/2+0)]/[x^1/2+y^1/2]^2
xdu/dx=1/2[(x+y)/(x^1/2+y^1/2)] .....(2)
Similarly, we find
ydu/dy=1/2[(x+y)/(x^1/2+y^1/2)].....(3)
Adding (2) and (3) we get
xdu/dx+ydu/dy=1/2 u(x,y)
Hence verify it.
If y = sin-1(x) prove that (1-x^2)Yn+2 - (2n +1) XYn+1 - n^2 Yn =0
ReplyDeletey=sin-1(x), diff. w.r.to x , we get
ReplyDeletey1=1/√(1-x^2) Therefore
√(1-x^2) y1=1 then again diff. w.r.to x, we get
√(1-x^2) y2-x/√(1-x^2) y1=0
Therefore multiplying by √(1-x^2), we get
(1-x^2)y2-xy1=0
Taking n th derivatives and by using Leibnitz's theorem we get
nc0yn+2(1-x^2)+nc1yn+1(-2x)+nc2yn(-2)-
nc0yn+1x-nc1yn(1)=0
Since nc0=1,nc1=n,nc2=n(n-1)/2, we have
(1-x^2)yn+2-(2n+1)xyn+1-n^2yn=0