amistre64
  • amistre64
What is the equation of the tangent plane to the function: 2x^2 +y^2 +2z^2 +2xz =7. At the point (-1,1,2) ?
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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amistre64
  • amistre64
it begins with the gradient....
anonymous
  • anonymous
yeah thatl giv u d direction ratios of the tangent plane
anonymous
  • anonymous
4x+2z, 2y, 4z+2x thats 0, 2, 6

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amistre64
  • amistre64
goody goody goody :)
anonymous
  • anonymous
but then how do u get the direction ratios of the normal?
amistre64
  • amistre64
the normal is <0,2,6>; so use the point to calibrate the equation
anonymous
  • anonymous
thats the ratios parallel to the plane
amistre64
  • amistre64
0(x+1)+2(y-1)+6(z-2)=0
anonymous
  • anonymous
thats wrong
anonymous
  • anonymous
those coefficients have to be in the direction of the normal, and not the tangent
amistre64
  • amistre64
can you explain?
amistre64
  • amistre64
n=; P=(x0,y0,z0) equation of the plane: a(x-x0)+b(y-y0)+c(z-z0)=0 is the way the textbook tells me it
anonymous
  • anonymous
ur using the standard eqn of a plane passing through a given pt a,b,c A(x-a) + B(x-b) +C(x-c)=0 where A,B,C are direction ratios of the NORMAL
amistre64
  • amistre64
x-x0; y-y0 and z-z0 are the components of the vector on the plane
anonymous
  • anonymous
u get the cross product of <0,2,6> and <-1, 1,2>
anonymous
  • anonymous
thatl give u a vector normal to the plane say
anonymous
  • anonymous
then use the standard equation
amistre64
  • amistre64
-1,1,2 is a point on the plane right? not a vector parallel to it tho.
anonymous
  • anonymous
yeah
amistre64
  • amistre64
the vector that is parallel to the plane is (x-x0,y-y0,z-z0) where P(x0,y0,z0) is on the plane right?
anonymous
  • anonymous
-2(x+1) +6(y-1) +2 (z-2) = 0
amistre64
  • amistre64
-2,6,2 is not the normal to the tangent plane tho. if anything it too is parallel to the plane
anonymous
  • anonymous
arent you suppose to find Fx(X,Y) and Fy(X,Y) use the cross product to find the N?
anonymous
  • anonymous
i told u get the vector product of the direction of the tangent and the point
anonymous
  • anonymous
weve already found the gradient, i suggest we cross that with the point to get a vector normal to the plane
amistre64
  • amistre64
the normal was found by the gradient; isnt the gradient and the normal parallel vectors?
anonymous
  • anonymous
no the gradient is parallel to the plane
anonymous
  • anonymous
If you found the N and have a point whats the problem?
amistre64
  • amistre64
hmmm.... whats the gradient of 0 = 2x -y +4
amistre64
  • amistre64
just trying to understand concepts :)
amistre64
  • amistre64
<2,-1,0> right?
amistre64
  • amistre64
is that perp or // to y=2x+4 ; z=0?
anonymous
  • anonymous
||
amistre64
  • amistre64
so the vector <2,-1>; whose slope is -1/2 is // with the plne y=2x+2; z=0?
amistre64
  • amistre64
(-1/2)*2=-1.....
anonymous
  • anonymous
u cant define a gradient vector for a line
amistre64
  • amistre64
i didnt; i defined a plane; hence the z=0
anonymous
  • anonymous
fine
anonymous
  • anonymous
bt m saying i think wt i did ws right
anonymous
  • anonymous
Do you guys know about green theorem?
amistre64
  • amistre64
i can tell :)
anonymous
  • anonymous
ive heard f it..nt studied it..
anonymous
  • anonymous
hey sorry..i confirmed..the gradient vector is normal to the surface..rly sorry
amistre64
  • amistre64
if we dot the gradient to a vector parallel to the plane y=2x+4; z=0, we will know if its perp or //
anonymous
  • anonymous
no i confirmed
amistre64
  • amistre64
g=<2,-1,0> v=<1, 2,0> -------- 2-2+0=0
anonymous
  • anonymous
yes m tellin u its normal to the surface
anonymous
  • anonymous
wt u were doing is right...i feel u alrdy know all this nd ur just testing us out
amistre64
  • amistre64
;) gotta keep entertained ya know lol
anonymous
  • anonymous
mean guy
amistre64
  • amistre64
id give you another medal if i could ;)
anonymous
  • anonymous
wts stopping u??
amistre64
  • amistre64
the only option available to me is the undo lol

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