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amistre64
 5 years ago
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) ?
amistre64
 5 years ago
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) ?

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it begins with the gradient....

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah thatl giv u d direction ratios of the tangent plane

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.04x+2z, 2y, 4z+2x thats 0, 2, 6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but then how do u get the direction ratios of the normal?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the normal is <0,2,6>; so use the point to calibrate the equation

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thats the ratios parallel to the plane

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.00(x+1)+2(y1)+6(z2)=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0those coefficients have to be in the direction of the normal, and not the tangent

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0n=<a,b,c>; P=(x0,y0,z0) equation of the plane: a(xx0)+b(yy0)+c(zz0)=0 is the way the textbook tells me it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ur using the standard eqn of a plane passing through a given pt a,b,c A(xa) + B(xb) +C(xc)=0 where A,B,C are direction ratios of the NORMAL

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0xx0; yy0 and zz0 are the components of the vector on the plane

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u get the cross product of <0,2,6> and <1, 1,2>

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thatl give u a vector normal to the plane say <A,B,C>

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then use the standard equation

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.01,1,2 is a point on the plane right? not a vector parallel to it tho.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the vector that is parallel to the plane is (xx0,yy0,zz0) where P(x0,y0,z0) is on the plane right?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02(x+1) +6(y1) +2 (z2) = 0

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.02,6,2 is not the normal to the tangent plane tho. if anything it too is parallel to the plane

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0arent you suppose to find Fx(X,Y) and Fy(X,Y) use the cross product to find the N?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i told u get the vector product of the direction of the tangent and the point

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0weve already found the gradient, i suggest we cross that with the point to get a vector normal to the plane

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the normal was found by the gradient; isnt the gradient and the normal parallel vectors?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no the gradient is parallel to the plane

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0If you found the N and have a point whats the problem?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm.... whats the gradient of 0 = 2x y +4

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0just trying to understand concepts :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0is that perp or // to y=2x+4 ; z=0?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0so the vector <2,1>; whose slope is 1/2 is // with the plne y=2x+2; z=0?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u cant define a gradient vector for a line

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0i didnt; i defined a plane; hence the z=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0bt m saying i think wt i did ws right

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Do you guys know about green theorem?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ive heard f it..nt studied it..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey sorry..i confirmed..the gradient vector is normal to the surface..rly sorry

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if we dot the gradient to a vector parallel to the plane y=2x+4; z=0, we will know if its perp or //

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0g=<2,1,0> v=<1, 2,0>  22+0=0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes m tellin u its normal to the surface

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wt u were doing is right...i feel u alrdy know all this nd ur just testing us out

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0;) gotta keep entertained ya know lol

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0id give you another medal if i could ;)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the only option available to me is the undo lol
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