Question

I'm unsure of how to start this question:

Find a scalar model for the plane tangent to the hyperboloid modeled implicitly by 3x^2 + y^2 - z^2 = 0 at the point (-2,2,4)

Answer 1

well taking partial derivatives we can find two vectors in the plane i think

Answer 2

almost; the partial construct of an equation gives the normal to the point

Answer 3

N=<Fx,Fy,Fz> then apply that to the given point

Answer 4

using the normal and the point we can construct the plane eq:

Fx(x-xo)+Fy(y-yo)+Fz(z-zo)=0

Answer 5

Fx, Fy, and Fz all being the derivative of the initial equation with respect to x, y, and z, respectively?

Answer 6

yes

Answer 7

so I got 6x^2 + 12x +2y^2 -4y - 2z^2 + 8z = 0. and since this is a scalar model I disgard the variables that are squared? leaving me with 12x - 4y + 8z = 0. Then I factor and disgard a 4 giving me the answer 3x - y +2z = 0?

Answer 8

im not sure how you got that ...

Answer 9

Fx = d/dx (3x^2 + y^2 - z^2 = 0) = 6x

Fy = d/dy (3x^2 + y^2 - z^2 = 0) = 2y

Fz = d/dz (3x^2 + y^2 - z^2 = 0) = -2z

N=<6x,2y,-2z>

Answer 10

correct thats what I got. then do the dot product with (x+2, y-2, z-4) and set equal to 0

Answer 11

applying the point values into the Normal; (-2,2,4)

we get N=<6(-2),2(2),-2(4)>
=<-12, 4, -8>

Answer 12

your dot product concept is good, giving us:

-12(x+2) +4(y-2) -8(z-4) = 0

of course we can scale the normal, say by -1/4

3(x+2) - (y-2) +2(z-4) = 0

Answer 13

right which gives 3x - y +2z =0

Answer 14

distributing we get:

3x-y+2z +2 = 0

Answer 15

which is what the given answer is. so thanks alot! that helps a whole bunch!

Answer 16

your welcome