Question

For the hyperboloid whose equation is x=4z^2-y^2+2y derive the scalar eqn for the plane that is tagent at pt (4,2,-1)?

Answer 1

Okay is this just as simple as I think it is using the tangent equation z=f(a,b,c) + f_x(x-a) + f_y(y-b)+f_z(z-c)

Answer 2

any ideas sirm3d?

Answer 3

first step is to rewrite the equation of the surface to F(x,y,z)=0, then evaluate the first partial derivatives at the specified point.

Answer 4

Okay I got f_x = -1, f_y=-2y-2 = -2 and f_z=8z = -8

Answer 5

compute these values \[\large F_x(4,2,-1),F_y(4,2,-1), F_z(4,2,-1)\] for the function \[\large F(x,y,z)=x-4z^2+y^2-2y\]. can you give me the result of your computation?

Answer 6

do I use the tangent equation now?

Answer 7

uhm, you mean \[\large f_y=-2y+2=-2\]

Answer 8

yeah sorry minor typo

Answer 9

if a plane has the vector representation <a, b, c>, the scalar form of the plane through (x0, y0, z0) is \[\large a(x-x_0)+b(y-y_0) + c(z-z_0) = 0\]

Answer 10

Ok so the vector representation in this case is that the f_x, f_y, f_z?

Answer 11

To give -(x-4) -2(y-2) -8(z+1)?

Answer 12

right

Answer 13

equated to zero, of course.

Answer 14

Do you know why though? Is it because since we're taking the derivative for each case

Answer 15

actually never mind I don't know why haha

Answer 16

vector representation is just the gradient?

Answer 17

this looks very similar to the tangent plane equation

Answer 18

the gradient (at a point) of the function is the vector representation of the tangent plane (at that point).

Answer 19

Oh gotcha, okay awesome explanation thanks a lot for your help! Very much appreciated