Question

Please help!
Show that the surfaces
z=1/2 (x^2 + y^2 - 1) and z=1/2(1 - x^2 - y^2) are orthoganal at all points of intersection

(We know two surfaces are said to be orthogonal to each other at a point P if the normals to their tangent planes are perpendicular at P. )

Answer 1

Take the gradient of each curve. That is the normal vector of their respective tangent planes.

Answer 2

Then take the dot product of the two gradients and show that you'll get 0.

Answer 3

Ok.. so how do I go about finding the normal vector of the tangent plane?

Answer 4

Have you taken up gradients?

Answer 5

No we haven't

Answer 6

How about partial derivatives?

Answer 7

yeah!

Answer 8

Ok the equation for the tangent plane for z=f(x,y) is

z = f(x0,y0) + fx (x-x0) + fy (y-y0)

where (x0,y0) is a point on the surface.

We can infer that the normal vector N = <fx, fy, -1>

so use that to find the normal vectors and show that they are orthogonal

Answer 9

Ok wait, what does that have to do with partial derivitives?!

Answer 10

oh wait, nvm. so, would i just randomly pick a point on the surface .. or leave it as f(x0,y0) ?

Answer 11

Pick a point. Actually you can compute vectorN=<fx, fy, -1> right away.

Answer 12

what do you mean right away?

Answer 13

Like what I said before, we can infer that the normal vector N=<fx,fy,-1>, no need to find a point and go through computing the equation of the plane.

Answer 14

that's the normal for the first equation right?

Answer 15

And the second is perpendicular to that? N = <1, -fy, -fx> ?

Answer 16

For any equation.
Let the first surface be z=f(x,y), then vector N1=<fx,fy,-1>
Let the second be z=g(x,y), then vector N2=<gx,gy,-1>

Answer 17

Ok, so z = f(x0, y0) + 2x (x-x0) + 2y (y-y0)
and then z= f(x0,y0) -2x (x-x0) - 2y (y-y0)

Answer 18

Use g instead of f for the 2nd surface to avoid confusion.

Remember the normal vector of a plane ax+by+cz=d is just <a,b,c> or ai+bj+ck?

Answer 19

Ok wait, I'm kind of confused now. So where do I go from here?
z = f(x0, y0) + 2x (x-x0) + 2y (y-y0)
z= g(x0,y0) -2x (x-x0) - 2y (y-y0)

Answer 20

It should be

1st plane: z = f(x0,y0) + x(x-x0) + y(y-y0)
2nd plane: z = g(x0,y0) - x(x-x0) - y(y-y0)

there is a (1/2) so fx=x, fy=y, gx=-x, gy=-y.

Rearrange it to the form ax+by+z=constant

1st plane: x (x-x0) + y (y-y0) - z = -f(x0,y0)
2nd plane: -x (x-x0) - y (y-y0) - z = -g(x0,y0)

So the components of the normal vectors are the coefficients of x,y,z.

N1 = xi + yj -k
N2 = -xi -yj -k

(Do you use the ijk notation?)

Answer 21

Take the dot/inner product.

N1 dot N2 = -x^2 -y^2 +1 --> show this is =0

Answer 22

so -x^2 - y^2 + 1 dot x^2 + y^2 - 1 = 0 ?

Answer 23

The problem says at all points of intersection.

so f(x,y,)=g(x,y) to get all the ponts (x,y) where the surfaces intersect.

(1/2)(x^2+y^2-1)=(1/2)(-x^2-y^2+1)

blah... blah... solve... solve..

we get

x^2+y^2=1

Answer 24

now N1 dot N2 = -x^2 -y^2 +1

= -(x^2 + y^2) +1

= -(1) + 1

= 0, therefore orthogonal at every point (x,y) of intersection.

Answer 25

Oh ok! Thanks!

Answer 26

dot product of the normal vectors.

Answer 27

what? dot product of the normal vectors? thats just what you were saying before right?

Answer 28

yeah. coz u said "so -x^2 - y^2 + 1 dot x^2 + y^2 - 1 = 0 ?"

It should be (xi+yj-k) dot (-xi-yj-k)