Question

How do I find the angle between the sphere x^2+y^2+z^2-9=0 and the paraboloid z-x^2-y^2+3=0 at point (2,-1,2) ?

Answer 1

@sidsiddhartha Could you help with this please? (:

Answer 2

tangent planes might be a start

Answer 3

Should I derive @amistre64

Answer 4

do you recall what a gradient is?

Answer 5

f(x,y)+fx(x,y)+fy(x,y)?

Answer 6

that sounds about right, the gradiant tells us the normal of the tangent plane at a given point ... it is those vectors that you need to be comparing

Answer 7

find f wrt.x and f wrt.y and f wrt.z

Answer 8

for both of them?

Answer 9

yep

Answer 10

uh so for the sphere <2x,2y,2z> and paraboloid <-2x,-2y,1>

Answer 11

yep

let:

f = x^2+y^2+z^2-9

and

g = z-x^2-y^2+3

find the partials for f and g, to define the normal vectors to their tangent planes and then we can determine the angles between them

Answer 12

apply the xyz from the point into the gradients to define them now

Answer 13

\[\nabla \phi_1=2x \hat{i}+2y \hat{j}+2z \hat{k}\\and\\ \nabla \phi_2=-2x \hat{i}-2y \hat{j}+\hat{k}\]
like this

Answer 14

uh right and then?

Answer 15

\[now~\angle~\between~them\\cos \theta=\frac{ \nabla \phi_1. \nabla \phi_2 }{ |\nabla \phi_1|.|\nabla \phi_2|}\]

Answer 16

recall the dotproduct trig definition

|u| |v| cos(a) = u.v

that will help us define the angle between the normals

Answer 17

Gotcha thank you guys so much

Answer 18

yes simple dot product