You are not logged in.
Pages: 1
Two surfaces are said to be tangential at a point P if they have the same tangent plane at P . Show that the surfaces z = √(2x²+2y²-1) and z = (1/3)√(x²+y²+4) are tangential at the point (1, 2, 3).
Offline
I tried to learn from luca, and of course the (1,2,3) is not
right if you just plug them in, the z's don't match.
So I set f(x,y,z) = 0 like luca,
and did partial deriv's.
Then set the df/dx ones equal to each other to see
what would happen, and I got
-145 = 34x^2 + 34y^2, which can't happen.
Don't know what this means.
Thanks to luca though for these amazing posts!!
I'm learning a lot!
igloo myrtilles fourmis
Offline
Differentiate....then evaluate both at 1,2,3 see if that helps
Offline
The df/dz is always -1, just a constant I think.
So the df/dx 's are different anyway: 1/9 and 2/3 at (1,2) for (x,y) and the z is not part of their slope equations.
igloo myrtilles fourmis
Offline
plugging in x=1,y=2
now the problem here though, is that although x_1,y_1 is related to x_2,y_2 by a multiple of 6, this is not the case for z_1 and z_2 so really, they are not colinear vectors, and the surfaces are not tangentenial?
either i did something wrong, or the question/answer is incorrect
Last edited by luca-deltodesco (2008-04-23 03:08:47)
The Beginning Of All Things To End.
The End Of All Things To Come.
Offline
Pages: 1