Question

Show that the curve traced parametrically by
\(x(t) = (cos(t - 1), t^3 - 1,\frac{1}{t} - 2)\) is tangent to the surface \(x^3 + y^3 + z^3 - xyz = 0 \) when t = 1.

Answer 1

could u find \(\Delta z\)
where z = x^2+y^3..... = 0
?

Answer 2

the idea is that x(t) will be tangent at t=2 when
\(\huge x'(2).\Delta z|_{t=2}=0\)

Answer 3

May I know what that delta z is?

Answer 4

And where does that "z = x^2+y^3..... = 0" come from?

Answer 5

i should have used other variable,
\(\Large w=x^3 + y^3 + z^3 - xyz = 0\)

Answer 6

\(\Delta \)
is the gradient function, heard of it ?

Answer 7

I'm sorry... I use \(\nabla\) instead of \(\Delta\), that's why I was confused.

Answer 8

eh, i don't remember what was used, you may be correct,
so could u find gradient ?

Answer 9

I can find ∇f, but do I have to substitute values into ∇ f?

Answer 10

when t=2,
x(2) = ...?

Answer 11

The problem is ∇f is in terms of x,y,z, so I don't know what to substitute there...

Answer 12

t=2 or t=1?

Answer 13

whatever 3 components u get for t=1 in x(t)

Answer 14

i guess, 1,0,-1
so put x=1,y=0 and z=-1

Answer 15

oh god, i used t=2 all the time

Answer 16

\(\huge x'(1).∇ w|_{x=1,y=-,z=-1}=0\)

Answer 17

y=0 i meant

Answer 18

0,3,-1 and
3,1,3
dot product comes out to be 0 ?
if yes, then tangent

Answer 19

Why would we find the dot product of x'(1) and ∇w(1,0,-1)?

Answer 20

the gradient is orthogonal to the surface at each point.
So, if x is tangent to the surface at some point, it should be orthogonal to the gradient of the level surface at that point.

Answer 21

Thanks a lot :)