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**Unformatted text preview: **3.5. SURFACES AND THEIR TANGENT PLANES 293 −
with →(0) = P . Then the velocity vector p ′ (0) lies in the plane tangent to
p
→
the surface L(f, c) at − (0).
p
Now on one hand, by the Chain Rule (3.3.6) we know that
d
dt t=0 →
−
→
[f (− (t))] = ∇ f (P ) · p ′ (0) ;
p →
→
on the other hand, since − (t) lies in the level set L(f, c), f (− (t)) = c for
p
p
all t, and in particular,
d
dt t=0 →
[f (− (t))] = 0.
p It follows that
→
−
∇ f (P ) · p ′ (0) = 0
for every vector tangent to L(f, c) at P ; in other words,10
Remark 3.5.4. If P is a regular point of f (x, y, z ), then the tangent plane
to the level set L(f, c) through P is the plane through P perpendicular to
→
−
the gradient vector ∇ f (P ) of f at P .
If we write this out in terms of coordinates, we ﬁnd that a point
(x, y, z ) = (x0 + △x, y0 + △y, z0 + △z ) lies on the plane tangent at
(x0 , y0 , z0 ) to the surface f (x, y, z ) = c = f (x0 , y0 , z0 ) if and only if
∂f
(x0 , y0 , z0 ) △x +
∂x ∂f
(x0 , y0 , z0 ) △y +
∂y ∂f
(x0 , y0 , z0 ) △z = 0,
∂z in other words, if
d(x0 ,y0 ,zso) f (x − x0 , y − y0 , z − z0 ) = 0.
Yet a third way to express this is to add c = f (x0 , y0 , z0 ) to both sides,
noting that the left side then becomes the linearization of f at P :
Tx0 ,y0 ,z0 f (x, y, z ) = f (x0 , y0 , z0 ) .
We summarize all of this in
10
Strictly speaking, we have only shown that every tangent vector is perpendicular to
→
−
→
−
∇ f ; we need to also show that every vector which is perpendicular to ∇ f is the velocity
vector of some curve in L(f, c) as it goes through P . See Exercise 14. ...

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