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Unformatted text preview: 3.6. EXTREMA 315 11. For each surface deﬁned implicitly, decide at each given point
whether one can solve locally for (a) z in terms of x and y ; (b) x in
terms of y and z ; y in terms of x and z . Find the partials of the
function if it exists.
(a) x3 z 2 − z 3 xy = 0 at (1, 1, 1) and at (0, 0, 0). (b) xy + z + 3xz 5 = 4 at (1, 0, 1) √
√
(c) x3 + y 3 + z 3 = 10 at (1, 2, 1) and at ( 3 5, 0, 3 5). (d) sin x cos y − cos x sin z = 0 at (π, 0, π ).
2 →
−
12. Prove that the gradient vector ∇ f is perpendicular to the level
surfaces of f , using the Chain Rule instead of Equation (3.18).
13. Mimic the argument for Theorem 3.5.3 to show that we can solve for
any variable whose partial does not vanish at our point. Challenge problem:
14. Suppose P (x0 , y0 , z0 ) is a regular point of the C 1 function f (x, y, z );
→
for deﬁniteness, assume ∂f (P ) = 0. Let − be a nonzero vector
v
∂z
→
−
perpendicular to ∇ f (P ). →
→
(a) Show that the projection − = (v1 , v2 ) of − onto the xy plane is
w
v
a nonzero vector. (b) By the Implicit Function Theorem, the level set L(f, c) of f
through P near P can be expressed as the graph z = φ(x, y ) of
some C 1 function φ(x, y ). Show that (at least for t < ε for some
→
ε > 0) the curve − (t) = (x0 + v1 t, y0 + v2 t, φ(x0 + v1 t, y0 + v2 t))
p
→
lies on L(f, c), and that p ′ (0) = − .
v
(c) This shows that every vector in the plane perpendicular to the
gradient is the velocity vector of some curve in L(f, c) as it goes
→
−
through P , at least if ∇ f (P ) has a nonzero z component. What
→
−
do you need to show this assuming only that ∇ f (P ) is a
nonzero vector? 3.6 Extrema Bounded Functions
Recall the following deﬁnitions from singlevariable calculus: ...
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This note was uploaded on 10/20/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.
 Fall '08
 ALL
 Calculus

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