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Unformatted text preview: ute = − + 2 + 2 and = − + 2 + − 1 into the PDE + =0 to rewrite the above PDE in terms of partial derivatives of
S with respect to and . According to the Chain Rule, we have
∂
+
∂ ∂
=
∂ ∂
∂
∂
=
+
∂
∂
∂ ∂
=
∂ ∂
+
∂ ∂
∂
∂
=2
−
∂
∂
∂ Now the ﬁrst derivatives are
∂
∂
+
∂
∂ = = + = 2 ∂
∂
−
∂
∂ =2 − Further,
= = ∂∂
= 2 ∂∂ ∂
∂ + + ∂∂
− ∂∂ ∂
∂ [ + ]=
[2 − ] = 2
[2 − ] = 4 +
+
+
=
+2 +
−
+2 −
=2
+
−
−2 −2 +
=4
−4 + Thus the equation is
2( +2 that is, 3 4 +
+ ) + (2 + − ) − (4 −4 = 0. + )+( + ) + (2 − )=0
:) Extra problems on Chain Rule √
√
√
√
Consider the surface given by the equation
+
+
=
, where
> 0 is some ﬁxed number. Let τ be the tangent plane to this surface at any point
( 0 0 0 ) and let
> 0 be the intercept, the intercept, and the intercept of the
plane τ . Prove that + + is constant.
E 5 S We have
1
=√
2 1
=√
2 1
=√
2 It follows that the equation of the tangent plane at the point (
(−
√ 0) + (−
√ 0 0) + (−
√ 0) =0 0 ⇔...
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This document was uploaded on 02/10/2014.
 Spring '13

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