Tutorial05-solutions

# Prove that is constant e 5 s we have 1 2 1 2 1 2 it

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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 &gt; 0 is some ﬁxed number. Let τ be the tangent plane to this surface at any point ( 0 0 0 ) and let &gt; 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.

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