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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ⇔...
View Full Document

This document was uploaded on 02/10/2014.

Ask a homework question - tutors are online