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Tutorial04-solutions - Problem Set 4 MH2100 Solutions E Let...

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Problem Set 4 MH2100 Solutions E Let : R 3 R 3 be a map given by = ( ) = ( ) = ( ) where , , and are polynomials in the variables . (a) Prove that is continuous on R 3 . (b) Is it true that is also differentiable on R 3 ? Justify your answer. S (a) Since all component functions are polynomials, they are elementary func- tions defined on the whole R 3 and are therefor continuous everywhere. (b) All partial derivatives of all component functions are also polynomials. They are elementary functions defined on the whole R 3 and continuous everywhere and hence the map is differentiable at any point. E What is the Jacobian matrix of a linear map ( x ) = A x , : R R . Here, A is an by matrix and x R is a vector-column. S It’s A itself at any point. Indeed, the definition of differentiability applied to ( x ) = A x with (supposedly) ( x ) = A says lim Δ x 0 A ( x + Δ x ) - A x - A Δ x Δ x = 0 which is true. :) 1
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1 Questions on calculation E Write down the equation of the tangent plane to each of the following sur- faces at given points (a) = 2 + 2 at the point (1 2 5) . (b) 2 + 2 + 2 = 169 at the point (3 4 12) . (c) = + ln at the point (1 1 1) . S (a) Here, = 2 , = 2 , so (1 2) = 2 and (1 2) = 4 . Thus the tangent plane is = 5 + 2( - 1) + 4( - 2) , which is 2 + 4 - - 5 = 0 . (b) Here, = ± 169 - 2 - 2 . Since at the given point we have = 12 > 0 , we use = 169 - 2 - 2 . Further, = - 169 - 2 - 2 = - = - 169 - 2 - 2 = - Hence (3 4 12) = - 1 4 and (3 4 12) = - 1 3 . Thus, the answer is = 12 - - 3 4 - - 4 3 3 + 4 + 12 = 169 (c) Here, it’s difficult to express as a function of , so we express in instead.
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