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

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Problem Set 4 MH2100 Solutions E±²³´µ¶² · Let ± : R 3 ²³´³µ R 3 ¶³·³¸ be a map given by = ¹ ( ²³´³µ ) ³ · = º ( ) ³ ¸ = » ( ) ³ where ¹ , º , 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¸¹º»µ¸¼ (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±²³´µ¶² ½ What is the Jacobian matrix of a linear map ± ( x ) = A x , ± : R ¼ R ½ . Here, A is an ½ by ¼ matrix and x R ¼ is a vector-column. S¸¹º»µ¸¼ It’s A itself at any point. Indeed, the definition of differentiability applied to ± ( x ) = A x with (supposedly) ± ¾ ( x ) = A says lim Δ x 0 ¿A ( x + Δ x ) - A x - A Δ x ¿ ¿ Δ x ¿ = 0 ³ which is true. :) 1
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1 Questions on calculation E±²³´µ¶² · Write down the equation of the tangent plane to each of the following sur- faces at given points (a) ± = ² 2 + ³ 2 at the point (1 ´ 2 ´ 5). (b) ² 2 + ³ 2 + ± 2 = 169 at the point (3 ´ 4 ´ 12). (c) ± = ³ + ln ² ± at the point (1 ´ 1 ´ 1). S¸¹º»µ¸¼ (a) Here, ± ² = 2 ² , ± ³ = 2 ³ , so ± ² (1 ´ 2) = 2 and ± ³ (1 ´ 2) = 4. Thus the tangent plane is ± = 5 + 2( ² - 1) + 4( ³ - 2), which is 2 ² + 4 ³ - ± - 5 = 0. (b) Here, ± = µ ½ 169 - ² 2 - ³ 2 . Since at the given point we have ± = 12 > 0, we use ± = ½ 169 - ² 2 - ³ 2 . Further, ± ² = - ² ½ 169 - ² 2 - ³ 2 = - ² ± ´ ± ³ = - ³ ½ 169 - ² 2 - ³ 2 = - ³ ± Hence ± ² (3 ´ 4 ´ 12) = - 1 4 and ± ³ (3 ´ 4 ´ 12) = - 1 3 . Thus, the answer is ± = 12 - ² - 3 4 - ³ - 4 3 3 ² + 4 ³ + 12 ± = 169 (c) Here, it’s difficult to express ± as a function of ²´³ , so we express ³ in ²´± instead. We have ³ = ±- ln ² ± = ln ² +ln ± , ³ ² = - 1 ² and ³ ² (1 ´ 1) = - 1. Further, ³ ± = 1+ 1 ± and ³ ± (1 ´ 1) = 2. Thus the answer is ³ = 1 - ( ² - 1) + 2( ± - 1), that is, ² + ³ - 2 ± = 0.
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Tutorial04-solutions - Problem Set 4 MH2100 Solutions E Let...

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