Tutorial10-sol(1) - Problem Set 10 MH2100/MTH211 Solutions...

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Problem Set 10 MH2100/MTH211 Solutions 1 Questions on understanding the lecture E±²³´µ¶² · The Newton-Leibniz formula ¸ b a ±² = ² ( b ) ( a ) was proved in lectures only for a smooth curve C . Does the formula hold for piecewise smooth curves? S¹º»¼µ¹½ Yes, we can just sum it over all smooth pieces. :) E±²³´µ¶² ¾ Prove that the work of a conservative vector field over any piecewise smooth closed curve is zero. S¹º»¼µ¹½ From Exercise 1, as we can assume the vector field equals grad( ² ) because it’s conservative. Then the work integral over a piecewise smooth closed curve C can be computed as ¿ C ±² = 0 ³ :) E±²³´µ¶² À How can one parametrize a plane ´µ + ¶· + ¸¹ + ± = 0 ³ S¹º»¼µ¹½ If ´ º = 0, we can let · = » , ¹ = ¼ and get µ = - ´ » - ¸ ´ ¼ - ± ´ . If ´ = 0 then either ¶ º = 0 or ¸ = º = 0 and we solve for that corresponding variable instead. :) 1
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2 Questions on calculation E±²³´µ¶² · Evaluate ¸ ( ± 2 ²³ 2 ²´ 2 ) ( ± 1 ²³ 1 ²´ 1 ) ±µ± + ³µ³ + ´µ´ ± 2 + ³ 2 + ´ 2 , where the point ( ± 1 ²³ 1 ²´ 1 ) lies on the sphere ± 2 + ³ 2 + ´ 2 = 2 and the point ( ± 2 2 2 ) is on the sphere ± 2 + ³ 2 + ´ 2 = · 2 for ¶ > 0 and · > 0. S¹º»¼µ¹½ Note that this integral can be regarded as a line integral of the vector field v = ± ± 2 + ³ 2 + ´ 2 i + ³ ± 2 + ³ 2 + ´ 2 j + ´ ± 2 + ³ 2 + ´ 2 k . Based on the expression of the integral, it suggests that the integral in path-independent hence we shall be able to find the potential of the given field. Let ¸ be the potential of the given vector field. Then ¸ ± = ± ¾ ± 2 + ³ 2 + ´ 2 ¹ ¸ ( ±²³²´ ) = ¾ ± 2 + ³ 2 + ´ 2 + º ( ³²´ ) » Notice that º ( ) = 0 works and hence ¸ ( ) = ¾ ± 2 + ³ 2 + ´ 2 is the potential. By Newton-Leibniz’s formula, the integral is ¸ ( ± 2 2 2 ) - ¸ ( ± 1 1 1 ) = · - ¶» :) E±²³´µ¶² ¿ (a) Let C be a closed curve. Prove that the area inside C equals À C ±µ³ = - À C ³µ± = 1 2 À C ±µ³ - ³µ±² (b) Apply this formula to find the area enclosed by the astroid ±
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Tutorial10-sol(1) - Problem Set 10 MH2100/MTH211 Solutions...

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