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Tutorial08-sol(1)

# Tutorial08-sol(1) - Problem Set 9 MH2100/MTH211 Solutions 1...

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Problem Set 9 MH2100/MTH211 Solutions 1 Questions on understanding the lecture E±²³´µ¶² · Given a curve C , what is ¸ C 1 ±² ? S¹º»¼µ¹½ The arc length of the curve. By the deﬁnition with Riemann sums, the sum of all Δ ² ³ is an approximation of the length of C , which converges to the actual length as the partition is made ﬁner. :) E±²³´µ¶² ¾ Deﬁne the average value of a function over a curve. A½¶¿²³ ¸ C ´±² ¸ C ±² . :) E±²³´µ¶² À Let C be a curve in R 3 given by a parametric equation x = r ( µ ) and let F = P i + Q j + R k be a vector ﬁeld. Deﬁning the work of the vector ﬁeld F along the curve C as ¸ C F · r ( µ ) ¸ r ( µ ) ¸ ±² , prove that it, indeed, equals Á C P±¹ + Q±º + R±»¼ S¹º»¼µ¹½ By deﬁnition, it is Á ½ ¾ ( µ ) + ( µ ) + ( µ ) ¸ r ( µ ) ¸ ¸ r ( µ ) ¸±µ = Á C + Q±º + R±» . :) 1

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2 Questions on calculation E±²³´µ¶² · Let ± : R 3 R be an arbitrary function having continuous partial derivatives. Evaluate curl(grad ± ). S¸¹º»µ¸¼ Consider ± ( ²³´³µ ). We have curl(grad ± ) = ½ ½ ½ ½ ½ ½ i j k ² ´ µ ± ² ± ´ ± µ ½ ½ ½ ½ ½ ½ = ( ± µ´ - ± ´µ ) i - ( ± µ² - ± ²µ ) j + ( ± ´² - ± ²´ ) k = 0 :) E±²³´µ¶² ¾ Let v = P i + Q j + R k be an arbitrary vector ﬁeld whose component functions have continuous partial derivatives. Evaluate div(curl v ). S¸¹º»µ¸¼ It is ( R ´ - Q µ ) ² + ( P µ - R ² ) ´ + ( Q ² - P ´ ) µ = R ´² - Q µ² + P µ´ - R ²´ + Q ²µ - P ´µ = 0 :) E±²³´µ¶² ¿ Evaluate À C ´ 2 ·¸ , where C is given by ² = ¹ ( º - sin º ), ´ = ¹ (1 - cos º ) for 0 ≤ º ≤ 2 π . Here, ¹ > 0 is some constant. S¸¹º»µ¸¼ First, ·¸ = Á Â ·² ·º Ã 2 + Â ·´ ·º Ã 2 ·º = ¹ Ä (1 - cos º ) 2 + sin 2 º·º = ¹ 2 1 - cos º·º¶ Thus, À C ´ 2 ·¸ = ¹ 3 2 À 2 π 0 (1 - cos º ) 5 2 ·º = ¹ 3 2 À 2 π 0 Â 2sin 2 º 2 Ã 5 2 ·º = 8 ¹ 3 À 2 π 0 ½ ½ ½ ½ sin 5 º 2 ½ ½ ½ ½ ·º = 16 ¹ 3 À π 0 sin 5 = 16 ¹ 3 À π 0 (1 - cos 2 º ) 2 sin = 16 ¹ 3 À 1 - 1 (1 - ² 2 ) 2 ·² = 32 ¹ 3 À 1 0 (1 - 2 ² 2 + ² 4 ) ·² = 32 ¹ 3 Â 1 - 2 3 + 1 5 Ã = 256 15 ¹ 3 :) 2
E±²³´µ¶² · Evaluate ¸ C (2 ± - ² ) ³´ + ´³² , where C is given by ´ = ± ( µ - sin µ ), ² = ± (1 - cos µ ) for 0 ≤ µ ≤ 2 π . Here, ± > 0 is some constant. S¹º»¼µ¹½ First, ³´ = ± (1 - cos µ ) ³µ¶ ³² = ± sin µ³µ¶ and hence ¸ C (2 ± - ² ) ³´ + = ¸ 2 π 0 ¾ ± (1 + cos µ ) ± (1 - cos µ ) + ± ( µ - sin µ ) ± sin µ ¿ ³µ = ± 2 ¸ 2 π 0 ¾ 1 - cos 2 µ + µ sin µ - sin 2 µ ¿ ³µ = ± 2 À Á cos µ Â 2 π 0 + ¸ 2 π 0

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Tutorial08-sol(1) - Problem Set 9 MH2100/MTH211 Solutions 1...

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