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Tutorial05-solutions

Tutorial05-solutions - Problem Set 5 MTH211 Solutions 1...

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Problem Set 5 MTH211 Solutions 1 Questions on understanding the lecture E±²³´µ¶² · Sketch level curves of the function ± ( ²³´ ) = 2 ² - 3 ´ and the vector ﬁeld grad ± . S¸¹º»µ¸¼ Level curves must be parallel equidistant straight lines and grad ± = (2 ³- 3) perpendicular to them. :) E±²³´µ¶² ½ Consider the function ± ( ) = ² 2 - ²´ and the initial point (2 ³ 1). In which direction does the function increase most rapidly? Decrease most rapidly? S¸¹º»µ¸¼ Increases most rapidly in the direction of grad ± = (2 ² - ´³-² ), which at the point (2 ³ 1) is (3 2). The unit vector of the direction is ¾ 3 13 2 13 ¿ . Decreases most rapidly in the direction - grad ± , and the unit vector of the direction is ¾ - 3 13 ³ 2 13 ¿ . :) E±²³´µ¶² À Let ± : R µ R be a function diﬀerentiable at ¶ ∈ R µ . Prove that D · u + ¸ v ± ( a ) = ·D u ± ( a ) + ¸D v ± ( a ). S¸¹º»µ¸¼ As ± is diﬀerentiable, we have D · u + ¸ v ± ( a ) = [ · u + ¸ v ] ¹ grad ± ( a ) = · u ¹ grad ± ( a )+ ¸ v ¹ grad ± ( a ) = ·D u ± ( a ) + ¸D v ± ( a ). :) 2 Questions on calculation E±²³´µ¶² Á Find the equation of the tangent plane to the surface 2 sin ² 2 ´ + º + sin ´ 2 ² + º + sin º 2 ² + ´ = 2 - 3 at the point ² = π , ´ = 2 π , º = 2 π . 1

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S±²³´µ±¶ Let ± ( ²³´³µ ) = 2 sin ² 2 ´ + µ + sin ´ 2 ² + µ + sin µ 2 ² + ´ - ( 2 - 3), then ± ² = 4 ² ´ + µ cos ² 2 ´ + µ - ´ 2 ( ² + µ ) 2 cos ´ 2 ² + µ - µ 2 ( ² + ´ ) 2 cos µ 2 ² + µ and thus ± ² ( π³ 2 2 π ) = 1 2 + 4 9 . Further, ± ´ = - 2 ² 2 ( ´ + µ ) 2 cos ² 2 ´ + µ + 2 ´ ² + µ cos ´ 2 ² + µ - µ 2 ( ² + ´ ) 2 cos µ 2 ² + µ and thus ± ´ ( 2 2 π ) = - 1 8 2 - 4 9 . Further, ± ( ) = ± ( ²³µ³´ ) and ´ 0 = µ 0 (= 2 π ), so ± µ ( 2 2 π ) = - 1 8 2 - 4 9 . Thus the equation of the tangent plane is · 1 2 + 4 9 ¸ ( ² - π ) - · 1 8 2 + 4 9 ¸ ( ´ - 2 π ) - · 1 8 2 + 4 9 ¸ ( µ - 2 π ) = 0 ³ which can be re-written as · 1 2 + 4 9 ¸ ² - · 1 8 2 + 4 9 ¸ ´ - · 1 8 2 + 4 9 ¸ µ = π 2 2 - 4 π 3 :) E¹º»¼µ½º ¾ Prove that any tangent plane to the surface ²´µ = · 3 ( · > 0) and the coor- dinate axes form a tetrahedron of a constant volume. S±²³´µ±¶ Let ( ² 0 ³´ 0 ³µ 0 ) be an arbitrary point on the surface. Notice that the equation doesn’t change if we change signs of two variables at the same time, hence it’s enough to assume that ² 0 0 0 > 0.
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