Sketch the fence and determine how much paint you

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Unformatted text preview: a circular fence with radius 10 m is given by x = 10 cos t, y = 10 sin t. The height of the fence varies from 3 m to 5 m such that, at position (x, y ), the height is given by the function h(x, y ) = 4 + 0.01(x2 − y 2 ). Suppose that 1 litre of paint covers 100 m2 . Sketch the fence and determine how much paint you will need if you paint both sides of the fence. The Berlin Wall in Germany32 Preliminaries Area Bounded by A Function y y c y=h(x) g(x) 0 a (b) b x • We know in single variable case, the area bounded by a function h(x) is the value of integration of the function b S= a 32 h(x) dx. One of the famous wall, check the history from wikipedia: http://en.wikipedia.org/wiki/Berlin wall 184 • For the single variable case, the base is contained in one dimensional space along xaxis (since we only have one the variable). If we change the base to a curve C in two dimensional space (this means we need two variables), then the integration along the curve C is still the area: h(x, y ) ds. S= C • Notice that, in this question, the area of the fence is given by a line integrating the top function h(x, y ), since h(x, y ) is the top of the fence. • Moreover, the base of the fence C is a circle with radius 10. Solution Before sketch the graph of the fence, we first take a look at the top function h(x, y ). We know that the function h(x, y ) is given by h(x, y ) = 4 + 0.01(x2 − y 2 ). So by second derivative test, we know that the point (0, 0) is a saddle point. (Go back to tutorial 6 if necessary.) It is easy to check that along x-axis (y = 0), (0, 0) is a minimal point (Concave Up): and along y -axis (x = 0), (0, 0) is a maximal point (Concave Down): The graph of the two variable function h(x, y ) = 4 + 0.01(x2 − y 2 ) is the following: 185 Thus, the fence with h(x, y ) as its top, and the circle x2 + y 2 = 102 as its base, can be sketched as follows: Sketch of the Fence The shape of the fence is somewhat like the Bird’s Nest Stadium of Beijing 2008 Olympic Games. One Application of This Function h(x, y ): Bird’s Nest Stadium of Beijing Olympics. 186 Now, let us see how to solve this question. To find the area bounded by the function h(x, y ), we only need to evaluate the line integral: h(x, y ) ds. C Note that h(x, y ) is a scalar function (the image of the function is a real number), so the formula is given by b h(x, y ) ds = C a h(x(t), y (t)) r (t) dt. Here, C is the base of the fence, which has vector equation: x(t) = 10 cos t, y (t) = 10 sin t =⇒ r(t) = 10 cos ti + 10 sin tj, Thus, r (t) = (10 cos t) i + (10 sin t) j = −10 sin t i + 10 cos t j, and so r (t) = − 10 sin t i + 10 cos t j = (−10 sin t)2 + (10 cos t)2 = 10. 0 ≤ t ≤ 2π. The area of the fence is then given by 2π h(x, y ) ds = C 0 2π h(x(t), y (t)) r (t) dt 4 +...
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This note was uploaded on 05/05/2009 for the course MA MA1505 taught by Professor Ma1505 during the Spring '09 term at National University of Juridical Sciences.

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