1702 - 203 Fall 2008 Group Final Solutions 1a: Take cross...

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1a: Take cross products of the normal to get the direction of the line: Cross @8 2, - 1, 1 < , 8 1, 0, - 3 <D 8 3, 7, 1 < Then parameterize the line using the point and the direction to get: x = 4 + 3 t , y = - 7 t , z = - 3 + t 1b: The desired plane contains the normals of the two specifed plane, so its normal is the same cross product as in 1a. That gives an equation of the plane as 3 H x - 4 L + 7 H y + 2 L + H z - 3 L = 0 2a: the gradient is: f = x 2 y + y 2 z x 2 y + y 2 z 8 D @ f, x D , D @ f, y D , D @ f, z D< 9 2 x y, x 2 + 2 y z, y 2 = 2b: the gradient will give a normal vector, so evaluating at the point gives: 9 2 x y, x 2 + 2 y z, y 2 = ê . 8 x Ø 2, y Ø - 1, z Ø 3 < 8 - 4, - 2, 1 < So an equation of the tangent plane is - 4 H x - 2 L + - 2 H y + 1 L + H z - 3 L = 0 2c: we just want D v f = “ f ÿ v to be negative, so there are many correct answers. For example, moving in the direction given by i gives: 8 - 4, - 2, 1 < . 8 1, 0, 0 < - 4 3: First, sketch the triangular region to reverse the order, giving an integral which goes easily by substitution: 0 1 0 x 1 + x 2 y x 1 3 K - 1 + 2 2 O 4a: Absolutely convergent by the ratio test 4b: Convergent by the alternating series test, and then absolutely convergent by the integral test:
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This note was uploaded on 10/29/2011 for the course MATH 190 taught by Professor Ocken during the Spring '11 term at CUNY City.

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1702 - 203 Fall 2008 Group Final Solutions 1a: Take cross...

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