HWSol-F09

HWSol-F09 - August 15, 2009 Mathematics of Physics and...

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August 15, 2009 Mathematics of Physics and Engineering II: Homework answers You are encouraged to disagree with everything that follows. Homework 1. Problem 1. (1) --→ PQ = --→ OQ - --→ OP = h- 1 , 0 , 2 i - h 1 , 1 , 1 i = h- 2 , - 1 , 1 i , -→ PR = h 0 , - 2 , - 2 i , -→ PS = h a, - 1 , - 2 a i . (2) The vertex of the angle is P , so you need -→ PR · -→ PS = 0, or 2+4 a = 0. Therefore, a = - 1 / 2 . (3) The area is (1 / 2) | --→ PQ × -→ PR | and --→ PQ × -→ PR = i j k - 2 - 1 1 0 - 2 - 2 = h 4 , - 4 , 4 i = 4 h 1 , - 1 , 1 i . Consequently, the area is 2 3 . (4) The normal vector to the plane is any vector parallel to --→ PQ × -→ PR , for example, h 1 , - 1 , 1 i . Taking P as the point on the plane, we get the equation ( x - 1) - ( y - 1) + ( z - 1) = 0 or x - y + z = 1 . (5) You are welcome to compute the scale triple product using the determinant, but, given the work you already did, you do not have to compute another determinant. The answer is 4 | 1 - a | (it has to be non-negative). (6) You want the coordinates of S to satisfy the equation of the plane through P,Q,R , that is, 1 + a - 0 + 1 - 2 a = 1 or a = 1 . You can also see it immediately from the volume formula. (7) The direction vector for the line is the normal vector to the plane, that is, h 1 , - 1 , 1 i . Then the equation of the line is r ( t ) = h 1 + t, - t, 1 + t i . At the point of intersection, (1 + t ) - ( - t ) + (1 + t ) = 1 or t = - 1 / 3, so the point is (2 / 3 , 1 / 3 , 2 / 3). Problem 2. (1) r (1) = h 0 , 1 , 2 i , so the coordinates of the point are (0 , 1 , 2) . (2) v ( t ) = ˙ r ( t ) = h- 2 t, 3 t 2 , 2 t i . (3) | v ( t ) | = 4 t 2 + 9 t 4 + 4 t 2 = t 8 + 9 t 2 . (4) a ( t ) = ˙ v ( t ) = h- 2 , 6 t, 2 i . (5) The particle is at (0 , 1 , 2) when 1 - t 2 = 0 or t = 1 (by assumption, t 0). Therefore, the equation of the tangent line is R ( u ) = h 0 , 1 , 2 i + u ˙ r (1). Next, ˙ r (1) = h- 2 , 3 , 2 i , and so the equation of the line is R ( u ) = h- 2 u, 1 + 3 u, 2 - 2 u i . (6) You want the coordinates of the particle to satisfy the equation of the plane. Then 1 + t 2 - (1 - t 2 ) = 2 or t = 1 (remember, t 0) So the point of intersection is (0 , 1 , 2) . (7) According to the formula, the distance is Z 1 0 | v ( t ) | dt = Z 1 0 p 8 + 9 t 2 tdt = (simply guess antiderivative) 1 27 (8 + 9 t 2 ) 3 / 2 | t =1 t =0 = 17 3 / 2 - 8 3 / 2 27 . Homework 2. Problem 1 . (1) f ( x,y ) = h f x ( x,y ) ,f y ( x,y ) i = h 4 x - y - 1 , 2 y - x + 1 i . (2) The direction vector is a = h- 1 , - 1 i . Therefore, the rate is f (1 , 1) · a | a | = h 2 , 2 i · h- 1 , - 1 i 2 = - 4 / 2 . The rate is negative, so the function is decreasing in that direction. (3) The direction is given by
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This note was uploaded on 10/06/2009 for the course MATH 445 taught by Professor Friedlander during the Fall '07 term at USC.

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HWSol-F09 - August 15, 2009 Mathematics of Physics and...

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