Midterm Exam 1 with Solution - Calculus and Vector Analysis | MATH 230, Exams of Mathematics

Download Midterm Exam 1 with Solution - Calculus and Vector Analysis | MATH 230 and more Exams Mathematics in PDF only on Docsity! Math 230, Fall 2008 Midterm I Solutions 1. (4 points) The radius of the circle is d((1, 2, 3), (0, 0, 0)) = √ 14, so its equation is (x − 1)2 + (y − 2)2 + (z − 3)2 = 14. 2. (4 points) The three points are collinear if and only if the displacement vectors −−→ AB and −→ AC are parallel. We have −−→ AB = 〈6,−3, 6〉, −→ AC = 〈2,−1, 2〉, which are scalar multiples of each other, and so A, B, and C lie on the same line. 3. (4 points) We need only find the direction of the desired line, since it goes through (5,−1, 2). We can read the direction vector of the second line from its symmetric equations: v = 〈2,−5, 0〉, where we recall that a zero component of v corresponds to a constant co- ordinate in the parametric equations. Thus the desired line has parametric equations x = 5 + 2t, y = −1 − 5t, z = 2. 4. (4 points) The direction vector of the given line is 〈1,−2,−1〉, so this is the normal vector to the desired plane. Thus the plane is given by the equation 1(x − 1) − 2(y + 1) − 1(z − 0) = 0, which simplifies to x − 2y − z − 3 = 0. 5. (4 points) Draw a picture of the xy-plane, including u and v, and observe that u = 〈−5, 0〉, v = 〈6, 6〉, thus u + v = 〈1, 6〉. 6. (4 points) 1 The curvature at a given point is 1/r, where r is the radius of the osculat- ing circle (the circle which most closely approximates the curve at the given point). The smallest of the osculating circles at A, B, C, and D is at C, and so this is where the curvature is greatest. 7. (9 points) We compute cos θ = u · v |u||v| = −3 − 15 − 3√ 32 + 32 + 32 √ 12 + 52 + 12 = −21 27 = −7 9 . The angle between the two planes is the angle between their normals, which are 〈−3, 3, 3〉 and 〈1,−5,−1〉, so the cosine of the angle is −7/9, as we just computed. 8. (4 points) The vector projection is proj a b = a · b |a|2 a = −18 − 8 + 36 62 + 22 + 32 〈6,−2, 3〉 = 10 49 〈6,−2, 3〉 = 〈 60 49 ,−20 49 , 30 49 〉 . 9. (8 points) For (I), the horizontal traces (z constant) are hyperbolas, while the ver- tical traces (x or y constant) are parabolas. Thus the shape is a hyperbolic paraboloid, or saddle, which is shown in (D). For (II), the traces with y or z constant are hyperbolas, while the traces with x constant are ellipses (which are never empty), so the surface is a hyperboloid of one sheet around the x-axis, which is shown in (B). 10. (12 points) The normal vector must be perpendicular to the normal vectors of both given planes, so we take their cross product: 〈1, 0,−1〉 × 〈0, 1, 4〉 = 〈1,−4, 1〉. Three adjacent edges of the parallelogram are given by the displacement vectors −−→ PQ = 〈2, 1, 1〉, −→ PR = 〈1,−1, 2〉, −→ PS = 〈0,−2, 3〉,