Page 1 Math 251 Spring 2015 copyright Joe Kahlig 1 compc a a c 6 c 3 2 center 2 4 0 and radius 45 3 a b a b sin and will have a max when sin 1 mas is a b 3 4 12 4 Note since we are identifying surfaces this means we are graphing in 3 dimensions cone x2 y 2 z 2 0 elliptic Paraboloid 5x 2y 2 3z 2 circular cylinder 6z 2 6y 2 8 Hyperboloid of two sheet x2 4y 2 2z 2 10 ellipsoid 3x2 2y 2 4z 2 9 5 Since the line is perpendicular to the plane this means the direction vector of the line is parallel to the normal vector of the plane so let the normal vector be some scalar multiple of the direction vector parametric equation of the line x 1 2t y 2 5t z 1 3t n v h2 5 3i Answer 2x 5y 3z 4 6 r t t t2 t3 and a p h1 4p 1 16p 1 52pi a x component t 1 4p y component t2 1 16p now solve 1 4p 2 1 16p to get 16p2 8p 0 or 8p 2p 1 0 or p 0 and p 1 2 The points are a 0 1 1 1 and a 1 2 3 9 27 b For the point 2 4 8 t 2 and r0 2 h1 4 12i y 4 z 8 x 2 4 12 7 either do a product rule or distribute the f t though the vector r t 3t2 et t3 et 5t4 3t2 3t2 sin 2t 2t3 cos 2t 8 r0 t 4t t2 0 and r0 t 16t2 t4 t 16 t2 R3 61 Length t 16 t2 dt use u sub with u 16 t2 3 0 D E 1 2 2t 3t2 4 4t2 9t4 b r0 t r00 t 6t2 12t 4 9 a T t 10 The angle between plane is the acute angle between the normal vectors x 2y z 4 has n1 h1 2 1i and n1 6 3x 6y 2z 12 has n2 h3 6 2i and n2 7 17 cos 1 7 49 degrees 7 6 Page 2 Math 251 Spring 2015 copyright Joe Kahlig 11 a Domain 4 1 1 4 b This limit does not exist since lim limit t 1 t2 sin t DNE Note this limit is not a L Hopital 6t 5 12 a The direction vectors for the lines are v1 h1 4 2i and v2 h2 4 0i Since there is not a number that can be multiplied by v1 to get v2 the lines are not parallel b first get the parametric equations for the second line Be sure to use a variable different from the variable for the first line Note If you use a t as the variable for both lines then you are trying to find out if both lines intersect at the same point at the same time This question doesn t ask this it only wants to know if the paths intersect Line 2 x 1 2s y 5 4s and z 10 z components equal gives 2t 10 or t 5 x components equal gives t 2 1 2s or 7 1 2s or x 3 Now check the y components for t 5 and s 3 Line 1 gives y 21 and Line 2 gives y 17 Since these are not the same the lines are skew 13 There was an error in this problem The given lines were actually skew not parallel and not intersecting and thus no plane will contain both lines This problem was graded as if the lines intersected Since this is what was intended Steps that would have been performed if the lines intersected step 1 find the direction vector of both lines step 2 find cross product of these vectors this is the normal vector of the desired plane step 3 use any point on one of these lines and create the desired plane 14 Method 1 If there is a plane P2 containing the line L where P2 is parallel to the given plane P then the normal vector of P and the normal vector of P2 are parallel Since the line lies in P2 then the dot product of the direction vector of the line and the normal vector of P2 will be zero In similar fashion the dot product of the direction vector of L and plane P will also be zero The direction vector of the line is v h3 1 5i and normal vector of plane P is n h1 2 1i v n 3 2 5 0 so the answer is yes there is a plane containing line L that will be parallel to plane P Method 2 If there is a plane P2 containing the line L where P2 is parallel to the given plane P then the line will not intersect the plane So check to see if there is a value of t will the line will intersect the plane x 2y z 5 1 3t 2 1 t 1 5t 5 5t 4 5t 5 4 5 This is a contradiction so there is not any value of t where the line intersects the plane Thus yes there is a plane containing line L that will be parallel to plane P
View Full Document
Unlocking...