Math32a 1 R Kozhan Midterm 1 Oct 24 2011 Name UID Section TA Instructions If you get stuck move on to the next question You don t have a lot of time Make sure to look at the last question It s short and easy Free points Show all work if you want to get full credit I reserve the right to take off points if I cannot see how you arrived at your answer even if your final answer is correct No books notes electronics incl calculators and cell phones are allowed Good luck Question Max Your score 1 12 2 8 3 14 4 10 5 10 6 14 Total 68 Problem 1 a 4 points Find the center and the radius of the circle x2 y 2 4x 10y 0 b 8 points Find parametric equation for the path of a particle that moves around this circle twice counterclockwise starting from the top point of the circle Solution a Note that x2 y 2 4x 10y x2 4x 4 4 y 2 10y 25 25 x 2 2 y 5 2 29 So we get equation x 2 2 y 5 2 29 which is the equation of the circle of radius 29 and center 2 5 b Note that x cos t y sin t 0 t 2 is a parametrization of the circle of radius 1 with center at the origin which goes around once counterclockwise starting with the rightmost point Note that we re in the topmost point at time t 2 Therefore x cos t t 2 2 2 gives the same circle which starts at the top and travels once counterclockwise To travel twice around we can just extend the time interval by another 2 y sin t x cos t t 4 2 2 Finally we need to scale this circle to have radius 29 and shift it to have center 2 5 so the parametrization becomes x 29 cos t 2 y 29 sin t 5 t 4 2 2 y sin t One could also start with x sin t y cos t 0 t 2 but then notice that this parametrization is clockwise so we ll need to put t instead of t to get clockwise orientation x sin t y cos t and the end result would be x 29 sin t 2 y 29 cos t 5 0 t 2 0 t 4 Problem 2 8 points Find the area of the triangle with vertices P 1 2 1 Q 2 3 0 R 1 0 1 Solution Note that P Q h1 1 1i and P R h 2 2 2i so the area of the parallelogram built on those two vectors is P Q P R 4 i 4 j 42 42 4 2 So the area of the triangle is 12 4 2 2 2 Problem 3 a 4 points Find parametric and symmetric equations of the line through the point A 2 3 1 which is perpendicular to the plane 2y z 0 b 2 points Prove that this line is parallel to the line x 0 y 2 4t z 1 2t c 8 points Find the equation of the unique plane passing through both of the above lines Solution a We are looking for the line through the point 2 3 1 parallel to h0 2 1i Its parametric equation is x 2 y 3 2t z 1 t Its symmetric equation is x 2 and y 3 z 1 2 1 note that you have to keep x 2 b The line in a is parallel to h0 2 1i The line in b is parallel to h0 4 2i Since these two vectors are scalar multiples of each other they are parallel Therefore the lines are parallel c In order to get the normal vector of the plane we need to take the cross product of any two non parallel vectors on the plane One vector could be h0 2 1i The other can t be h0 4 2i since it s parallel to the previous one Instead we can take the vector joining the two points on the two lines A 2 3 1 and B 0 2 1 So AB h 2 1 2i and therefore a normal vector to the plane is h0 2 1i h 2 1 2i 5 i 2 j 4 k So now we can just write the equation of the plane with normal vector h 5 2 4i through the point A 2 3 1 or B 0 2 1 if you wish 5 x 2 2 y 3 4 z 1 0 i e 5x 2y 4z 0 Problem 4 at time 10 points Find the unit tangent vector T and the principle normal vector N t t t t 2 of the curve r t he sin t e cos t e i Solution r t het sin t et cos t et i so r 0 t het sin t et cos t et cos t et sin t et i Its length r 0 t p et sin t et cos t 2 et cos t et sin t 2 e2t p 2e2t sin2 t 2e2t cos2 t e2t 3et So unit tangent vector is r 0 t T t 0 r t sin t cos t cos t sin t 1 3 3 3 The derivative of T t is T t 0 cos t sin t sin t cos t 0 3 3 and its length is 1 p 1 T 0 t cos t sin t 2 sin t cos t 2 2 3 3 so the principle normal vector is T 0 t cos t sin t sin t cos t N t 0 2 2 T 0 t Finally at time t 2 1 1 1 3 3 3 1 1 2 0 N 2 2 T 2 Problem 5 10 points Suppose r 00 t 2 j 6t k r 0 0 5 k r 0 2 i Find r 1 Solution Integrating r 00 t 2 j 6t k we get r 0 t 2t j 3t2 k c for some vector c hc1 c2 c3 i Plugging in t 0 we obtain c r 0 0 So c 5 k and r 0 t 2t j 5 3t2 k Integrating this equation we get r t t2 j 5t t3 k w for some vector w hw1 w2 w3 i Plugging in t 0 we get w r 0 2 i so r t 2 i t2 j 5t t3 k Finally at t 1 r 1 2 i j 4 k Problem 6 a 2 points Write the formula for the vector projection of a onto b Answer proj b a a b b b 2 points Let x t cos t y t sin t Find Answer so at t 4 we get dy dx b b at t 4 dy dy dt cos t dx dx dt sin t 1 2 1 2 1 Answer if the following statement is always TRUE or if it …