Calc 3 Exam 1 Sin - k < 1. For k > 1,...

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CALCULUS AND ANALYTIC GEOMETRY III SOLUTIONS TO SAMPLE TEST 1 Please let me know if you find errors or typos. 1. (a) The volume is given by the (absolute value of the) scalar triple product a · ( b × c ). This triple product is equal to the determinant ± ± ± ± ± ± 1 1 0 0 1 2 1 0 1 ± ± ± ± ± ± = 3 . (b) The points will lie in the same plane if and only if the vectors -→ PQ = h 1 , 4 , 5 i , -→ PR = h 2 , - 1 , 1 i and -→ PS = h 4 , 7 , 11 i have triple scalar product equal to zero. The calculation is similar to the that in (a) and the answer is indeed zero. 2. You are supposed to draw pictures, but I will just describe the answers in words. In the xy -plane the trace is a hyperbola. In the yz -plane we have a circle. The intersections with the planes z = k are as follows. For k = 1 we have a pair of lines; for k > 1 and for k < 1 we get hyperbolae having the previous lines as asymptotes, with the branches of the hyperbolae for k > 1 being on the opposite sides of the asymptotic lines from those for
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Unformatted text preview: k &lt; 1. For k &gt; 1, as k increases, the hyperbolae move further from the asymptotes. 3. The point (1 , 1 , 1) corresponds to t = 1. For a normal vector to the normal plane we can take r (1) = h 1 , 2 , 3 i , so the equation is ( x-1) + 2( y-1) + 3( z-1) = 0. As a normal vector for the osculating plane we can take r (1) r 00 (1). This is because B ( t ) is perpendicular to both T ( t ) and N ( t ), and the plane of T ( t ) and N ( t ) is the same as the plane of r ( t ) and r 00 ( t ). Thus, the normal vector we want is h 1 , 2 , 3 i h , 2 , 6 i = h 6 ,-6 , 2 i , so we can write the equation of the osculating plane as 3( x-1)-3( y-1) + ( z-1) = 0. 4. The endpoints correspond to t = 0 and t = 2 , so the arc length is given by Z 2 | r ( t ) | dt = Z 2 p sin 2 t + cos 2 t + 1 dt = 2 2 ....
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This note was uploaded on 10/20/2011 for the course MAC 2313 taught by Professor Keeran during the Fall '08 term at University of Florida.

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