MT soln - Math 53 Section 002 Midterm Solution 1. Find the...

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Unformatted text preview: Math 53 Section 002 Midterm Solution 1. Find the limit or show the limit does not exist. a ) lim ( x , y ) ( 1 , 2 ) x 4 y y 2 + 3 x 8 , b ) lim ( x , y ) ( , ) x 4 y y 2 + 3 x 8 . Solution. a) Since the function x 4 y y 2 + 3 x 8 is a rational function, it is continuous on its domain, which is the set D = { ( x , y ) | ( x , y ) negationslash = ( , ) } . Now ( 1 , 2 ) D , so we may find the limit by direct substitution: lim ( x , y ) ( 1 , 2 ) x 4 y y 2 + 3 x 8 = 1 4 ( 2 ) ( 2 ) 2 + 3 ( 1 8 ) = 2 4 + 3 = 2 7 . b) Let f ( x , y ) = x 4 y y 2 + 3 x 8 . When ( x , y ) ( , ) along the x-axis, we have f ( x , y ) = f ( x , ) = / ( y 2 ) = so f ( x , y ) as ( x , y ) ( , ) along the x-axis . We now let ( x , y ) ( , ) along the curve y = x 4 , we get f ( x , y ) = f ( x , x 4 ) = x 4 x 4 ( x 4 ) 2 + x 8 = x 8 4 x 8 = 1 4 hence f ( x , y ) 1 4 as ( x , y ) ( , ) along y = x 4 . Since different paths lead to different limiting values, the given limit does not exist. 2. Given a space curve C with parametric equations x = t , y = cos3 t , z = sin3 t . Let P be the point corresponding to t = 6 . Please find a) The unit tangent vector of the curve at P ; b) The equation of the tangent line passing through P ; c) The equation of the normal plane at P , i.e, the plane passing through P and perpendicular to the tangent line there. 1 Solution. The vector equation of the curve (a helix) is r ( t ) = < t , cos3 t , sin3 t > , so r ( t ) = < 1 , 3sin3 t , 3cos3 t > . The point corresponding to t = 6 is parenleftbigg 6 , cos ( 3 6 ) , sin ( 3 6 ) parenrightbigg = ( 6 , , 1 ) and the tangent vector there is r ( / 6 ) = < 1 , 3 , > . Hence the unit tangent vector there is < 1 , 3 , > | < 1 , 3 , > | = < 1 , 3 , > 1 2 + 3 2 = (bigg 1 10 , 3 10 , )bigg The parametric equation of the tangent line at ( / 6 , , 1 ) is x = 6 + t , y = 3 t , z =...
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MT soln - Math 53 Section 002 Midterm Solution 1. Find the...

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