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Fall 2007 - Cioaba's Class - Final Exam (Version 2)

Fall 2007 - Cioaba's Class - Final Exam (Version 2) - Math...

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Math 20C - Final (Lecture D, Winter 2007) Duration: 3 hours Please close your books and turn off your phones. You can use one page of handwritten notes. To get full credit, you should support your answers. 1. Let P = (3 , 2 , 3) , Q = (1 , 1 , 2) and R = (1 , 2 , 1). a) (3 points) Find the angle θ between the displacement vectors vector PQ and vector PR and the area of the triangle PQR . Solution. It is easy to see that vector PQ = (− 2 , 1 , 1 ) and vector PR = (− 2 , 0 , 2 ) . From the properties of the dot product, we get that cos θ = vector PQ · vector PR | vector PQ || vector PR | = 6 6 8 = 3 2 Thus, cos θ = 3 2 which implies θ = π 6 or 30 degrees. The area of the triangle PQR equals half of the length of the cross product vector PQ × vector PR . This cross product is vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vector i vector j vector k 2 1 1 2 0 2 vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle vextendsingle = 2 vector i 2 vector j 2 vector k The length of vector AB × vector AC is radicalbig 2 2 + ( 2) 2 + ( 2) 2 = 2 3. Thus, the area is 3. b) (2 points) Find an unit vector that is normal to the plane formed by the points P, Q and R . Find the equation of the plane determined by the points P, Q and R . Solution. A vector perpendicular to the plane formed by P, Q and R is the vector vector PQ × vector PR = ( 2 , 2 , 2 ) . A unit vector perpendicular to the plane formed by P, Q and R is obtained by dividing the previous vector by its length. Thus, the answer is ( 1 3 , 1 3 , 1 3 ) . Also, a correct answer is the vector (− 1 3 , 1 3 , 1 3 ) . The plane formed by P, Q and R is the plane passing through P = (3 , 2 , 3) which is perpendicular on the vector vector PQ × vector PR = ( 2 , 2 , 2 ) . An equation of this plane is 2( x 3) 2( y 2) 2( z 3) = 0 which simplifies to x y z = 2.

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2. Let f ( x, y ) = radicalbig 25 ( x 3) 2 ( y 1) 2 . a) (3 points) Find an equation of the tangent plane to the graph of f ( x, y ) at the point (3 , 2 , 4). Use the tangent plane at (3 , 2 , 4) to find an approximation for f (2 . 99 , 1 . 99). Is f (2 . 99 , 1 . 99) greater or less than your approximation ? Explain your answer. Hint: Use the graph of z = f ( x, y ). Solution. An equation of the tangent plane is z = f (3 , 2) + f x (3 , 2)( x 3) + f y (3 , 2)( y + 2) Since f x ( x, y ) = 2( x 3) 2 25 ( x 3) 2 ( y 1) 2 = x 3 25 ( x 3) 2 ( y 1) 2 and f y ( x, y ) = 2( y 1) 2 25 ( x 3) 2 ( y 1) 2 = y 1 25 ( x 3) 2 ( y 1) 2 , it follows that f x (3 , 2) = 0 and f y (3 , 2) = 3 4 . We deduce that the equation of the plane is z = 4 + 3( y +2) 4 which simplifies to z =
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