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

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

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Unformatted text preview: 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 , , − 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 − 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. 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...
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This note was uploaded on 04/29/2008 for the course MATH 20C taught by Professor Helton during the Fall '08 term at UCSD.

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

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