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Unformatted text preview: Midterm Exam I, Calculus III, Sample A 1. (10 points) Show that the 4 points P 1 = (0 , , 0) ,P 2 = (2 , 3 , 0) ,P 3 = (1 , 1 , 1) ,P 4 = (1 , 4 , 1) are coplanar (they lie on the same plane), and find the equation of the plane that contains them. Solution: u =→ P 1 P 2 = h 2 , 3 , i , v =→ P 1 P 3 = h 1 , 1 , 1 i , w = h 1 , 4 , 1 i , and the scalar triple product is equal to u • ( v × w ) = 2 3 1 1 1 1 4 1 = 2 1 1 4 1 3 1 1 1 1 + 0 1 1 1 4 = 0 , so the volume of the parallelepiped determined by u , v , and w is equal to 0. This means that these vectors are on the same plane. So, P 1 , P 2 , P 3 , and P 4 are coplanar. 2. (10 points) Find the equation of the plane that is equidistant from the points A = (3 , 2 , 1) and B = ( 3 , 2 , 1) (that is, every point on the plane has the same distance from the two given points). Solution: The midpoint of the points A and B is the point C = 1 2 ( 3 , 2 , 1)+( 3 , 2 , 1) = (0 , , 0). A normal vector to the plane is given by→ AB = h 3 , 2 , 1 i  h 3 , 2 , 1 i = h 6 , 4 , 2 i . So, the equation of the plane is 6( x 0) + 4( y 0) + 2( z 0) = 0, that is, 3 x + 2 y + z = 0. 3. (6 points) Find the vector projection of b onto a if a = h 4 , 2 , i and b = h 1 , 1 , 1 i . Solution: Since  a  2 = 4 2 + 2 2 = 20, the vector projection of b onto a is is equal to proj a b = a • b  a  2 a = h 4 , 2 , i • h 1 , 1 , 1 i 20 = 6 20 h 4 , 2 , i = 3 5 h 2 , 1 , i . 4. (12 points) Consider the curve r ( t ) = √ 2cos t i + sin t j + sin t k . (a) (8 points) Find the unit tangent vector function T ( t ) and the unit normal vector function N ( t ) . (b) (4 points) Compute the curvature κ . Solution: (a) r ( t ) = √ 2sin t i + cos t j + cos t k and  r ( t )  = p 2sin 2 t + cos 2 t + cos 2 t = √ 2. So, the unit tangent vector T ( t ) is is equal to T ( t ) = r ( t )  r ( t )  = sin t i + √ 2 2 cos t j + √ 2 2 cos t k . Since T ( t ) = cos t i √ 2 2 sin t j √ 2 2 sin t k and  T ( t )  = q cos 2 t + 1 2 sin 2 t + 1 2 sin 2 t = 1, the normal vector is equal to N ( t ) = cos t i √ 2 2 sin t j √ 2 2 sin t k . 1 (b) One can use the formula κ ( t ) =  T ( t )   r ( t )  and  r ( t )  = √ 2 form part (a) and  T ( t )  = 1 form part (b) to get κ ( t ) = 1 √ 2 = √ 2 2 . One can also use the formula κ ( t ) =  r ( t ) × r 00 ( t )   r ( t )  3 . Since r 00 ( t ) = √ 2cos t i sin t j sin t k ,  r 00 ( t )  = 2, and r ( t ) × r 00 ( t ) = i j k √ 2sin t cos t cos t √ 2cos t sin t sin t = √ 2 j + √ 2 k = h , √ 2 , √ 2 i . Then  r ( t ) × r 00 ( t )  = 2 and κ ( t ) = 2 ( √ 2) 3 = √ 2 2 . 5. (10 points) Find the length of the curve with parametric equation: r ( t ) = h e t ,e t sin t,e t cos t i , between the points (1 , , 1) and ( e 2 π , ,e 2 π ) ....
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This note was uploaded on 10/13/2010 for the course MATH 230 at Penn State.
 '08
 WEINERMICHAELDA
 Math, Calculus

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