math 230 sample exam 1

# math 230 sample exam 1 - Midterm Exam I Calculus III Sample...

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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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math 230 sample exam 1 - Midterm Exam I Calculus III Sample...

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