# fmtsol - t = 0 so T(0 = h 1 1 1 i √ 3 Furthermore N t = T...

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Instructor: Dr. Yuli B. Rudyak MAC 2313, Calculus 3 with Analytic Geometry, Fall 2011 FIRST MIDTERM EXAM, OCTOBER 19, 2011 Problem 1. Find the area of the triangle with vertices P (1 , 1 , 0) ,Q (1 , 0 , 1) and R (0 , 0 , 1). Solution: The area is equal to the (1 / 2) | PQ × PR | = | j + k | = 2 / 2. Problem 2. Find the parametric and symmetric equations of the line through P (2 , - 3 , 4) and perpendicular to the plane 2 x - 3 y + 3 z = 4. Solution: A direction vector v of the line is a normal vector to the plane, and so we put v = h 2 , - 3 , 3 i . So, we use the point-vector formula and give the parametric equations: x = 2 t + 2 ,y -- 3 t - 3 ,z = 3 t + 4. The symmetric equations are x - 2 2 = y + 3 - 3 = z - 4 3 . Problem 3. Find a vector function that represents the curve of inter- section of the sphere 2 x 2 + y 2 + z 2 = 8 and the plane y = z . Solution: We put y = z in the ﬁrst equation and get x 2 + y 2 = 4. Now, x = 2 cos t,y = 2 sin t,z = 2 sin t . Problem 4. Find the vectors T , N , B for the curve r ( t ) = h e t cos t,e t sin t,e t i at P (1 , 0 , 1). Solution: First, r 0 ( t ) = e t h cos t - sin t, sin t + cos t, 1 i , and | r 0 ( t ) | = e t 3. We have T ( t ) = r ( t ) | r 0 ( t ) = h (cos t - sin t, sin t + cos t, 1 i 3 . The point P (1 , 0 , 1) correspond the value of parameter

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Unformatted text preview: t = 0, so, T (0) = h 1 , 1 , 1 i / √ 3. Furthermore, N ( t ) = T ( t ) / | T ( t ) | = h cos ,t-sin t,-sin t-cos t, i / √ 2 , and N (0) = h 1 ,-1 , i / √ 2. Finally, B (0) = T (0) × N (0) = h 1 ,-1 , 2 i / √ 6. 1 Problem 5. Does the limit lim ( x,y ) → (0 , 0) x-y x 2 + y 2 exist? Why? Solution: The limit does not exists, because the limit =0 on the line y = x , while the limit = ∞ on the line y = 0. Problem 6. For f ( x,y,z ) = √ x + yz , ﬁnd the rate of change of at P (1 , 3 , 1) in the direction of the vector v = h 2 , 3 , 6 i . Solution: The rate of change is equal to ∇ f ( P ) · u where u is the unit vector in the direction v . Now, ∇ f (1 , 3 , 1) = h 1 , 1 , 3 i / 4 and u = h 2 , 3 , 6 i / 7. Thus, the answer is 23/28....
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## This note was uploaded on 12/15/2011 for the course MAC 2313 taught by Professor Keeran during the Spring '08 term at University of Florida.

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fmtsol - t = 0 so T(0 = h 1 1 1 i √ 3 Furthermore N t = T...

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