# Solution when r 1 t h 1 0 1 i then t 1 when r 2 s h 1

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Solution: When r 1 ( t ) = h 1 , 0 , 1 i , then t = 1. When r 2 ( s ) = h 1 , 0 , 1 i , then s = 1. Calculating derivatives, we obtain: r 0 1 ( t ) = h- sin( t - 1) , 2 t , 4 t 3 i r 0 1 (1) = h 0 , 2 , 4 i r 0 2 ( s ) = h 1 s , 2 s - 2 , 2 s i r 0 2 (1) = h 1 , 0 , 2 i . Hence, cos θ = r 0 1 (1) · r 0 2 (1) | r 0 1 (1) || r 0 2 (1) | = h 0 , 2 , 4 i · h 1 , 0 , 2 i 20 5 = 1 100 (0 · 1 + 2 · 0 + 4 · 2) = 8 10 = 4 5 .
Problem 24(b) - Fall 2006 Suppose a particle moving in space has velocity v ( t ) = h sin t , cos 2 t , e t i and initial position r (0) = h 1 , 2 , 0 i . Find the position vector function r ( t ).
Problem 24(b) - Fall 2006 Suppose a particle moving in space has velocity v ( t ) = h sin t , cos 2 t , e t i and initial position r (0) = h 1 , 2 , 0 i . Find the position vector function r ( t ). Solution: The position vector function r ( t ) is the integral of its derivative r 0 ( t ) = v ( t ): r ( t ) = Z v ( t ) dt = Z h sin t , cos 2 t , e t i dt = h- cos( t ) + x 0 , 1 2 sin(2 t ) + y 0 , e t + z 0 i . Now use the initial position r (0) = h 1 , 2 , 0 i to solve for x 0 , y 0 , z 0 . - cos(0) + x 0 = - 1 + x 0 = 1 = x 0 = 2 . 1 2 sin(0) + y 0 = 0 + y 0 = 2 = y 0 = 2 . e 0 + z 0 = 1 + z 0 = 0 = z 0 = - 1 . Hence, r ( t ) = h- cos( t ) + 2 , 1 2 sin(2 t ) + 2 , e t - 1 . i
Problem 25(a) - Fall 2006 Let f ( x , y ) = e x 2 - y + x p 4 - y 2 . Find partial derivatives f x , f y and f xy . Problem 25(b) - Fall 2006 Find an equation for the tangent plane of the graph of f ( x , y ) = sin(2 x + y ) + 1 at the point (0 , 0 , 1). Problem 26(a) - Fall 2006 Let g ( x , y ) = ye x . Estimate g (0 . 1 , 1 . 9) using the linear approximation of g ( x , y ) at ( x , y ) = (0 , 2). Solutions to these problems: These types of problems might not be on this exam (check web site).
Problem 26(b) - Fall 2006 Find the center and radius of the sphere x 2 + y 2 + z 2 + 6 z = 16 .
Problem 26(b) - Fall 2006 Find the center and radius of the sphere x 2 + y 2 + z 2 + 6 z = 16 . Solution: Complete the square in order to put the equation in the form: ( x - x 0 ) 2 + ( y - y 0 ) + ( z - z 0 ) 2 = r 2 . We get: x 2 + y 2 + ( z 2 + 6 z ) = x 2 + y 2 + ( z 2 + 6 z + 9) - 9 = 16 . This gives the equation ( x - 0) 2 + ( y - 0) 2 + ( z + 3) 2 = 25 = 5 2 . Hence, the center is C = (0 , 0 , - 3) and the radius is r = 5.
Problem 26(c) - Fall 2006 Let f ( x , y ) = p 16 - x 2 - y 2 . Draw a contour map of level curves f ( x , y ) = k with k = 1 , 2 , 3. Label the level curves by the corresponding values of k .
Problem 26(c) - Fall 2006 Let f ( x , y ) = p 16 - x 2 - y 2 . Draw a contour map of level curves f ( x , y ) = k with k = 1 , 2 , 3. Label the level curves by the corresponding values of k . Solution: A problem of this type might not be on this exam (check web site).
Problem 27 Consider the line L through points A = (2 , 1 , - 1) and B = (5 , 3 , - 2). Find the intersection of the line L and the plane given by 2 x - 3 y + 4 z = 13.
Problem 27 Consider the line L through points A = (2 , 1 , - 1) and B = (5 , 3 , - 2). Find the intersection of the line L and the plane given by 2 x - 3 y + 4 z = 13. Solution: The vector part of L is -→ AB = h 3 , 2 , - 1 i and the point A is on the line. The vector equation of L is: L = ~ A + t -→ AB = h 2 , 1 , - 1 i + t h 3 , 2 , - 1 i = h 2 + 3 t , 1 + 2 t , - 1 - t i . Plugging x = 2 + 3 t , y = 1 + 2 t and z = - 1 - t into the equation of the plane gives: 2(2 + 3 t ) - 3(1 + 2 t ) + 4( - 1 - t ) = - 4 t - 3 = 13 = ⇒ - 4 t = 16 = t = - 4 . So, the point of intersection is: L ( - 4) = h 2 - 12 , 1 - 8 , - 1 - ( - 4) i = h- 10 , - 7 , 3 i .
Problem 28(a) Two masses travel through space along space curve described by the two vector functions r 1 ( t ) = h t , 1 - t , 3 + t 2 i , r 2 ( s ) = h 3 - s , s - 2 , s 2 i where t and s are two independent real parameters.