Math252Sols2F08

Math252Sols2F08 - QUIZ 2 (CHAPTER 15, 16.1, 16.2) SOLUTIONS...

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Unformatted text preview: QUIZ 2 (CHAPTER 15, 16.1, 16.2) SOLUTIONS MATH 252 – FALL 2008 – KUNIYUKI SCORED OUT OF 125 POINTS MULTIPLIED BY 0.84 105% POSSIBLE Clearly mark vectors, as we have done in class. I will use boldface, but you don’t! When describing vectors or vector-valued functions, you may use either or “ i j k ” notation. Assume we are in our usual 2- and 3-dimensional Cartesian coordinate systems. Give exact answers, unless otherwise specified. 1) Find the length of the curve parameterized by: x = t sin t + cos t , y = t cos t sin t , z = 32 t, 0 t 2 4. Leave your answer as a simplified exact answer; do not approximate it using a calculator. Distance is measured in meters. Show all work! (20 points) L= = 4 0 4 0 dx dt (D t 2 dy + dt 2 2 dz + dt t sin t + cos t dt ) + (D 2 t t cos t sin t Use the Product Rule for = = = = = = 4 0 4 0 4 0 4 0 4 0 4 0 = 10 ( sin t + t cos t (t cos t ) 2 + ( ) 2 dx dy and . dt dt ) + ( cos t 2 sin t t sin t ) 2 cos t dt ) + (3t ) dt 2 t sin t 2 3 + Dt t 2 2 2 + 9t 2 dt t 2 cos 2 t + t 2 sin 2 t + 9t 2 dt ( ) t 2 cos 2 t + sin 2 t + 9t 2 dt t 2 + 9t 2 dt 10t 2 dt 4 0 t dt Observe: t 2 = t = t , since t 0 on the interval 0, 4 . Now, apply the Fundamental Theorem of Calculus. t2 = 10 2 4 0 ( 4) = 10 (0) 2 2 2 2 () = 10 8 = 8 10 meters This is about 25.2982 meters. ( ) 2) Find parametric equations for the tangent line to C at the point 25, 0, 478 , ( t), z = 2t 5 It helps that x is a one-to-one function of t. Solve t 3 2 = 25 . where C is parameterized by: x = t 3 2, y = sin t2 + 1. (12 points) () r t = t3 2, sin ( t ), 2t 5 t2 + 1 . Find the t-value corresponding to the given point. 2 = 25 t3 t 3 = 27 t=3 Check: ( ) (3) r3= 3 ( ) , 2 (3) (3) 5 2, sin 3 2 +1 = 25, 0, 478 Find a direction vector for the desired tangent line. () r (t ) = ( t ) , 2t t + 1 , 10t 2t 3t , cos ( t ) = 3t , cos ( t ) , 10t 2t r ( 3) = 3( 3) , cos ( 3 ) , 10 ( 3) 2 ( 3) = 27, ( 1) , 804 = 27, , 804 ( You can use any non-0 scalar multiple of this.) r t = t3 2, sin 5 2 2 4 2 4 2 4 Write parametric equations for the desired tangent line. x = 25 + 27u y= 0 u (u in R ) z = 478 + 804u x = 25 + 27u y= u (u in R ) z = 478 + 804u 3) A curve C in 3-space is smoothly parameterized by the position vector-valued function (VVF) rule r(t ) . The position vector r(t ) and the tangent vector r (t ) () () are orthogonal for all real t. Simplify Dt r t • r t for all real t. Use a differentiation rule discussed in class. (5 points) () () Dt r t • r t () () = r t •r t =0 by orthogonality () () + r t •r t =0 by orthogonality ( by a Product Rule for VVFs: (iii) on p.756) =0 () ( )( Find the position vector-valued function (VVF rule) r(t ) if r ( 0 ) = 2i v ( 0 ) = i + 3j . (15 points) ) 4) The acceleration of a moving particle is given by a t = 3sin t i + 5cos t j . () () v t = a t dt 6 j and ( one member ) = 3sin t , 5cos t dt = 3cos t , 5sin t + C () Solve for C by plugging in t = 0 and using the initial condition v 0 = i + 3j, or 1, 3 . () 1, 3 = () () 3(1) , 5( 0 ) + C 1, 3 = 3, 0 + C v0= 3cos 0 , 5sin 0 + C C = 4, 3 Therefore, vt = () v (t ) = Now, 3cos t , 5sin t + 4, 3 3cos t + 4, 5sin t + 3 () () ( one member ) r t = v t dt = 3cos t + 4, 5sin t + 3 dt = 3sin t + 4t , 5 cos t + 3t + D () Solve for D by plugging in t = 0 and using the initial condition r 0 = 2i 6 j, or 2, 6 . () r0= () () 3sin 0 + 4 0 , 2, 6 = 0, () () 5 cos 0 + 3 0 + D 5 +D D = 2, 1 Therefore, () 3sin t + 4t , () 3sin t + 4t + 2, rt = rt = ( 5 cos t + 3t + 2, 1 5 cos t + 3t 1 , or )( 3sin t + 4t + 2 i + ) 5 cos t + 3t 1 j () t3 t2 , 5) The curve C is determined by r t = , where t > 0 . Show all work, 32 simplify radicals, and simplify completely, as we have done in class. Do not eliminate the parameter. Messy and/or undisciplined work may not be graded! Note: When differentiating, avoid using the Product Rule as an alternative to the Quotient Rule, unless you simplify your result to the most compact form! (18 points total) a) Find the unit tangent VVF (rule) T(t ) for C. () rt = t3 t2 , 32 () r t = t 2 , t , or t t , 1 () rt = t2, t =t =t =t , or t t , 1 t, 1 ( because we can assume t > 0) (t ) + (1) t, 1 2 2 = t t2 + 1 () Tt = = = () Tt = () r (t ) rt t t, 1 t t2 + 1 t, 1 t2 + 1 t, 1 t2 + 1 () , or T t = t t2 + 1 , 1 t2 + 1 Note: If we had eliminated the parameter, we would have obtained 3 y= 9x2 , which can be analyzed directly. 2 () b) Find the VVF (rule) T t for C. () t Remember, T t = t2 + 1 () Tt= , () t2 + 1 Dt t ( 1 or t2 + 1 ( t2 + 1 ) Dt t 2 + 1 t t2 + 1 t ) ( ) , t +1 2 1 2 1/ 2 ( 2 ) ( 2t) 12 t +1 2 , 3/ 2 ( for the first component, we used the Quotient Rule for Diff.) t2 + 1 1 = ( t +1 2 t2 + 1 , = (t + 1) (t + 1) (t + 1) t2 + 1 ( t2 3/ 2 1 ) t2 + 1 3/ 2 t2 + 1 t2 (t + 1) 2 2 t t2 + 1 2 2 = ) (t + 1) (t ) , 2 2 t2 + 1 = t2 + 1 1/ 2 t2 t2 + 1 = ) ( 2t) ( 12 t +1 2 t (t + 1) 3/ 2 2 t , 3/ 2 t , (t + 1) 3/ 2 2 , ( t ) t2 + 1 3/ 2 , or ( 1 ) t2 + 1 3/ 2 1, t 3/ 2 () You didn’t have to do this, but here’s how we can find N t : () Tt = = = ( ( 1 (1) + ( t ) 2 (t + 1) 3/ 2 1 1+ t2 (t + 1) 3/ 2 1 (t + 1) 1/ 2 2 (t + 1) 3/ 2 2 1 t +1 2 () T (t ) () Tt Nt = = 3/ 2 1 2 = 1, t ) t2 + 1 2 = ) 1 2 = 1, t 3/ 2 t2 + 1 ( 1 1, t ) t2 + 1 3/ 2 1 t +1 2 t +1 1 = 1, t 3/ 2 1 t2 + 1 2 ( = 1 (t + 1) 2 = 1/ 2 1 t2 + 1 ) 1, t , t t2 + 1 Observe: For all t > 0 , N(t ) • T(t ) = 0 , which reflects the fact that () Nt () Tt. 6) Assume that r is a position VVF of t in 3-space that is twice differentiable everywhere (i.e., second derivatives exist for all real t). Write a curvature formula we discussed for (t ) that involves a cross product. (4 points) ( ) a (t ) v (t ) () vt () t= 3 () rt rt or () rt 3 () 7) A twisted cubic curve C is determined by r t = t , t 2 , t 3 , where t > 0 . (19 points total) a) Find a general curvature formula, (t ) , for every point on C. Use your formula from Problem 6), and simplify your answer completely. Show all work! (15 points) () Then, r ( t ) or v ( t ) = r ( t ) or a ( t ) = Let r t = t , t 2 , t 3 . ( ) a (t ) = 1, 2t , 3t 2 , and 0, 2, 6t . 1, 2t , 3t 2 vt 0, 2, 6t ijk = 1 2t 3t 2 0 2 6t = ( 2t 3t 2 i 2 6t = 12t 2 1 2t 1 3t 2 j+ k 02 0 6t ) (6t 0) j + ( 2 0) k 6t 2 i = 6t 2 i 6t j + 2k = 6t 2 , 6t , 2 ( ) a (t ) vt or 2 3t 2 , 3t , 1 = 6t 2 , 6t , 2 = (6t ) + ( 6t ) + ( 2) 2 2 2 = 36t 4 + 36t 2 + 4 ( ) = 4 9t 4 + 9t 2 + 1 = 2 9t 4 + 9t 2 + 1 2 = 2 3t 2 , 3t , 1 or = 2 (3t ) + ( 3t ) + (1) 2 2 = 2 9t 4 + 9t 2 + 1 2 2 () 3 vt = (1) + ( 2t ) + (3t ) 2 = = 3 1, 2t , 3t 2 2 3 2 ( 1 + 4t + 9t ) 3 2 4 ( ) = 9t 4 + 4t 2 + 1 (t ) = 2 ( ) a (t ) v (t ) vt 3 3/ 2 2 9t 4 + 9t 2 + 1 = (9t 4 ) + 4t 2 + 1 3/ 2 b) Use your formula in part a) to find the curvature of the twisted cubic curve C at the point 2, 4, 8 . Approximate your answer to four decimal ( ) places. (4 points) ( ) The point 2, 4, 8 corresponds to t = 2 . ( 2) = 2 9t 4 + 9t 2 + 1 (9t 4 ) + 4t 2 + 1 3/ 2 = t=2 () () (9(2) + 4(2) + 1) 4 2 2 9 2 + 9 2 +1 4 0.0132 2 3/ 2 () 8) Sketch the level curves of f x , y = y x 2 for k = 3, 0, 3 on the grid below. Label the curves with their corresponding k-values. Be reasonably accurate. (8 points) Let f ( x , y ) = k . For any real k: k=y x2 x2 + k = y y = x2 + k The graph of this is a parabola that opens upward and that has the point (0, k ) as its vertex and its y-intercept. () has vertex ( 0, 0 ) . + 3 has vertex ( 0, 3) . k = 3 : The parabola y = x 2 k = 0 : The parabola y = x 2 k = 3 : The parabola y = x 2 5 3 has vertex 0, 3 . y 4 3 3 2 0 1 -5 -4 -3 -2 -1 -1 -2 1 2 3 4 -3 -3 -4 -5 Here is the corresponding surface, the graph of z = y x2 : 5 x 9) Matching. (9 points total) Fill in each blank with the best choice (A-K) in the list below to indicate the level surface of f for the given value of k. A. B. C. D. E. F. G. H. I. J. K. A Sphere or Ellipsoid A Hyperboloid of One Sheet A Hyperboloid of Two Sheets A Cone A Circular or Elliptic Paraboloid A Hyperbolic Paraboloid A Right Circular or Elliptic Cylinder A Plane A Line (a “degenerate” surface) A Point (a “degenerate” surface) NONE (no surface) ( ) a) The level surface of f x , y , z = x 2 + z 2 , k = 7 is __G__. Analyze: 7 = x 2 + z 2 . Its graph is a right circular cylinder in xyz-space. Its axis is the y-axis. Imagine taking a circle in the xz-plane and sweeping it parallel to the y-axis. ( ) b) The level surface of f x , y , z = 3x + 4 y 5z , k = 0 is __H__. Analyze: 0 = 3x + 4 y 5z . Its graph is a plane in xyz-space that passes through the origin. ( ) c) The level surface of f x , y , z = x 2 + 4 y 2 + 9 z 2 , k = 25 is __A__. Analyze: 25 = x 2 + 4 y 2 + 9 z 2 . Its graph is an ellipsoid centered at the origin in xyz-space. Below are graphs of a), b), and c), respectively. (Courtesy Mathematica.) 10) 3x 2 4 y 2 Show that lim does not exist. (8 points) ( x , y ) (0 ,0) 2 x 2 + 3 y 2 Let ( x , y ) (0,0) along the y-axis (x = 0): () () 2 30 4 y2 3x 2 4 y 2 lim = lim ( x , y ) (0 ,0) 2 x 2 + 3 y 2 ( x , y ) (0 ,0) 2 0 2 + 3 y 2 = lim y = Let ( x , y ) 0 4 y2 3y2 4 3 (0,0) along the x-axis (y = 0): () () 3x 2 4 0 3x 2 4 y 2 lim = lim ( x , y ) (0 ,0) 2 x 2 + 3 y 2 ( x , y ) (0 ,0) 2 x 2 + 3 0 2 2 3x 2 = lim 2 x 0 2x 3 = 2 () We have found two paths approaching 0, 0 that yield different limit values for 3x 2 4 y 2 , so the indicated limit does not exist by the Two-Path Rule. 2x2 + 3y2 Here is a graph of z = 3x 2 4 y 2 : 2x2 + 3y2 11) x3 y3 Use polar coordinates to find lim . (7 points) ( x , y ) (0 ,0) x 2 + y 2 ( )( ) 3 3 r cos r sin x3 y3 lim = lim 2 2 2 r0 ( x , y ) (0 ,0) x + y r ( r cos )( r sin ) = lim 3 r 3 3 r2 sin 3 0 r 6 cos3 = lim r0 r2 = lim r 4 cos3 sin 3 r 0 3 ( ) =0 Justifying the last step: For all real , 1 cos As a result, 1 cos3 1 and 1 sin 1 and 1 sin 3 Therefore, 1 cos3 sin 3 1. 1. 1. () () Observe that r 4 > 0 along any path approaching 0, 0 that avoids 0, 0 , itself. Multiply all three parts by r 4 and apply the Sandwich / Squeeze Theorem: As r 0, r4 r 4 cos3 sin 3 0 r4 . 0 So, 0 Informally, the limit of something approaching 0 times something that is bounded is 0. Here is a graph of z = x3 y3 : x2 + y2 ...
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