Sec103homeworksolns.Pdf

Answers and Solutions to Section 10.3 Homework Problems 1, 3, 7-27 (odd), 35, 37, and 39 S. F. Ellermeyer

1. The curve rt  2sint,5t,2cost, 10  t  10 (which is a piece of a helix) has length 10 10  |rt|dt   2cost2  52  2sint2 dt 10 10 10   29 dt 10  20 29 . 3. The curve rt  2 t,et,et ,0 t  1 has length 1 1 2  |rt|dt   2  et 2  et 2 dt 0 0 1   et 2  2etet  et 2 dt 0 1   et  et 2 dt 0 1   et  et dt 0 t t t1  e  e |t0  e1  e1   e0  e0   e  e1   1  1  e  e1. 7. For rt  et sinti  et costj, we have rt  etcost  sinti  etsint  costj and |rt|  etcost  sint2  etsint  cost2  2e2t  2 et. Thus, the arc length function starting from t  0 and measured in the direction of increasing t is t s   |ru|du 0 t   2 eu du 0  2 et  1. Solving the equation 2 et  1  s for t, we obtain

1 t  ln 1  1 s . 2 Thus, using arc length (s) as our parameter, we can write the equation of this curve as

qs  1  1 s sin ln 1  1 s i 2 2

 1  1 s cos ln 1  1 s j. 2 2 9. For rt  3sinti  4tj 3costk, we have rt  3costi  4j  3sintk and |rt|  3cost2  42  3sint2  5 Thus, the arc length function starting from t  0 and measured in the direction of increasing t is t s   |ru|du 0 t   5du 0  5t. Thus, using arc length (s) as our parameter, we can write the equation of this curve as q s  3sin 1 s i  4 sj 3cos 1 s k   5 5 5 11. For rt  2sint,5t,2cost, the tangent vector is rt  2cost,5,2sint and the length of this tangent vector is |rt|  2cost2  52  2sint2  29 . The unit tangent vector is Tt  1 rt  1 2cost,5,2sint. |rt| 29 Note that Tt  1 2sint,0,2cost 29 and the length of this vector is |Tt|  1 2sint2  02  2cost2  2 . 29 29 Thus, the unit normal vector is

2 Nt  1 Tt |Tt| 29  1 2sint,0,2cost 2 29  sint,0,cost. In order to find the curvature using the formula   dT , ds we must first find the relationship between the parameter t and the arc length s. This is similar to the work done in problem 1 of this exercise set. We obtain s  29 t. Now note (by the Chain Rule) that dT  ds dT dt dt ds which means that dT dT  dt ds ds dt

 1 1 2sint,0,2cost 29 29  1 2sin t ,0,2cos t 29        2 sin t ,0,cos t . 29      We now obtain   dT  2 . ds 29 Since this curve is a helix, it makes sense that it has constant curvature. Also, since we did all of the work of computing |Tt| and |rt|, it is much easier to compute the curvature using the formula 2  |T t| 29     2 |rt| 29 29 since that way we don’t have to deal with s at all. 1 3 2 13. For rt  3 t ,t ,2t , the tangent vector is rt  t2,2t,2 and the length of this tangent vector is

3 2 |rt|  t2   2t2  22  t4  4t2  4 2  t2  2  t2  2. The unit tangent vector is Tt  1 rt |rt|  1 t2,2t,2 t2  2 2  t , 2t , 2 . t2  2 t2  2 t2  2 Note that

2  4t 4  2t 4t T t  2 , 2 , 2 t2  2 t2  2 t2  2 and the length of this vector is 2 2 2 2  4t 4  2t 4t |T t|  2  2  2 t2  2 t2  2 t2  2

1 2 2 4 2  2 16t  16  16t  4t   16t t2  2 1 4 2  2 4t  16t  16 t2  2

2 2 2  2 t  2 t2  2  2 . t2  2 Thus, the unit normal vector is Nt  1 Tt |Tt| t2  2 4t 4  2t2 4t  2 , 2 , 2 2 t2  2 t2  2 t2  2 2  2t , 2  t , 2t . t2  2 t2  2 t2  2 The curvature using the formula  2 |T t| t22 2    2  2 . |rt| t  2 t2  2 15. For rt  t2i  tk we have

5 rt  2 i  etj  etk 2 |rt|  2  et 2  et 2

 et 2  2etet  et 2

 et  et 2  et  et rt  etj  etk ijk rt  rt  2 et et 0 et et

 2i  2 etj  2 etk. 2 2 |rt  rt|  22   2 et  2 et

 4  2et 2  2et 2

 2 et 2  2etet  et 2

 2 et  et 2  2 et  et . Thus, the curvature is |rt  rt| 2 et  et  2     . |rt|3 et  et 3 et  et 2 Since the point 0,1,1 on this curve corresponds to the parameter value t  0, we see that the curvature at this point is 2 2 0  2  . e0  e0  4 21. For the curve fx  x3, we have fx  3x2 fx  6x so the curvature is |fx| 6|x| x  3/2  3/2 . 1  fx2 1  9x4  23 For the curve fx  4x5/2, we have fx  10x3/2 fx  15x1/2

6 so the curvature is |fx| 15x1/2 x  3/2  3/2 1  fx2 1  100x3  . 25. For the curve fx  ex, we have fx  ex fx  ex so the curvature is |fx| ex x  3/2  3/2 1  fx2 1  e2x  . We would like to find at what value of x the curvature is maximum: Since 3/2 1/2 1  e2x  ex  ex 3 1  e2x  2e2x   2  x  3 1  e2x  3/2 1/2 1  e2x  ex  ex 31  e2x  e2x   3 1  e2x  1/2 ex1  e2x  1  2e2x   3 1  e2x  ex1  2e2x   5/2 1  e2x  we see that x  0 when 1  2e2x  0. Solving this equation for x gives us e2x  1/2 or ex 2  1/2 or ex  2 /2 or 2 x  ln . 2 Since this x gives us the only critical point of  and since it is clear that the curvature must be maximum at some point for the graph of fx  ex,we conclude that the curvature is maximum at the point 2 2 ln ,  0.34637,0.70711 . 2 2  

7 27. The curvature of the curve (drawn on page 724 of the textbook) seems to be greater at point P. 2 2 3 35. For rt  t , 3 t ,t , we have rt  2t,2t2,1 |rt|  4t2  4t4  1

 2 t4  t2  1 4 2  2 t2  1 2  2 t2  1 2  2t2  1 Tt  1 rt |rt| 2  2t , 2t , 1 2t2  1 2t2  1 2t2  1 2  2  4t 4t 4t T t  2 , 2 , 2 2t2  1 2t2  1 2t2  1 2 The point 1, 3 ,1 corresponds to the parameter value t  1 so

8 T 1  2 , 2 , 1   3 3 3 N1  1 T1 |T1|  1  2 , 4 , 4 2 2 4 2 4 2 9 9 9 9  9  9   1 , 2 , 2 3 3 3 and B1  T1  N1 ijk 2 2 1  3 3 3 1 2 2  3 3  3   2 i  1 j  2 k. 3 3 3 37. For rt  2sin3t,t,2cos3t, we have rt  6cos3t,1,6sin3t |rt|  37 Tt  1 6cos3t,1,6sin3t 37 Tt  1 18sin3t,0,18cos3t 37 The point 0,,2 corresponds to the parameter value t   so T  1 6,1,0 37 N  1 T |T|  0,0,1 and B  T  N ijk  6 1 0  37 37 001

 1 i  6 j. 37 37 Since the binormal vector is orthogonal to the osculating plane, the osculating plane has equation

9 x  6y    0 and since the unit tangent vector is orthogonal to the normal plane, the normal plane has equation  6x  y    0. 39. The ellipse 9x2  4y2  36 can be written as 2 y2 x   1. 4 9 Parametric equations of this ellipse are x  2cost y  3sint. For this parametric curve (r) we have rt  2sint,3cost |rt|  4sin2t  9cos2t rt  2cost,3sint ijk rt  rt  2sint 3cost 0  6k 2cost 3sint 0 |rt  rt|  6 |rt  rt| 6   3  3/2 . |rt| 4sin2t  9cos2t The point 2,0 on the ellipse corresponds to the parameter value t  0 so at this point the curvature is 6 2   3/2  4sin20  9cos20 9 The osculating circle at this point thus has radius 9/2 and is centered at the 5 point  2 ,0 . An equation for the osculating circle is 2 x  5  y2  81 . 2 4 Parametric equations of this circle (used to graph the circle) are x   5  9 cos t 2 2   y  9 sin t . 2   The point 0,3 on the ellipse corresponds to the parameter value t  /2 so at this point the curvature is   6  3 2  2  3/2 4 4sin 2  9cos 2 The osculating circle at this point thus has radius 4/3 and is centered at the

10 5 point 0, 3 . An equation for the osculating circle is 2 x2  y  5  16 . 3 9 Parametric equations of this circle (used to graph the circle) are x  4 cos t 3   y  5  4 sin t . 3 3  