SKKU-Calculus-Sec-13-7 Tangent Plane and Differentiability-New


   13.7     Tangent Plane      by SGLee - HSKim, YJLim, 서용태, JHLee, 이원준

 http://youtu.be/uOf-5YHKGI4                                  http://youtu.be/GDkE8OqUvsk  

 

13장 7절은 Tangent Plane(접평면)에 관한 내용이다.

연습문제의 경우 Tangent Plane을 구하고, 그것의 Normal Line을 구하는 것이 주를 이룬다.

(Lecture)  http://youtu.be/uOf-5YHKGI4

(Exercises)  http://youtu.be/GDkE8OqUvsk

1. Find the equation of the tangent plane at the point  to the surface .

 

 

      

      

      

      => The equation of the tangent plane is

      

     

 




시각화 해보자.




2. Find the equations of the tangent plane and the normal line at the point  to the surface .

 

 

      

      =>   and 

      Hence at the point , the tangent plane is  and

      the normal line is  (that is, ).

by Sage.

1. Tangent Plane




2. Normal Line




Note : x-1=y=-z 를 Sage로 표현할 수가 없었다. 따라서 매개변수 방정식으로 나타내었다. Revised 이원준

3. Find  so that all tangent planes to the surface  pass through

    the origin  for all .

 

 

    Let  . Then we have that

              and .

    Since every tangent plane intersects the origin, the equation of the tangent plane

    at a point  on the surface should satisfies

            .

   We then have

           .

f_x 를 구하면




f_y 를 구하면




따라서, 접평면의 방정식은




이 식을 정리하면, (1-2k)(x^2 + y^2)^k = 0 이 되므로,

k = 1/2 

    Simplifying the equation, we obtain , which implies .

4. Find the equation of the tangent plane of the surface  at .

 

      Let . Due to implicit differentiation, we have

         and .

    In particular, at  we have .

    Thus, the tangent plane is . Simplifying it, we have .




Note :  접평면을 구하는것은 계속 같은 방식이므로 직관적으로 확인하기 위한 그림만 그려보았다.

5. Find an equation of tangent plane and normal line to the surface at .

 

 

      Equation of tangent plane:

                       

                     .

                      or .

      Equation of normal line:

                     

1. Tangent Plane




2. Normal Line




이를 시각화 해보자




Note : Gradint 가 0이 나왔는데, Parametric Equation을 이용하여, Normal line을 표현했다. 이원준

6. Let  be the surface whose equation in cylindrical coordinates is .

   Find the equations of the tangent plane and the normal line to  at the point  in rectangular coordinates.

 

 

    The equation of   in rectangular coordinates is given by 

    If we regard  as a level surface of  at ,

    then the normal vector of the tangent plane is , where  and

    hence

             

    Therefore, the equations of the tangent plane is

         or

                              .

     And the normal line is

                                .

 




7. Let  if  and let  

    In the direction of what unit vectors does the directional derivative of  at  exist?

 

 

    Suppose we wish to compute the directional derivative of  in the direction of a unit vector  Then

     

     and the limit does not exist unless  or 

     Hence  the directional derivative of  at  exists in the directions of  and  or their unit scalar multiples.

 




8-9. Find the equation of the tangent plane to the given surface at the indicated point.

 

8.   at 

     http://matrix.skku.ac.kr/cal-lab/cal-12-3-4.html  

 

 




9.  at 

    http://matrix.skku.ac.kr/cal-lab/cal-12-3-11.html

 

 

      




 10. Find the equations of the tangent plane and normal line at the point  to the paraboloid .

 

 




























So tangency normal vector is clear.

11. Find the equations of the tangent plane and the normal line at  the point  to the surface .

 

 

      We have

       .

      Hence, .

      Therefore by  the equation of the tangent plane to the surface  at the point  is . Or, .

      By  the equation of the normal line at the point  to the surface  is

      .

1. Tangent Plane




2. Normal Line




12. The surface  and   meet in an ellipse .

     Find parametric equations for the line tangent to  at the point .

 

 

      Let  be a point on the intersection of these surfaces.

      Then the tangent line at  to the curve  is orthogonal to  and .

      In particular, it is parallel to .

      The components of  and the coordinates of  give us equations for the line.

      We have 

                                 ,

                  .

      Therefore

          .

      Thus the tangent to the curve  at  is parallel to the vector .

      Therefore, its equation is .

 




13. Sketch a level curve (ellipse) of  passing through the pont .

     Find a vector perpendicular to this ellipse at the point .

 

       The value of  at the point  is .

       Therefore, the level curve of  passing through  is 

       ,     which is an ellipse. 

       Vector perpendicular to this ellipse at the point  is .

                                                            




Note : 

                                               Back to Part II




                                                    Back to Part II