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
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 :
at 
