Eq. of tangent plane at a point.

catch22
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Homework Statement


upload_2015-10-29_21-37-8.png


Homework Equations

The Attempt at a Solution


what is the difference between the form

z - zo = fx(x-xo) + fy(y-yo)

and:

Fx(xo,yo,zo) + Fy(xo,yo,zo) + Fz(xo,yo,zo)=0
____________________________________________________________________________using the first eq :
upload_2015-10-29_21-48-20.png


while the latter gives:
upload_2015-10-29_21-49-44.png


Am I doing something wrong? shouldn't they give the same answer?
 
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catch22 said:

Homework Statement


View attachment 91052

Homework Equations

The Attempt at a Solution


what is the difference between the form

z - zo = fx(x-xo) + fy(y-yo)

and:

Fx(xo,yo,zo) + Fy(xo,yo,zo) + Fz(xo,yo,zo)=0
____________________________________________________________________________using the first eq :
View attachment 91053

while the latter gives:
View attachment 91054

Am I doing something wrong? shouldn't they give the same answer?

You're doing something wrong, when you solve for ##z-z_0##, what happens to the term in front of ##(z-z_0)##, and what about your signs?
 
Student100 said:
You're doing something wrong, when you solve for ##z-z_0##, what happens to the term in front of ##(z-z_0)##, and what about your signs?
are you talking about the first equation z - zo = fx(x-xo) + fy(y-yo)? I thought there aren't suppose to be any terms in front of (z−z0)

actually, in order to get z - zo = fx(x-xo) + fy(y-yo), didn't we have to divide everything by z so we had something like A/C = Fx and B/C = Fy
 
I think I found the problem.

if I use gradient, the Fz has to equate to -1, right?
 
Student100 said:
You're doing something wrong, when you solve for ##z-z_0##, what happens to the term in front of ##(z-z_0)##, and what about your signs?
is this correct?
upload_2015-10-29_22-48-52.png
 

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  • upload_2015-10-29_22-48-32.png
    upload_2015-10-29_22-48-32.png
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The difference between "z= f(x, y)" and "F(x, y, z)= 0" is that you are taking F(x, y, z)= f(x, y)- z. Taking the gradient \nabla F= <f_x, f_y, -1>[.

It is correct that F(1, -1, 4\sqrt{2})= \frac{1}{4}+ \frac{1}{4}+ \frac{32}{64}= 1 but that does NOT say "z= 1". You are told that at this point z= 4\sqrt{2}. The fact that F(1, -1, 4\sqrt{2})= 1 just verifies that the given point is on the given surface.

The tangent plane is z- z_0= f_x(x_0, y_0)(x- x_0)+ f_y(x_0, y_0)(y- y_0) only applies if you are given z= f(x, y) which is NOT the case here. You are given, rather, F(x, y, z)= x^2/4+ y^2/4+ z^2/64= 1. In order to convert that to the "z= f(x,y)" form you need to solve for z: z^2/64= 1- x^2/4- y^2/4, z/8= \sqrt{1- x^2/4- y^2/4}, z= 8\sqrt{1- x^2/4- y^2/4}.

Much simpler is to use the fact that if F(x, y, z)= 1 (or any constant) then \nabla F is perpendicular to the surface at any point. Since for any point (x, y, z) on the tangent plane to the surface at (x_0, y_0, z_0), the vector (x- x_0)\vec{i}+ (y- y_0)\vec{j}+ (z- z_0)\vec{k} is in the tangent plane, we must have \nabla F\cdot [(x- x_0)\vec{i}+ (y- y_0)\vec{j}+ (z- z_0)\vec{k}]= 0 which is the same as F_x(x_0, y_0, z_0)(x- x_0)+ F_y(x_0, y_0, z_0)(y- y_0)+ F_z(x_0, y_0, z_0)(z- z_0)= 0 as the equation of the tangent plane.
 
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HallsofIvy said:
The difference between "z= f(x, y)" and "F(x, y, z)= 0" is that you are taking F(x, y, z)= f(x, y)- z. Taking the gradient \nabla F= <f_x, f_y, -1>[.

It is correct that F(1, -1, 4\sqrt{2})= \frac{1}{4}+ \frac{1}{4}+ \frac{32}{64}= 1 but that does NOT say "z= 1". You are told that at this point z= 4\sqrt{2}. The fact that F(1, -1, 4\sqrt{2})= 1 just verifies that the given point is on the given surface.

The tangent plane is z- z_0= f_x(x_0, y_0)(x- x_0)+ f_y(x_0, y_0)(y- y_0) only applies if you are given z= f(x, y) which is NOT the case here. You are given, rather, F(x, y, z)= x^2/4+ y^2/4+ z^2/64= 1. In order to convert that to the "z= f(x,y)" form you need to solve for z: z^2/64= 1- x^2/4- y^2/4, z/8= \sqrt{1- x^2/4- y^2/4}, z= 8\sqrt{1- x^2/4- y^2/4}.

Much simpler is to use the fact that if F(x, y, z)= 1 (or any constant) then \nabla F is perpendicular to the surface at any point. Since for any point (x, y, z) on the tangent plane to the surface at (x_0, y_0, z_0), the vector (x- x_0)\vec{i}+ (y- y_0)\vec{j}+ (z- z_0)\vec{k} is in the tangent plane, we must have \nabla F\cdot [(x- x_0)\vec{i}+ (y- y_0)\vec{j}+ (z- z_0)\vec{k}]= 0 which is the same as F_x(x_0, y_0, z_0)(x- x_0)+ F_y(x_0, y_0, z_0)(y- y_0)+ F_z(x_0, y_0, z_0)(z- z_0)= 0 as the equation of the tangent plane.
upload_2015-10-30_6-11-52.png


looks correct?
 
What are x_0, y_0 and z_0?
 
HallsofIvy said:
What are x_0, y_0 and z_0?
I used r⋅n = n ⋅ (1,-1,4sqrt(2))
to find my previous answer.

r being (x,y,z)
 
  • #10
So you do not know what x_0, y_0, and z_0 are? Then you need to talk to your teacher.
 
  • #11
HallsofIvy said:
So you do not know what x_0, y_0, and z_0 are? Then you need to talk to your teacher.
HallsofIvy said:
So you do not know what x_0, y_0, and z_0 are? Then you need to talk to your teacher.
they are the point given, (1,-1,4sqrt(2))
 
  • #12
Yes. I asked that in post #8 and you did not answer. Now, my question is "why do you not have (x- 1), (y+ 1), and (z- 4\sqrt{2}) in your answer"?
 
  • #13
upload_2015-10-29_21-49-44-png.91054.png


that was what I initially had
 
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