Tangent Plane Equation for Surface x^2 + y^2 - xyz = 1 at Point (2,3,2)

In summary, the conversation is about finding the equation of the tangent plane at a given point on a surface. The surface can be viewed as the graph of a function of two variables or as a level surface for a function of three variables. The provided solutions for both perspectives are the same, but one is being marked as incorrect on a web assign problem. Despite this discrepancy, it is believed that the provided solutions are correct.
  • #1
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Homework Statement



Find the equation of the tangent plane at (2,3,2) to the surface below.
x^2 + y^2 - xyz = 1

The question asks me to do this in two way, one is to view the surface as the graph of a function of two variables z = g(x,y). and the other one is to view the surface as a level surface for a function f(x,y z).

Homework Equations





The Attempt at a Solution



For the first part, I already got an answer of z = (-x+y+5)/3, and so the answer to the second one I assume is x-y+3z-5 = 0. but why is it wrong? am I doing something wrong here?
 
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  • #2
z = (-x+y+5)/3 and x-y+3z-5 = 0 are the same plane. How can one be right and the other one be wrong?
 
  • #3
That's why I am confused as well... I thought I misunderstand the question for a reason... as one asks to answer it to view the surface as the graph of a func of two variables and the other one as three...
 
  • #4
One is probably suggesting you use a cross product and the other to use a gradient to find the normal. But they should both give you the same answer. And they do. Why do you think it's wrong?
 
  • #5
I don't think it's wrong, I think it's correct, but it's a web assign problem and when I submit the answer above it marks it as wrong. However the z = (-x+y+5)/3 is accepted
 
  • #6
That's web assign's problem. Who do you believe? I think you are correct.
 
  • #7
ok then... hoping for some more inputs
 

1. What is the equation for a tangent plane?

The equation for a tangent plane is given by:
x(a-x) + y(b-y) + z(c-z) = 0
where (a,b,c) is a point on the plane and (x,y,z) is any point on the plane.

2. How is the equation for a tangent plane derived?

The equation for a tangent plane is derived using the partial derivatives of a function. The normal vector of the plane is given by the gradient of the function at the point (a,b,c). The equation of the plane is then found by setting the dot product of the normal vector and the position vector (x-a, y-b, z-c) equal to 0.

3. Can the equation for a tangent plane be used to find the slope of a curve?

Yes, the equation for a tangent plane can be used to find the slope of a curve. The slope of the curve at a point is given by the gradient of the function at that point, which is the same as the normal vector of the tangent plane at that point.

4. What is the significance of the tangent plane in calculus?

The tangent plane is significant in calculus as it helps in understanding the behavior of a function at a specific point. It also allows us to approximate the function at that point and find the slope of the curve at that point. The tangent plane is also used in optimization problems and to find critical points of a function.

5. How is the equation for a tangent plane related to the concept of differentiability?

The equation for a tangent plane is closely related to the concept of differentiability. A function is said to be differentiable at a point if there exists a tangent plane at that point. The equation for the tangent plane gives us a way to check if a function is differentiable at a given point by finding the normal vector and checking if it is continuous at that point.

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