So what exactly is an osculating plane? r'(t) X r (t)? Am I right?

• Hellken
In summary, the osculating plane is a plane that is tangent to a curve at a specific point and is normal to the cross product of the first and second derivatives of the curve. It can be found by using the equation of a plane, which requires a point on the plane and a normal vector to the plane. Finding the cross product of r'(t) and r"(t) can save time compared to finding the tangent and normal vectors separately.
Hellken
So what exactly is an osculating plane? r'(t) X r"(t)? Am I right?

From what I know it is a plane that kisses a point on the curve at point t of curve f(t) and it is a plane normal to r'(t) X r"(t)? Am I right?

So to look for this osculating plane I simply have to find the cross product of r'(t) and r"(t)?

What does it look like on a xyz graph?

====================
EDIT: By the way per definition of the equation of a plane: a(x-x0) + b(y-y0 + c(z-z0) I need to find

a) a point on the plane (x0,y0,z0)
b) a vector normal to the plane <a,b,c>

Hence if I am looking for the osculating plane I would need:

a) a point on the osculating plane (x0,y0,z0)
b) B(t) which happens to be the vector normal to my osculating plane B(t) = T(t) X N(t), from this I find my desired <a,b,c>

I put it all together and I get my osculating plane of form: a(x-x0) + b(y-y0 + c(z-z0). (Right or wrong?)
==============================
So what exactly is the cross product r'(t) X r"(t) for? Cutting down time so I don't have to go through the tedious process of finding T(t) and N(t)?

Last edited:

You are right. Since the osculating plane passes through the curve at some point, you can take x0, y0, z0 as a point on the curve.
You don't have to find T and N (by the way, what do you mean by N?) to find a, b and c, but just r' and r''. What's so tedious on differentiating two times the curve?

1. What is an osculating plane?

An osculating plane is a geometric concept that refers to the unique plane that best approximates the curvature of a curve at a specific point. It is also known as the tangent plane and is perpendicular to the curve at that point.

2. How is the osculating plane calculated?

The osculating plane is calculated using the first and second derivatives of the curve at the given point. The first derivative gives the slope of the tangent line, and the second derivative gives the rate of change of the slope, which is used to determine the orientation of the osculating plane.

3. What is the significance of the osculating plane?

The osculating plane helps us understand the behavior of a curve at a specific point. It provides important information about the curvature, direction, and rate of change of the curve at that point, which is crucial in many real-world applications such as engineering, physics, and computer graphics.

4. Can the osculating plane change along the curve?

Yes, the osculating plane can change along the curve as the curvature and direction of the curve change. This means that the osculating plane is not constant and varies depending on the point on the curve.

5. How is the osculating plane related to the cross product of r'(t) and r(t)?

The osculating plane is perpendicular to the cross product of the first derivative (r'(t)) and the position vector (r(t)). This means that the osculating plane is determined by the direction of the curve (r'(t)) and the position of the point on the curve (r(t)).

• Calculus
Replies
3
Views
1K
• Calculus
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
1K
• Calculus
Replies
4
Views
2K
• Introductory Physics Homework Help
Replies
12
Views
227
• Calculus
Replies
5
Views
1K
• Calculus
Replies
3
Views
1K
• Calculus
Replies
1
Views
485
• Calculus
Replies
3
Views
1K
• Calculus
Replies
11
Views
3K