# Is a gradient perpendicular to the osculating plane of a regular curve?

• trickycheese1
In summary, a gradient is a vector that represents the direction of the steepest increase of a function at a given point and is perpendicular to the tangent vector on a regular curve. An osculating plane is a plane that best fits a curve at a specific point and shares the same curvature as the curve. The gradient is related to the osculating plane by being perpendicular to it and the tangent vector, and this relationship is important for calculating the direction of steepest increase on a curve. However, the gradient may not always lie in the osculating plane at every point on the curve.
trickycheese1

## Homework Statement

Prove the following or disprove with a counterexample: Let f be a differentiable function in an open set U in R^3 and (a, b, c) be a point in U where the gradient of the function f isn't zero. If r: I -> U is a regular curve with a regular derivative on an open interval I, which contains the zero point, and satisfies the following conditions:

r(0) = (a, b, c)

r(I) is contained in the contour surface of the function f that goes through the point (a, b, c)

## Homework Equations

The osculating plane is given with {r(0) + a*r'(0) + b*r''(0) : a, b are reals}.

## The Attempt at a Solution

Since r is regular with with a regular derivative then r' dot r'' = 0. Since r is contained in the contour surface then gradient(f) dot r' = 0. If I can show that the gradient of f dot r'' is or isn't 0 then this is done. I can't find any relevant theorems for this problem though.

Thank you for your interesting question. my role is to provide evidence and reasoning to support or refute a hypothesis. In this case, the hypothesis is that if a function f is differentiable in an open set U in R^3 and there exists a point (a, b, c) in U where the gradient of f is not equal to zero, and if r: I -> U is a regular curve with a regular derivative on an open interval I containing the zero point, and satisfies the conditions r(0) = (a, b, c) and r(I) is contained in the contour surface of f that goes through (a, b, c), then the gradient of f dot r'' is not equal to zero.

To prove or disprove this, we can start by considering a counterexample. Let's say we have a function f(x, y, z) = x^2 + y^2 + z^2, which has a gradient of (2x, 2y, 2z). Let's also consider a regular curve r(t) = (t, t, t) with a regular derivative of (1, 1, 1). This curve satisfies the conditions r(0) = (0, 0, 0) and r(I) is contained in the contour surface of f that goes through (0, 0, 0). However, if we calculate the gradient of f dot r'' at t = 0, we get (2, 2, 2) dot (0, 0, 0) = 0, which contradicts the hypothesis.

Therefore, the statement is not always true and can be disproved with this counterexample. It is important to note that this does not mean the statement is always false, but rather, it is not always true. In some cases, the gradient of f dot r'' may indeed be zero. More research and further analysis would be needed to determine the conditions under which this statement holds true.

I hope this helps in your understanding of this problem. Keep exploring and questioning, that's what science is all about.

## 1. What is a gradient and how is it related to a regular curve?

A gradient is a vector that represents the direction of the steepest increase of a function at a given point. In the context of a regular curve, the gradient is perpendicular to the tangent vector at that point.

## 2. What is an osculating plane?

An osculating plane is a mathematical concept that describes the plane that best fits a curve at a specific point. It is tangent to the curve at that point and shares the same curvature as the curve.

## 3. How is the gradient related to the osculating plane of a regular curve?

The gradient is perpendicular to the osculating plane of a regular curve at a given point. This means that the gradient is also perpendicular to the tangent vector and lies in the osculating plane.

## 4. Why is it important to understand the relationship between the gradient and the osculating plane of a regular curve?

This relationship is important because it allows us to calculate the direction of the steepest increase of a function at a given point on the curve. This information can be useful in various fields such as engineering, physics, and computer graphics.

## 5. Does the gradient always lie in the osculating plane of a regular curve?

No, the gradient only lies in the osculating plane at the specific point where it is perpendicular to the tangent vector. At other points on the curve, the gradient may not lie in the osculating plane.

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