Is a gradient perpendicular to the osculating plane of a regular 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.

 

[mighty2000]

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.