Differentiable curve

Let f (x, y) = x² + y² and let λ (t) = (x (t), y (t)) be a differentiable curve whose image is contained in the level curve f (x, y) = 1, that is, for all t in the domain of λ, f (x (t), y (t)) = 1 (give an example of such a curve). Let λ (t₀) = (x₀, y₀). Prove that λ '(t₀), ∇f (x₀, y₀) = 0. Interpret geometrically.

Suggestion: for all t in the domain of λ, (x (t))²+ (y (t))² = 1; derive from t and make t = t₀)

Someone? I really need some help with this

Note by Go!Game Rj
3 months ago

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To start, we require the following:

x2+y2=1 x^2 + y^2 = 1

A parametric solution to this is:

x=costy=sintx = \cos t \\ y = \sin t

Steven Chase - 3 months ago

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Why would it be a solution?

Go!Game RJ - 3 months ago

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Because for any value of the parameter tt , the function f=1 f = 1

Steven Chase - 3 months ago

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How can I use this information to resolve the issue?

Go!Game RJ - 3 months ago

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5) Derive x (t) ^ 2 + y (t) ^ 2 = 1; so we have x (t) x '(t) + y (t) y' (t) = 0, this is λ (t) .λ '(t) = 0 (Scalar product in reais ^ 2). λ (t) describes the position of a moving object in a circle; λ '(t) = (x' (t), y '(t)) is its velocity. The velocity is tangent to the circle and the velocity and position are perpendicular.

Jefferson Nascimento - 3 months ago

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Yes, this answers the second part of the question

Steven Chase - 3 months ago

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I want someone to correct or complement me if possible

Jefferson Nascimento - 3 months ago

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