# 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
1 year, 10 months ago

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To start, we require the following:

$x^2 + y^2 = 1$

A parametric solution to this is:

$x = \cos t \\ y = \sin t$

- 1 year, 10 months ago

Why would it be a solution?

- 1 year, 10 months ago

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

- 1 year, 10 months ago

How can I use this information to resolve the issue?

- 1 year, 10 months ago

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.

- 1 year, 10 months ago

Yes, this answers the second part of the question

- 1 year, 10 months ago

I want someone to correct or complement me if possible

- 1 year, 10 months ago