Determine the coordinates of points on the unit circle where the tangent line is horizontal.

To find the coordinates on the unit circle where the tangent line is horizontal, we first recall that the unit circle is defined by the equation:

$ x^2 + y^2 = 1 $

The slope of the tangent line to the circle at any point (x, y) is given by the derivative of y with respect to x. Differentiating implicitly, we get:

$ 2x + 2y \x0crac{dy}{dx} = 0 $

Solving for $\x0crac{dy}{dx}$, we find:

$ \x0crac{dy}{dx} = -\x0crac{x}{y} $

For the tangent line to be horizontal, the slope $\x0crac{dy}{dx}$ must be zero. This occurs when:

$ -\x0crac{x}{y} = 0 $

Thus, x must be zero. On the unit circle, the points with x = 0 are (0, 1) and (0, -1). Therefore, the coordinates are (0, 1) and (0, -1).

To find the coordinates on the unit circle where the tangent line is horizontal, consider the unit circle equation:

$ x^2 + y^2 = 1 $

Differentiating implicitly, we get:

$ 2x + 2y x0crac{dy}{dx} = 0 $

Solving for $x0crac{dy}{dx}$ gives us:

$ x0crac{dy}{dx} = -x0crac{x}{y} $

For a horizontal tangent line, $x0crac{dy}{dx} = 0$, which implies x = 0. The solutions on the unit circle are (0, 1) and (0, -1).

For horizontal tangents, $x0crac{dy}{dx} = 0$. On the unit circle, this occurs at (0, 1) and (0, -1).

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