Find the exact coordinates of the point(s) on the unit circle where the tangent line is vertical

The equation of the unit circle is given by:

$ x^2 + y^2 = 1 $

We find the tangent line to be vertical when the derivative is undefined. Thus, we need to find the points where $ \x0crac{dy}{dx} $ is undefined.

Implicitly differentiate the unit circle equation with respect to $ x $:

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

Simplify and solve for $ \x0crac{dy}{dx} $:

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

The derivative is undefined when $ y = 0 $. Thus, we solve for $ x $:

When $ y = 0 $, substituting back into the original equation:

$ x^2 = 1 $

So, $ x = 1 $ or $ x = -1 $.

Therefore, the points are:

$(1,0)$ and $(-1,0)$.

The equation of the unit circle is:

$ x^2 + y^2 = 1 $

We want the tangent lines to be vertical. This occurs when $ x0crac{dy}{dx} $ is undefined.

Implicitly differentiate:

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

So, $ x0crac{dy}{dx} = -x0crac{x}{y} $.

The derivative is undefined when $ y = 0 $. Substituting $ y = 0 $:

$ x^2 = 1 $

Thus, $ x = 1 $ or $ x = -1 $.

Therefore, the points are:

$(1,0)$ and $(-1,0)$.

The equation of the unit circle is:

$ x^2 + y^2 = 1 $

Tangent lines are vertical when $ x0crac{dy}{dx} $ is undefined.

Differentiate implicitly:

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

So, $ x0crac{dy}{dx} = -x0crac{x}{y} $.

Undefined when $ y = 0 $. So:

$ x^2 = 1 $

Thus, the points are:

$(1,0)$ and $(-1,0)$.

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