Find the tangent line to the unit circle at the point $(1, 0)$.

The equation of the unit circle is $x^2 + y^2 = 1$.

The point (1,0) is on the unit circle.

The slope of the tangent line at any point (a,b) on the circle can be found by implicit differentiation:

$2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = – \frac{x}{y}$

At the point (1,0), the slope is undefined (vertical line).

Therefore, the equation of the tangent line is $ x = 1 $

The unit circle is defined by $ x^2 + y^2 = 1 $.

Let’s determine the tangent line at the point (1, 0).

By differentiating $ x^2 + y^2 = 1 $, we get $ 2x + 2y frac{dy}{dx} = 0 $

Simplifying gives $ frac{dy}{dx} = – frac{x}{y} $

For the point (1, 0), the slope is undefined because we cannot divide by zero.

Therefore, the tangent line is a vertical line passing through x = 1, so the equation is $ x = 1 $

Given the unit circle $ x^2 + y^2 = 1 $,

we find the tangent line at (1, 0).

Differentiating, $ 2x + 2y frac{dy}{dx} = 0 $

$ frac{dy}{dx} = – frac{x}{y} $

At (1, 0), the slope is undefined,

thus the tangent line is vertical at $ x = 1 $

Start Using PopAi Today