Determine the Tangent Slope at a Given Point on the Unit Circle

Let the given point on the unit circle be $(a, b)$, where $a^2 + b^2 = 1$. We need to determine the slope of the tangent line at this point.

The equation of the unit circle is given by:

$x^2 + y^2 = 1$

To find the slope of the tangent line at $(a, b)$, we first implicitly differentiate both sides of the equation with respect to $x$:

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

Solving for $\frac{dy}{dx}$:

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

Substituting the point $(a, b)$ into the derivative:

$\frac{dy}{dx}\bigg|_{(a,b)} = -\frac{a}{b}$

Therefore, the slope of the tangent line at the point $(a, b)$ is $-\frac{a}{b}$.

Let’s consider the point $(a, b)$ on the unit circle, where $a^2 + b^2 = 1$. We are to find the slope of the tangent line at this point.

The unit circle equation is:

$x^2 + y^2 = 1$

Implicit differentiation with respect to $x$ yields:

$2x + 2yfrac{dy}{dx} = 0$

Solving for $frac{dy}{dx}$, we get:

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

Evaluating at $(a, b)$:

$frac{dy}{dx}igg|_{(a,b)} = -frac{a}{b}$

Thus, the slope of the tangent line at $(a, b)$ is $-frac{a}{b}$.

Given a point $(a, b)$ on the unit circle where $a^2 + b^2 = 1$, we find the tangent slope.

The unit circle is described by:

$x^2 + y^2 = 1$

Implicit differentiation gives:

$2x + 2yfrac{dy}{dx} = 0$

Simplifying:

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

At $(a, b)$:

$frac{dy}{dx}igg|_{(a,b)} = -frac{a}{b}$

The tangent slope at $(a, b)$ is $-frac{a}{b}$.

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