Find the slope of the tangent line to the unit circle at the point $left( frac{1}{2}, frac{sqrt{3}}{2}
ight)$.

To find the slope of the tangent line to the unit circle at a given point, we need to differentiate the equation of the unit circle. The unit circle is defined by:

$x^2 + y^2 = 1.$

We implicitly differentiate with respect to $x$:

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

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

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

Next, we substitute the point $\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)$ into the equation:

$\frac{dy}{dx} = -\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}} = -\frac{\sqrt{3}}{3}.$

Therefore, the slope of the tangent line at the point $\left( \frac{1}{2}, \frac{\sqrt{3}}{2} \right)$ is $-\frac{\sqrt{3}}{3}$.

First, recall the equation of the unit circle:

$x^2 + y^2 = 1.$

To find the slope of the tangent line, we differentiate implicitly:

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

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

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

Inserting the coordinates $left( frac{1}{2}, frac{sqrt{3}}{2}
ight)$, we find:

$frac{dy}{dx} = -frac{frac{1}{2}}{frac{sqrt{3}}{2}} = -frac{1}{sqrt{3}} = -frac{sqrt{3}}{3}.$

Thus, the slope of the tangent line at the point $left( frac{1}{2}, frac{sqrt{3}}{2}
ight)$ is $-frac{sqrt{3}}{3}$.

Given the unit circle:

$x^2 + y^2 = 1,$

differentiate implicitly:

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

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

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

For the point $left( frac{1}{2}, frac{sqrt{3}}{2}
ight)$:

$frac{dy}{dx} = -frac{frac{1}{2}}{frac{sqrt{3}}{2}} = -frac{1}{sqrt{3}} = -frac{sqrt{3}}{3}.$

The slope of the tangent line at $left( frac{1}{2}, frac{sqrt{3}}{2}
ight)$ is $-frac{sqrt{3}}{3}$.

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