1. **State the problem:** We are given a unit circle with equation $$x^2 + y^2 = 1$$ and a point on the circle with coordinates $\left(\frac{1}{3}, y\right)$. We need to find the value of $y$ at this point. 2. **Recall the formula:** The unit circle equation is $$x^2 + y^2 = 1$$ which means any point $(x,y)$ on the circle satisfies this. 3. **Substitute the known $x$ value:** Plug in $x = \frac{1}{3}$ into the equation: $$\left(\frac{1}{3}\right)^2 + y^2 = 1$$ 4. **Simplify:** $$\frac{1}{9} + y^2 = 1$$ 5. **Isolate $y^2$:** $$y^2 = 1 - \frac{1}{9}$$ 6. **Simplify the right side:** $$y^2 = \frac{9}{9} - \frac{1}{9} = \frac{8}{9}$$ 7. **Take the square root:** $$y = \pm \sqrt{\frac{8}{9}} = \pm \frac{\sqrt{8}}{3} = \pm \frac{2\sqrt{2}}{3}$$ 8. **Determine the sign of $y$:** Since the point is in the first quadrant (where both $x$ and $y$ are positive), we take the positive root: $$y = \frac{2\sqrt{2}}{3}$$ **Final answer:** $$y = \frac{2\sqrt{2}}{3}$$