The path of a total solar eclipse is modeled by $f(t)=0.00301 t^{2}-0.497 t+39.668$, where $f(t)$ is the latitude in degrees south of the equator at $t$ minutes after the start of the total eclipse. What is the latitude closest to the equator, in degrees, at which the total eclipse will be visible. The latitude closest to the equator at which the total eclipse will be visible is ${ }^{\circ} \mathrm{S}$ (Round the final answer to two decimal places as needed. Round all intermediate values to four decimal places as needed.)

Step 1 ：The latitude closest to the equator at which the total eclipse will be visible is the minimum value of the function \(f(t)\). Since \(f(t)\) is a quadratic function, its graph is a parabola.

Step 2 ：The minimum value of a parabola with a positive leading coefficient (like this one, where the leading coefficient is 0.00301) occurs at its vertex.

Step 3 ：The \(t\)-coordinate of the vertex of a parabola given by \(f(t) = at^2 + bt + c\) is \(-\frac{b}{2a}\).

Step 4 ：So, we need to substitute \(a = 0.00301\) and \(b = -0.497\) into this formula to find the time at which the minimum latitude occurs.

Step 5 ：Then we substitute this time into the function \(f(t)\) to find the minimum latitude.

Step 6 ：Final Answer: The latitude closest to the equator at which the total eclipse will be visible is \(\boxed{19.15}^{\circ} \mathrm{S}\).