Question

The path of a total solar eclipse is modeled by f (t )equals 0.00223 t squared minus 0.558 t plus 41.395​, 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.

Answer 1

Answer:

Therefore the latitude closest to the equator is 6.49°(approx).

Step-by-step explanation:

Given the path of a solar ellipse is modeled by

$f(t)=0.00223t^2-0.558t+41.395$

where f(t) is the latitude in degrees south of the equator at t minutes.

We know that ,

The minimum or maximum value of a function y(t) =at²+bt+c

is when $t=-\frac{b}{2a}$

In this case a= 0.00223, b= - 0.558  and c= 41.395.

The minimum value of f(t)

when $t=-\frac{( - 0.558)}{2\times 0.00223}$

           =125.11

Therefore the latitude closest to the equator is

f(125.11)=0.00223(125.11)²-0.558×125.11+41.395

            ≈6.49