Answer
(a) $s=\sqrt{d^2+36}$
(b) $d=30t$
(c) $(f \circ g)(t) = \sqrt{900t^2+36}$.
This function represents the distance between the lighthouse and the ship as a function of the time elapsed since noon.
Work Step by Step
(a) We can find the distance $s$ using the Pythagorean theorem, since the distance the ship has traveled since noon and its 6 km distance from shore form two sides of a triangle, where $s$ is the hypotenuse. Thus:
$s^2 = 6^2 + d^2$
$s = \sqrt{6^2 + d^2} = \sqrt{d^2+ 36}$
(b) Since speed is distance travelled divided by travel time, we know that:
$30 = \frac{d}{t}$
Which can be rearranged to:
$d = 30t$
(c) $(f \circ g)(t) = \sqrt{(30t)^2+36} = \sqrt{900t^2+36}$
We know $f(t)$ is the distance between the lighthouse and the ship as a function of $d$, the distance the ship has travelled since noon, and $d$ is a function of $t$, the time elapsed since noon.
Thus, we can conclude that the function represents the distance between the lighthouse $s$ and the ship as a function of the time elapsed since noon $t$.
