## Calculus 8th Edition

(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.
(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 $t$.