Answer
a. $s(t)=\sqrt {(8t)^2+(12-12t)^2}=4\sqrt {13t^2-18t+9}$
b. $s'(0)=-12$ knots. $s'(1)=8$ knots.
c. The ships will not sight each other.
d. See graph and explanations.
e. $4\sqrt {13}$ knots, see explanations.
Work Step by Step
a. Assume that boat-B is at the origin at $t=0$; then boat-A has coordinates of $A(0,y), y=12-12t$ and boat-B has coordinates of $B(x,0), x=8t$. The distance between the two boats is given by $s(t)=\sqrt {x^2+y^2}=\sqrt {(8t)^2+(12-12t)^2}=4\sqrt {13t^2-18t+9}$
b. Take the derivative to get $s'=\frac{2(26t-18)}{\sqrt {13t^2-18t+9}}=\frac{4(13t-9)}{\sqrt {13t^2-18t+9}}$. At noon, $s'(0)=\frac{4(13(0)-9)}{\sqrt {13(0)^2-18(0)+9}}=-12$ knots. One hour later $s'(1)=\frac{4(13(1)-9)}{\sqrt {13(1)^2-18(1)+9}}=8$ knots.
c. The minimum distance can be found when $s'=0$, which gives $t=\frac{9}{13}\approx0.69$ and $s(\frac{9}{13})=4\sqrt {13(\frac{9}{13})^2-18(\frac{9}{13})+9}=\frac{24\sqrt {13}}{13}\approx6.66$ nautical miles (this value is not a maximum as the staring point is larger at 12 nautical miles). Since $6.66\gt5$, the ships will not sight each other.
d. See graph; function $s(t)$ has a minimum at $x=0.69$ corresponding to a zero in $s'(t)$. When $t\lt0.69$, $s'(t)\lt0$ and $s(t)$ is decreasing. When $t\gt0.69$, $s'(t)\gt0$ and $s(t)$ is increasing. The graph agrees with the results above.
e. $\lim_{t\to\infty}s'(t)=\lim_{t\to\infty}\frac{4(13t-9)}{\sqrt {13t^2-18t+9}}=\lim_{t\to\infty}\frac{4(13-9/t)}{\sqrt {13-18/t+9/t^2}}=4\sqrt {13}$ knots. As $s'(t)$ represents the relative speed between the two boats, this limit should be the sum of the two individual speeds (amplitude of the sum of two velocity vectors). That is $\vec v_s=\vec v_A+\vec v_B$, $|\vec v_s|=\sqrt {12^2+8^2}=\sqrt {208}=4\sqrt {13}$.
