Hello, Harryhit4!

$\displaystyle \text{Ship A is traveling due west toward Blunt Rock at a speed of 15 kph.}$

$\displaystyle \text{Ship B is traveling due north away from Blunt Rock at a speed of 10 kph.}$

$\displaystyle \text{Let }x\text{ be the distance between Ship A and Blunt Rock at time }t.$

$\displaystyle \text{Let }y\text{ be the distance between Ship B and Blunt Rock at time }t.$

$\displaystyle \text{1. Find the distance, in km, between Ship A and Ship B}$

. . . $\displaystyle \text{when }x=4\text{ and }y=3.$

$\displaystyle \text{(seems simple, use Pythagorus to get hypotenuse = 5 km)}$ .Good!

$\displaystyle \text{2. Find the rate of change, in kph, of the distance between the ships}$

. . . $\displaystyle \text{when }x=4 \text{ and }y=3.$

$\displaystyle \text{(Think I got this, using implicit on Pythagorus I get: -6 kph)}$ .Yes!

$\displaystyle \text{3. Let }\theta\text{ be the angle at ship A, between the }x\text{-axis and hypotenuse.}$

$\displaystyle \text{Find the rate of change of }\theta\text{, in radians/hr, when }x =4\text{ and }y=3.$ Code:

B *
: | *
: | *
y | *
: | *
: | @ *
R *-----------------*
: - - - x - - - A

Blunt Rock is at $\displaystyle \,R.$

$\displaystyle \text{Ship A is at }\,A\!:\;x = AR,\;\dfrac{dx}{dt} = \text{-}6\text{ kph}$

$\displaystyle \text{Ship B is at }\,B\!:\;y = BR,\;\dfrac{dy}{dt} = 10\text{ kph}$

$\displaystyle \,\theta \,=\,\angle A$

Note: .when $\displaystyle x = 4,\;y = 3\!:\;\cos\theta \:=\:\frac{4}{5}$

$\displaystyle \text{We have: }\;\tan\theta \:=\:\dfrac{y}{x} $

$\displaystyle \text{Then: }\;\sec^2\!\theta\,\dfrac{d\theta}{dt} \;=\;\dfrac{x\frac{dy}{dt} - y \frac{dx}{dt}}{x^2}$

. . . . . . . . . .$\displaystyle \dfrac{d\theta}{dt} \;=\;\dfrac{x\frac{dy}{dt} - y\frac{x}{dt}}{x^2}\cos^2\!\theta $

. . . . . . . . . .$\displaystyle \dfrac{d\theta}{dt} \;=\;\dfrac{4(10) - 3(\text{-}15)}{4^2}\cdot \left(\frac{4}{5}\right)^2 $

. . . . . . . . . .$\displaystyle \displaystyle \frac{d\theta}{dt} \;=\;\frac{85}{16}\cdot\frac{16}{25} \;=\;\frac{17}{5} \;=\;3\tfrac{2}{5}\text{ radians/hour}$