▶️ Answer/Explanation
Solution
(a) Distance between ships:
Using Pythagorean theorem: \[ \text{Distance} = \sqrt{x^2 + y^2} = \sqrt{4^2 + 3^2} = 5 \text{ km} \]
(b) Rate of change of distance:
1. Relate variables: \[ r^2 = x^2 + y^2 \]
2. Differentiate with respect to t: \[ 2r\frac{dr}{dt} = 2x\frac{dx}{dt} + 2y\frac{dy}{dt} \]
3. Substitute known values (note: \(\frac{dx}{dt} = -15\) km/hr, \(\frac{dy}{dt} = 10\) km/hr): \[ \frac{dr}{dt} = \frac{x\frac{dx}{dt} + y\frac{dy}{dt}}{r} = \frac{4(-15) + 3(10)}{5} = -6 \text{ km/hr} \]
(c) Rate of change of angle θ:
1. Relate angle to distances: \[ \tan θ = \frac{y}{x} \]
2. Differentiate with respect to t: \[ \sec^2θ \frac{dθ}{dt} = \frac{x\frac{dy}{dt} – y\frac{dx}{dt}}{x^2} \]
3. When x = 4, y = 3: \[ \sec θ = \frac{r}{x} = \frac{5}{4} \]
4. Solve for \(\frac{dθ}{dt}\): \[ \frac{dθ}{dt} = \cos^2θ \left( \frac{4(10) – 3(-15)}{16} \right) = \left(\frac{16}{25}\right)\left(\frac{85}{16}\right) = \frac{17}{5} \text{ rad/hr} \]