22. divergent paths two boats leave a port at the same time; one travels west at 20 mi/hr and the other…

Explanation:

Step1: Calculate distance for part a

Use the formula (d = vt). For the first boat (west - traveling), (v = 20) mi/hr and (t=\frac{30}{60}=0.5) hr. So (d_1=20\times0.5 = 10) mi. For the second boat (south - traveling), (v = 15) mi/hr and (t = 0.5) hr. So (d_2=15\times0.5=7.5) mi.

Step2: Set up related - rates for part b

Let (x) be the distance of the west - traveling boat from the port, (y) be the distance of the south - traveling boat from the port, and (z) be the distance between the two boats. By the Pythagorean theorem (z^{2}=x^{2}+y^{2}). Differentiate both sides with respect to time (t): (2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}), so (\frac{dz}{dt}=\frac{x\frac{dx}{dt}+y\frac{dy}{dt}}{z}). We know (\frac{dx}{dt}=20) mi/hr, (\frac{dy}{dt}=15) mi/hr. From part a, (x = 10) mi, (y = 7.5) mi. Then (z=\sqrt{10^{2}+7.5^{2}}=\sqrt{100 + 56.25}=\sqrt{156.25}=12.5) mi. Substitute (x = 10), (y = 7.5), (z = 12.5), (\frac{dx}{dt}=20), (\frac{dy}{dt}=15) into (\frac{dz}{dt}) formula: (\frac{dz}{dt}=\frac{10\times20+7.5\times15}{12.5}=\frac{200 + 112.5}{12.5}=\frac{312.5}{12.5}=25) mi/hr.

Answer:

a. The west - traveling boat is (10) mi from the port and the south - traveling boat is (7.5) mi from the port. b. The rate at which the distance between the boats is changing is (25) mi/hr.