two boats leave a dock at the same time. one boat travels south at 16 mi/hr and the other travels east at 12…

Explanation:

Step1: Define variables

Let $x$ be the distance of the boat moving east, $y$ be the distance of the boat moving south, and $z$ be the distance between the two boats. By the Pythagorean - theorem, $z^{2}=x^{2}+y^{2}$.

Step2: Differentiate with respect to time $t$

Differentiating both sides of $z^{2}=x^{2}+y^{2}$ with respect to $t$, we get $2z\frac{dz}{dt}=2x\frac{dx}{dt}+2y\frac{dy}{dt}$, which simplifies to $z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$.

Step3: Calculate $x$, $y$, and $z$ after half - hour

The boat moving east has a speed $\frac{dx}{dt}=12$ mi/hr, so after $t = 0.5$ hr, $x=\frac{dx}{dt}\times t=12\times0.5 = 6$ miles. The boat moving south has a speed $\frac{dy}{dt}=16$ mi/hr, so after $t = 0.5$ hr, $y=\frac{dy}{dt}\times t=16\times0.5 = 8$ miles. Then, by the Pythagorean theorem, $z=\sqrt{x^{2}+y^{2}}=\sqrt{6^{2}+8^{2}}=\sqrt{36 + 64}=\sqrt{100}=10$ miles.

Step4: Solve for $\frac{dz}{dt}$

We know that $\frac{dx}{dt}=12$ mi/hr, $\frac{dy}{dt}=16$ mi/hr, $x = 6$ miles, $y = 8$ miles, and $z = 10$ miles. Substitute these values into $z\frac{dz}{dt}=x\frac{dx}{dt}+y\frac{dy}{dt}$: $10\frac{dz}{dt}=6\times12 + 8\times16$. $10\frac{dz}{dt}=72+128$. $10\frac{dz}{dt}=200$. $\frac{dz}{dt}=20$ mi/hr.

Answer:

20