an angler hooks a trout and reels in his line at 4 in/s. assume the tip of the fishing rod is 11 ft above…

Explanation:

Step1: Establish a right - triangle relationship

Let $y = 11$ ft (height of the fishing rod tip above water), $x$ be the horizontal distance of the fish from the angler, and $z$ be the length of the fishing line. By the Pythagorean theorem, $x^{2}+y^{2}=z^{2}$. Since $y = 11$, we have $x^{2}+121=z^{2}$.

Step2: Differentiate with respect to time $t$

Differentiating both sides of $x^{2}+121 = z^{2}$ with respect to $t$ gives $2x\frac{dx}{dt}=2z\frac{dz}{dt}$, which simplifies to $x\frac{dx}{dt}=z\frac{dz}{dt}$.

Step3: Find $z$ when $x = 16$

When $x = 16$, using $x^{2}+121=z^{2}$, we substitute $x$: $z=\sqrt{16^{2}+121}=\sqrt{256 + 121}=\sqrt{377}\approx19.42$.

Step4: Determine $\frac{dz}{dt}$

The angler reels in the line at a rate of $4$ in/s. Convert to ft/s: $\frac{dz}{dt}=-\frac{4}{12}=-\frac{1}{3}$ ft/s (negative because $z$ is decreasing).

Step5: Solve for $\frac{dx}{dt}$

Substitute $x = 16$, $z\approx19.42$, and $\frac{dz}{dt}=-\frac{1}{3}$ into $x\frac{dx}{dt}=z\frac{dz}{dt}$. We get $16\frac{dx}{dt}=19.42\times(-\frac{1}{3})$. Then $\frac{dx}{dt}=\frac{19.42\times(-\frac{1}{3})}{16}\approx - 0.40$ ft/s. The negative sign indicates the fish is moving towards the angler.

Answer:

$0.40$ ft/s