a dinghy is pulled toward a dock by a rope from the bow through a ring on the dock 8 ft above the bow. the…

Explanation:

Step1: Define variables and establish relationship

Let $x$ be the horizontal distance between the dinghy and the dock, and $l$ be the length of the rope. By the Pythagorean theorem, $x^{2}+8^{2}=l^{2}$. Differentiating both sides with respect to time $t$, we get $2x\frac{dx}{dt}=2l\frac{dl}{dt}$, or $x\frac{dx}{dt}=l\frac{dl}{dt}$.

Step2: Find $x$ when $l = 10$

When $l = 10$, using $x^{2}+8^{2}=l^{2}$, we have $x=\sqrt{l^{2}-64}=\sqrt{100 - 64}=\sqrt{36}=6$.

Step3: Solve for $\frac{dx}{dt}$

We know that $\frac{dl}{dt}=- 3$ (negative because the length of the rope is decreasing). Substituting $x = 6$, $l = 10$ and $\frac{dl}{dt}=-3$ into $x\frac{dx}{dt}=l\frac{dl}{dt}$, we get $6\frac{dx}{dt}=10\times(-3)$. Then $\frac{dx}{dt}=\frac{-30}{6}=-5$ ft/sec.

Step4: Establish relationship for the angle $\theta$

We know that $\sin\theta=\frac{8}{l}$. Differentiating both sides with respect to $t$ gives $\cos\theta\frac{d\theta}{dt}=-\frac{8}{l^{2}}\frac{dl}{dt}$.

Step5: Find $\cos\theta$ when $l = 10$

When $l = 10$, $\sin\theta=\frac{8}{10}=\frac{4}{5}$, so $\cos\theta=\frac{6}{10}=\frac{3}{5}$.

Step6: Solve for $\frac{d\theta}{dt}$

Substitute $\cos\theta=\frac{3}{5}$, $l = 10$ and $\frac{dl}{dt}=-3$ into $\cos\theta\frac{d\theta}{dt}=-\frac{8}{l^{2}}\frac{dl}{dt}$. We have $\frac{3}{5}\frac{d\theta}{dt}=-\frac{8}{100}\times(-3)$. Then $\frac{3}{5}\frac{d\theta}{dt}=\frac{24}{100}$, and $\frac{d\theta}{dt}=\frac{24}{100}\times\frac{5}{3}=\frac{2}{5}$ rad/sec.

Answer:

a. -5 ft/sec b. $\frac{2}{5}$ rad/sec