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: Establish a right - triangle relationship

Let the length of the rope be $l$, the horizontal distance between the dinghy and the dock be $x$. By the Pythagorean theorem, $x^{2}+8^{2}=l^{2}$.

Step2: Differentiate the equation with respect to time $t$

Differentiating both sides of $x^{2}+64 = l^{2}$ with respect to $t$ gives $2x\frac{dx}{dt}=2l\frac{dl}{dt}$, so $\frac{dx}{dt}=\frac{l}{x}\cdot\frac{dl}{dt}$.

Step3: Find $x$ when $l = 10$

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

Step4: Calculate $\frac{dx}{dt}$

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

Step5: Establish a trigonometric relationship for part (b)

We know that $\sin\theta=\frac{8}{l}$.

Step6: Differentiate the trigonometric equation with respect to time $t$

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

Step7: Find $\cos\theta$ when $l = 10$

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

Step8: Calculate $\frac{d\theta}{dt}$

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

Answer:

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