6. a fish is being reeled in at a rate of 30 cm/s from a bridge 4 m above water. at what rate is the angle…

Explanation:

Step1: Define variables

Let $y = 4$ m (constant height of bridge), $l$ be the length of the line, and $\theta$ be the angle between the line and the water. We know $\sin\theta=\frac{y}{l}$.

Step2: Differentiate with respect to time $t$

Differentiating $\sin\theta=\frac{y}{l}$ with respect to $t$ using the chain - rule. We get $\cos\theta\frac{d\theta}{dt}=-\frac{y}{l^{2}}\frac{dl}{dt}$.

Step3: Find $\cos\theta$ when $l = 8$ m

Since $y = 4$ m and $l = 8$ m, and $\sin\theta=\frac{y}{l}=\frac{4}{8}=\frac{1}{2}$, then $\cos\theta=\sqrt{1 - \sin^{2}\theta}=\sqrt{1-\left(\frac{1}{2}\right)^{2}}=\frac{\sqrt{3}}{2}$. Also, $\frac{dl}{dt}=- 30$ cm/s=-0.3 m/s.

Step4: Solve for $\frac{d\theta}{dt}$

Substitute $y = 4$ m, $l = 8$ m, $\cos\theta=\frac{\sqrt{3}}{2}$, and $\frac{dl}{dt}=-0.3$ m/s into $\cos\theta\frac{d\theta}{dt}=-\frac{y}{l^{2}}\frac{dl}{dt}$. $\frac{\sqrt{3}}{2}\frac{d\theta}{dt}=-\frac{4}{8^{2}}\times(-0.3)$ $\frac{\sqrt{3}}{2}\frac{d\theta}{dt}=\frac{4\times0.3}{64}$ $\frac{d\theta}{dt}=\frac{4\times0.3\times2}{64\times\sqrt{3}}=\frac{0.24}{64\sqrt{3}}=\frac{3}{800\sqrt{3}}=\frac{\sqrt{3}}{800}\text{ rad/s}$

Answer:

$\frac{\sqrt{3}}{800}\text{ rad/s}$