A dinghy is pulled toward a dock by a rope from the bow through a ring on the dock 6 ft. above the bow. The rope is hauled in a rate of 2 ft/s.
At what rate is the angle $\theta$ changing when 10 ft. of rope is out?
I understand related rates problems, but the trig and angle part of the question is confusing me. How do I solve this?
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$\begingroup$You have the relation $\cos\theta=\dfrac{6}{r}$ so $-\sin(\theta)\dfrac{d\theta}{dt}=-\dfrac{6}{r^2}\dfrac{dr}{dt}$, or
\begin{equation} \sin(\theta)\frac{d\theta}{dt}=\frac{6}{r^2}\frac{dr}{dt} \end{equation}
When $r=10$ we know that $\dfrac{dr}{dt}=-2$ and $\sin(\theta)=\dfrac{8}{10}=\dfrac{4}{5}$ so
\begin{equation} \dfrac{d\theta}{dt}=\frac{5}{4}\cdot\frac{6}{100}(-2)=-0.15 \text{ per seconds} \end{equation}
Corrected as per comment by @TonyK
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