A lighthouse is .8 miles away from shore. The light rotates 6 times a minute. How fast in MPH is the spot of light moving along the shore when it is .6 miles away?

So I did the problem and I'm not sure but I think I got 1/.6. is that right?

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- Dec 16th 2012, 10:32 PMcamjensonRelated Rate
A lighthouse is .8 miles away from shore. The light rotates 6 times a minute. How fast in MPH is the spot of light moving along the shore when it is .6 miles away?

So I did the problem and I'm not sure but I think I got 1/.6. is that right? - Dec 16th 2012, 11:41 PMMarkFLRe: Related Rate
Let's let $\displaystyle D$ be the perpendicular distance from the lighthouse to the shore (a straight line) and $\displaystyle x$ be the distance of the spot of light from the point on the shore intersected by the distance line segment. Let $\displaystyle \theta$ be the angle subtended by the distance line segment and the beam of light. Let $\displaystyle \omega=\frac{d\theta}{dt}$. We may then state:

$\displaystyle \tan(\theta)=\frac{x}{D}$

Differentiating with respect to time $\displaystyle t$, we find:

$\displaystyle \omega\sec^2(\theta)=\frac{1}{D}\cdot\frac{dx}{dt}$

Solving for $\displaystyle \frac{dx}{dt}$ and using $\displaystyle \sec(\theta)=\frac{\sqrt{x^2+D^2}}{D}$ we have:

$\displaystyle \frac{dx}{dt}=\frac{\omega(x^2+D^2)}{D}$

For this problem, we are given:

$\displaystyle \omega=6\cdot2\pi\frac{\text{rad}}{\text{min}} \cdot\frac{60\text{ min}}{1\text{ hr}}=720\pi\,\frac{1}{\text{hr}}$

$\displaystyle x=\frac{3}{5}\,\text{mi}$

$\displaystyle D=\frac{4}{5}\,\text{mi}$

Now, plug 'n' chug.