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 March 7th, 2013, 09:45 AM #1 Newbie   Joined: Dec 2012 Posts: 10 Thanks: 0 a problem on a moving searchlight beam; implicit diff. PROBLEM Moving searchlight beam The figure shows a boat 1 km offshore, sweeping the shore with a searchlight. The light turns at a constant rate, $\frac{d\theta}{dt}=-0.6 \frac{rad}{sec}$ a. How fast is the light moving along the shore when it reaches point A? b. How many revolutions per minute is 0.6 rad/sec? (There is a figure provided. It shows a right triangle with it's hypothenuse facing the left with points BCA reading clockwise. Point B represents the location of the boat. Point C and A are along the shore. There is an angle $\theta$ attributed to point B. The hypothenuse represents the searchlight. Line BA equals 1 km.) ATTEMPT So, we are tasked to find $\frac{dCA}{dt}$, which represents the speed of the searchlight moving along the shore. Part A asks us to find the speed "when it reaches point A", but the speed should be consistent throughout because $\frac{d\theta}{dt}$ is consistent (?). I look for a relation to describe $\frac{dCA}{dt}=f(?)$ and felt that $tan\theta=\frac{CB}{BA}$ was sufficient. When differentiated, we arrive with the form: $\frac{dtan\theta}{dt}=\frac{d}{dt}(\frac{CB}{BA})$ $sec^2(\theta)\frac{d\theta}{dt}=\frac{BA\frac{dCB} {dt}-CB\frac{dBA}{dt}}{(BA)^2}$ Rearrangement to solve for $\frac{dCB}{dt}$, $\frac{dCA}{dt}=\frac{(BA)^2sec^2(\theta)\frac{d\th eta}{dt}+CB\frac{dBA}{dt}}{BA}$ Now, the problem is that I do not know the value for $CB$ and $\theta$, and when I seek to evaluate these variables; it turns out that the triangle cannot be fully described with just an angle and a side. Is there something I'm overlooking?
 March 7th, 2013, 10:11 AM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: a problem on a moving searchlight beam; implicit diff. You may find these topics helpful: http://www.mymathforum.com/viewtopic...thouse#p129431 http://www.mathhelpboards.com/f52/ka...ed-rates-3395/ You asked if the speed of the beam along the shore should be constant...can you see that this would only be true if the shoreline was circular, and the beam was located at the center of the circle?
March 7th, 2013, 02:14 PM   #3
Math Team

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From: Lexington, MA

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Re: a problem on a moving searchlight beam; implicit diff.

Hello, "Lambin!

Your work is correct, but some of it is unnecessary.

Quote:
 Moving searchlight beam The figure shows a boat 1 km offshore, sweeping the shore with a searchlight. The light turns at a constant rate, $\frac{d\theta}{dt}\,=\,-0.6 \frac{rad}{sec}$ a. How fast is the light moving along the shore when it reaches point A?

Code:
                            * B
* @|
*     |
*         | 1
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C * - - - - - - - - - - * A
x
$\text{W\!e have: }\:\tan\theta \,=\,\frac{x}{1} \;\;\;\Rightarrow\;\;\;x \:=\:\tan\theta$

$\text{Differentiate with respect to time: }\:\frac{dx}{dt} \:=\:\sec^{^2}\theta\,\frac{d\theta}{dt}$

$\text{If the light is at point }A\text{, we have: }\,\theta= 0,\;x = 0,\;\frac{d\theta}{dt} = -0.6$

$\text{Therefore: }\:\frac{dx}{dt} \:=\\sec^{^2}0)(-0.6) \:=\:-0.6\text{ km/sec}" />

Quote:
 b. How many revolutions per minute is 0.6 rad/sec?

$\frac{0.6\text{ rad}}{\text{1 sec}} \,\times\, \frac{\text{1 rev}}{2\pi\text{ rad}} \,\times\,\frac{\text{60 sec}}{\text{1 min}} \;=\; \frac{18}{\pi}\text{ rev/min} \;\approx\;5.73\text{ rpm.}$

 Tags beam, diff, implicit, moving, problem, searchlight

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# a search light from a light house 2km offshore

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