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Mathematics 22 Online
OpenStudy (anonymous):

RELATED RATES QUESTION!! A rotating beacon is located 4 miles out in the water. Let A be the point on the shore that is closest to the beacon. As the beacon rotates at 8 rev/min, the beam of light sweeps down the shore once each time it revolves. Assume that the shore is straight. How fast is the point where the beam hits the shore moving along the shore at an instant when the beam is lighting up a point 4 miles along the shore from the point A?

OpenStudy (anonymous):

Hint: Because this is rotational motion the velocity will be the (8)circumference. Multiply this by 60 if you want this in miles per hour. Now for the related rate. The distance from the beacon to the shore changes as it rotates therefore the velocity changes. How is the radius going to change as this beam rotates. You can find this by using the Pythagorean formula. I hope this helps.

OpenStudy (anonymous):

actually... there's an easier way to do this:

OpenStudy (anonymous):

|dw:1350526714463:dw|

OpenStudy (anonymous):

differentiate with respect to time, find theta when x=4, plug that and d(theta)/dt (in rad.s/sec) into the expression for dx/dt

OpenStudy (anonymous):

any questions?

OpenStudy (anonymous):

i'm not getting it

OpenStudy (anonymous):

which part?

OpenStudy (anonymous):

the sketch? the differentiation? finding and plugging in theta and d(theta)/dt?

OpenStudy (anonymous):

making the equation and differentiating it i'm not sure where theta comes in, is it the rev/ min?

OpenStudy (anonymous):

but they don't give you a theta so you can't derive it with theta

OpenStudy (anonymous):

x= 4 tan(theta) is the equation... it's just trig.

OpenStudy (anonymous):

theta we can find ... (but don't use it until *after* you do the differentiation) ...because they ask for dx/dt when x=4

OpenStudy (anonymous):

so i get dx/dt = (dtheta,dt)4tan(theta)

OpenStudy (anonymous):

*dtheta/dt

OpenStudy (anonymous):

use the chain rule theta is a function of time: (x= 4tan( f(t) ) )'

OpenStudy (anonymous):

what's the derivative of tan(x)

OpenStudy (anonymous):

sec^2x

OpenStudy (anonymous):

yes. so what's the derivative of tan( f(x) ) ?

OpenStudy (anonymous):

sec^2 ( f(x) ) * f ' (x)

OpenStudy (anonymous):

so sec^ 2 (theta) * d(theta)/dt (whole thing times 4)

OpenStudy (anonymous):

dx/dt = 4*sec^2(theta) * d(theta)/dt

OpenStudy (anonymous):

now we need to find theta when x=4 and we need to find d(theta)/dt in appropriate units.

OpenStudy (anonymous):

what's theta going to be (from the sketch) if x=4 ?

OpenStudy (anonymous):

sqrt. 2/ 2

OpenStudy (anonymous):

|dw:1350528393429:dw|

OpenStudy (anonymous):

45 degrees

OpenStudy (anonymous):

:)

OpenStudy (anonymous):

now for d(theta)/dt

OpenStudy (anonymous):

you want your answer in miles per hour?

OpenStudy (anonymous):

if you want your answer in miles per hour, convert the RPM to rad.s/hour...

OpenStudy (anonymous):

hello?

OpenStudy (anonymous):

480rev/hr

OpenStudy (anonymous):

so 480= 4 (pi/4)^2 d(theta)/dt

OpenStudy (anonymous):

480 rev/hour * 2pi rad / rev = 960 pi rad.s / hour = d(theta) /dt

OpenStudy (anonymous):

so dx/dt =4*sec^2(45) *960pi

OpenStudy (anonymous):

the answer is supposed in rads/ min

OpenStudy (anonymous):

I just asked that...

OpenStudy (anonymous):

so why look for dx/dt

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