Posted by **Tori** on Saturday, March 5, 2011 at 4:36pm.

Consider the illustration, which shows a rotating beam of light located 0.5 mile from a shoreline. The beam rotates at a rate of 4 revolutions per minute. How fast (in miles per minute) is the distance between the beam and the point where it strikes the shore changing at the instant when x = .25 miles?

- calculus -
**Reiny**, Saturday, March 5, 2011 at 6:14pm
Believe it or not, my crystal ball is working to today and I can "see" your illustration.

let the angle formed be Ø, then

dØ/dt = 4(2π) or 8π rad/min

tan Ø = x/(1/2)

tanØ = 2x

sec^2 Ø dØ/dt = 2 dx/dt

dx/dt = (sec^2 Ø)(dØ/dt)/2

when x = 2.5, tan Ø = 2.5/.5 = 5/1

then cosØ = 1/√26

secØ = (3x^4 + 6x^2 - 48x + 8)26

sec^2 Ø = 26

dx/dt = (26)(8π)/2 = 1004π miles/min

- calculus -
**Reiny**, Saturday, March 5, 2011 at 6:17pm
third last line should have been

sec Ø = √26

(on my Mac, I create √ by pressing "option" V, but I must have pressed "command" V, giving me a "cut-and -paste" of some previous post)

- calculus -
**H H Chau**, Friday, August 22, 2014 at 6:19pm
tan(Ø)=x/0.5

cos(Ø)=0.5/sqrt(x^2+0.5^5)

sec^2(Ø)=(x^2+0.5^2)

dØ/dt=4 * 2π = 8π rad/min

x=0.5 tan(Ø)

dx/dt=0.5 sec^2(Ø) dØ/dt

dx/dt=(x^2+0.5^2)/0.5 * 8π

when x=0.25 miles

dx/dt=(0.25^2+0.5^2)/0.5 * 8π = 5π = 15.7 miles/min

(note to about attempt: if it were x=2.5, then dx/dt=104π miles/min)

- calculus -
**H H Chau**, Friday, August 22, 2014 at 6:21pm
Oops! Third line should be:

ec^2(Ø)=(x^2+0.5^2)/0.5^2

- calculus -
**H H Chau**, Friday, August 22, 2014 at 6:23pm
Um ...

sec^2(Ø)=(x^2+0.5^2)/0.5^2

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