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?

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

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)

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)

Oops! Third line should be:

ec^2(Ø)=(x^2+0.5^2)/0.5^2

Um ...

sec^2(Ø)=(x^2+0.5^2)/0.5^2

To find the rate at which the distance between the beam of light and the point where it strikes the shoreline is changing, we can use the concept of related rates. We need to determine the derivative of the distance with respect to time.

Let's denote the distance between the beam and the point where it strikes the shore as y (in miles) and the time as t (in minutes).

We are given that the beam rotates at a rate of 4 revolutions per minute. This means that its angle of rotation, θ, changes at a rate of 4 revolutions per minute.

Since there are 2π radians in one revolution, the angle of rotation in radians, θ (in radians), changes at a rate of 4 * 2π = 8π radians per minute.

Now, we can set up a relationship between the variables x, y, and θ using trigonometry. Consider a right triangle formed by the horizontal line connecting the beam and the shoreline, the vertical line from the point where the beam strikes the shore, and the line connecting the beam and the point where it strikes the shore.

We have the following trigonometric relationship:
tan(θ) = y / x

Taking the derivative of both sides with respect to time t, we can use the chain rule to differentiate the tangent function and the variables with respect to time:

sec²(θ) * dθ/dt = (dy/dt * x - y * dx/dt) / x²

We are interested in finding dy/dt, the rate at which the distance y is changing with respect to time.

Now we can plug in the given values:
θ = 8π radians/min
x = 0.25 miles
dx/dt = ?

Remember that x is the distance from the beam to the shore and, in this case, it is given as a constant (0.25 miles), so dx/dt, the rate of change of x with respect to time, is zero.

Plugging these values into the equation:

sec²(8π) * (8π) = (dy/dt * 0.25 - y * 0) / (0.25)²

We can simplify this equation to solve for dy/dt:

dy/dt = (0.25² * sec²(8π) * (8π)) / 0.25

Finally, we can evaluate this expression to find the rate at which the distance between the beam and the point where it strikes the shore is changing at the instant when x = 0.25 miles.