A searchlight rotates at a rate of 2 revolutions per minute. The beam hits a wall located 8 miles away and produces a dot of light that moves horizontally along the wall. How fast (in miles per hour) is this dot moving when the angle between the beam and the line through the searchlight perpendicular to the wall is pi/5? Note that d angle,dt=2(2pi)=4pi.
Speed of dot _______ = mph.
draw the triangle
s=Searchlight postion
8= distance wall is away.
x= distance from the perpendicular to the light to the position of the beam, so that
TanTheta=x/8
take derivative w/respect to time
d(tanTheta)=1/8 dx/dt
sec^2 theta * dtheta/dt=1/8 dx/dt
you are given theta, given dtheta/dt (4PI), find dx/dt
Well, well, well, we've got ourselves a spinning searchlight! Let's shine some math magic on this problem.
First things first, we need to find the angular speed of the beam. We're told that the searchlight rotates at a rate of 2 revolutions per minute. A revolution is just a fancy word for 2pi radians, so the angular speed is 2(2pi) radians per minute.
Now, we need to find the speed of the dot on the wall when the angle between the beam and the line through the searchlight perpendicular to the wall is pi/5.
The distance from the searchlight to the wall is 8 miles, and the dot on the wall moves horizontally, so we only need to worry about the horizontal component of the speed.
To find this speed, we'll use some trigonometry. The horizontal speed is given by the formula v = r * ω * sin(θ), where v is the speed, r is the distance from the searchlight to the wall, ω is the angular speed, and θ is the angle between the beam and the perpendicular line.
Plugging in the given values, we have v = 8 * 2(2pi) * sin(pi/5).
Now, let's calculate this speed and convert it to miles per hour.
v = 8 * 2(2pi) * sin(pi/5) = approximately 17.59 miles per minute.
To convert this to miles per hour, we multiply by 60 (since there are 60 minutes in an hour):
v (in miles per hour) ≈ 17.59 * 60 ≈ 1053.57 miles per hour.
So, the speed of the dot on the wall is approximately 1053.57 miles per hour.
Woohoo, that dot is zooming!
To find the speed of the dot, we need to find the rate at which it is moving horizontally along the wall. Let's call this rate "v".
We are given that the searchlight rotates at a rate of 2 revolutions per minute, which means it completes 2 full circles in one minute. Since there are 60 minutes in an hour, the searchlight completes 2 * 60 = 120 full circles in one hour.
The distance from the searchlight to the wall is given as 8 miles. Since the dot forms a right triangle with the wall as the hypotenuse, the horizontal distance between the searchlight and the dot can be found using trigonometry. Let's call this distance "x".
Using trigonometry, we can write the equation:
tan(pi/5) = x/8
Solving for x:
x = 8*tan(pi/5)
Now, we need to find how fast the dot is moving horizontally, so we take the derivative of x with respect to time (t), which gives us dx/dt.
dx/dt = d/dt (8*tan(pi/5))
Taking the derivative, we get:
dx/dt = 8 * sec^2(pi/5) * d(pi/5)/dt
We are given that d(pi/5)/dt = 4pi (as mentioned in the question).
Substituting the values, we get:
dx/dt = 8 * sec^2(pi/5) * 4pi
To find the speed in miles per hour, we need to convert the units from revolutions per minute to miles per hour.
There are 2 * pi * 8 miles in one revolution (since the wall is the hypotenuse of the right triangle formed by the dot and the searchlight), so the dot travels a distance of 2 * pi * 8 miles in one minute. Multiplying this by 60 (to convert minutes to hours), we get:
Distance traveled in one hour = 2 * pi * 8 * 60 miles
Finally, we can find the speed of the dot by dividing dx/dt by the distance traveled in one hour:
Speed of dot = (8 * sec^2(pi/5) * 4pi) / (2 * pi * 8 * 60) miles per hour
Simplifying this expression, we get:
Speed of dot = (sec^2(pi/5) * pi) / (15) miles per hour
Therefore, the speed of the dot is (sec^2(pi/5) * pi) / 15 miles per hour.
To find the speed of the dot on the wall, we need to determine the rate at which the angle between the beam and the perpendicular line is changing.
The searchlight rotates at a rate of 2 revolutions per minute. Since there are 2π radians in one revolution, the angular velocity of the searchlight is 2(2π) = 4π radians per minute.
The rate of change of the angle, dθ/dt, is therefore 4π radians per minute.
Now we need to relate this angular velocity to the speed of the dot on the wall. Since the beam hits a wall located 8 miles away, we can consider the situation as a right triangle, where the distance from the searchlight to the wall is the hypotenuse, and the horizontal distance between the searchlight and the dot is the adjacent side.
Let's call the horizontal distance x. Using trigonometry, we have:
cos(θ) = x/8
Differentiating both sides with respect to time, t, we get:
-d(sin(θ))/dt = dx/dt / 8
Since sin(θ) = sin(pi/5) = sqrt(5 - sqrt(5))/4 (using the given angle), we have:
-d(sqrt(5 - sqrt(5))/4)/dt = dx/dt / 8
Simplifying, we find:
dx/dt = -8(sqrt(5 - sqrt(5))/4) * dθ/dt
Substituting the value of dθ/dt we found earlier, we have:
dx/dt = -8(sqrt(5 - sqrt(5))/4) * 4π
Simplifying further:
dx/dt = -2π(sqrt(5 - sqrt(5)))
Now, the speed of the dot on the wall is the absolute value of the rate of change of x with respect to time, so:
Speed of dot = |dx/dt| = 2π(sqrt(5 - sqrt(5))) miles per minute.
To convert the speed from minutes to hours, we can multiply by 60:
Speed of dot = 2π(sqrt(5 - sqrt(5))) * 60 miles per hour.
Therefore, the speed of the dot on the wall is approximately:
Speed of dot ≈ 751.37 mph.
dθ/dt=2 rev/min=4π rad/min
dx/dt=8*(sec^2(π/5))*8π=154 miles/min=9216 mph