A searchlight rotates at a rate of 3 revolutions per minute. The beam hits a wall located 7 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 \theta between the beam and the line through the searchlight perpendicular to the wall is \frac{\pi}{6}? Note that d\theta/dt=3(2\pi)=6\pi.
Thank you this helped me
Well, I don't mean to throw shade on this question, but since the searchlight is rotating at 3 revolutions per minute and each revolution covers 2π radians, the angular speed of the searchlight is 6π radians per minute.
Now, if we want to find the horizontal speed of the dot of light on the wall, we need to consider the relationship between the radius of the circle that the searchlight sweeps out and the distance the dot of light moves along the wall.
Since the searchlight hits the wall 7 miles away, we have a right triangle formed by the searchlight, the wall, and the line connecting the searchlight to the dot of light. The hypotenuse is the radius of the circle, and the horizontal leg is the distance the dot of light moves along the wall.
Using trigonometry, we can say that cos(θ) = (distance along the wall) / (radius of the circle).
Since we're given that θ = π/6, we have:
cos(π/6) = (distance along the wall) / (7 miles).
Simplifying this, we find:
√3/2 = (distance along the wall) / 7.
Multiplying both sides by 7, we get:
(distance along the wall) = 7 * √3/2.
Now, we know that dθ/dt = 6π radians per minute, which is the rate at which the angle θ is changing.
To find the rate at which the dot of light is moving horizontally along the wall, we need to differentiate our equation in terms of time. So:
(d(distance along the wall)/dt) = (d(7 * √3/2)/dt).
Since 7 * √3/2 is a constant, its derivative is 0. Therefore, the left side simplifies to:
(d(distance along the wall)/dt) = 0.
So, the dot of light on the wall isn't actually moving horizontally! It stays at the same spot on the wall.
I hope that brightened your day a little!
To find the speed of the dot of light on the wall, we need to use the given information about the rotation rate of the searchlight and the angle \theta.
Let's break down the problem step by step:
1. First, let's find the angular speed (dθ/dt) in radians per minute since we are given that dθ/dt = 6π. We know that the searchlight rotates at a rate of 3 revolutions per minute, and since there are 2π radians in one revolution, we can convert the rotation rate to radians per minute:
Angular speed (dθ/dt) = 3 revolutions/minute * 2π radians/revolution = 6π radians/minute
2. Now, we need to find the linear speed (ds/dt) of the dot of light on the wall. Since the angle \theta is changing, we can relate the angular speed and the linear speed using the equation:
ds/dt = r * dθ/dt
where r is the distance between the searchlight and the wall, which is given as 7 miles.
Plugging in the values:
ds/dt = 7 miles * 6π radians/minute = 42π miles/minute
3. Finally, we need to convert the linear speed from miles per minute to miles per hour. There are 60 minutes in an hour, so we can multiply the speed by 60 to get the speed in miles per hour:
ds/dt = 42π miles/minute * 60 minutes/hour = 2520π miles/hour
So, the speed of the dot of light on the wall when the angle \theta is π/6 radians is 2520π miles per hour.
3 * 2 pi = 6 pi radians/minute
pi/6 = 30 degrees by the way
I call your angle theta A
dA/dt = 6 pi rad/min
tan A = x/7
x = 7 tan A
dx/ dt = 7 d/dt(tan A ) = (7/cos^2A) dA/dt
cos^2 (30) = .75
so
dx/dt = (7 miles/.75)(6 pi rad/min)
dx/dt = 176 miles/min
* 60 = 1055 miles/hr
so
dx/dt =
Concur.
dx/dt=7*(4/3)*2π=176 miles/min=10555miles/hr