A police cruiser hunting for a suspect pulls over and stops at a point 20 ft from a straight wall. The flasher on top of the cruiser revolves at a constant rate of 90 deg/sec, and the light beam casts a spot of light as it strikes the wall. How fast is the spot of light moving along the wall at a point 50 ft from the point on the wall closest to the cruiser?
Im having difficulties on what to do with the theta, only thing i know is it might include SOHCAHTOA, thank you if anyone can help
Draw a diagram.
Review your basic trig functions. See which ones involve the quantities you are given.
When the beam is at a distance x from the closest spot on the wall,
x/20 = tanθ
so,
dx/dt = 20 sec^2θ dθ/dt
so, find sec^2θ when x=50, and you know that dθ/dt = π/2
so, plug and chug
To solve this problem, we can use trigonometry, specifically the tangent function (tan).
Let's start by defining some variables:
- Let θ be the angle in radians (not degrees) between the light beam and the wall.
- Let d be the distance between the cruiser and the spot of light on the wall.
- Let x be the distance between the point on the wall closest to the cruiser and the spot of light on the wall.
We are given that the flasher on top of the cruiser revolves at a constant rate of 90 degrees per second. To convert this to radians per second, we multiply by the conversion factor (π/180) since there are π radians in 180 degrees:
ω = (90 degrees/sec) * (π/180) = π/2 radians/sec
To find the relationship between θ and t (time), we can use the equation:
θ = ω * t
Now, let's define a right triangle with the hypotenuse being a line from the cruiser to the spot of light on the wall, and the opposite side being x. The adjacent side (the wall) is fixed at a distance of 20 ft. From this triangle, we can write:
tan(θ) = x / 20
Taking the derivative of both sides with respect to time t, we get:
sec²(θ) * dθ/dt = (dx/dt) / 20
Now, we need to find dx/dt, the rate at which the spot of light is moving along the wall. The problem gives us the value of d, which is 50 ft. Therefore, we have:
x = 50 ft
Now, we need to find the value of sec²(θ). Since we know the tangent function, we can use the Pythagorean identity:
sec²(θ) = 1 + tan²(θ)
Plugging this into our equation, we have:
(1 + tan²(θ)) * dθ/dt = (dx/dt) / 20
We can solve this equation for dθ/dt, which represents the rate at which the angle θ is changing with respect to time. Finally, we can substitute the value of dθ/dt into the equation to get the desired rate dx/dt.
I hope this explanation helps you understand the process for solving this problem.