A vehicle moves along a straight path with a speed of 4m/s. A searchlight is located on the ground 20m from the path and is kept focused on the vehicle. At what rate (in rad/s) is the searchlight rotating when the vehicle is 15 from the point on the path cloest to the searchlight?

What does the diagram look like? I switched the numbers 20, and 15 around a couple of times and still can't get the correct answer.

dθ/dt= ?
dθ/dt=(dx/dt)(dθ/dx)
sinθ=15/x
x=15/sinθ
1=(15cosθ/sin²θ)(dθ/dx)
dx/dθ=sin²θ/15cosθ

dx/dt=(dθ/dt)(dx/dθ)
4sin²θ/15cosθ=dθ/dt
dθ/dt=0.135rad/s

I got 25m for hypotenuse, 20m for adjaecent, 15m for opposite, dx/dt is on the opposite line, and θ between 20 and 25m. Thank you in advance.

Why are you switching things around? What is wrong with a diagram?

TanTheta=x/20
d(TanTheta)/dt=1/20 * dx/dt

sec^2 Theta * dTheta/dt=1/20 * dx/dt

dx/dt given as 4m/s
at 15m, sec Theta=20/sqrt(15^2+20^2)
solve for dTheta/dt

I'm not good with drawing diagrams from word problems, so I switched the numbers around to see if I could get the correct answer.

4/sec²θ=dθ/dt

cosθ=20/25

4cos²θ/20=dθ/dt
dθ/dt=0.160rad/s
but the answer is 0.128rad/s.

Why did you not continue with bobpursley's solution, he gave you almost 90% of it.

He gave you
sec^2 Theta * dTheta/dt=1/20 * dx/dt

then d(theta)/dt = dx/dt /(20sec^2 theta)
= 4/(20(25/20)^2
= 80/625
= .128

I did try to continue with it. I brought sec²θ to the top -> cos²θ, so I plugged in cos(20/25). How come you typed 25/20 instead of sec(25/20)²?

I'm not understanding how you're getting tanTheta as x/20.

Tan is opposite/adjacent is it not? Which would make it tanTheta = 15/20

To better understand the problem, let's consider a diagram:

searchlight
^
|
|
| 20 m
|
|
|------------------------
| | /
| | /
| | /
| | / vehicle
| | /
| | /
| | /
| | /
15 m |------ | -----/
| \ | /
| \ | /
| \| /
|-------|/

In the above diagram, the searchlight is situated on the ground, 20 meters away from the path of the vehicle. The point on the path closest to the searchlight is where the vehicle is at a distance of 15 meters from the searchlight. The angle between the line connecting the searchlight and the point on the path closest to the searchlight, and the path of the vehicle is denoted by θ.

To find the rate at which the searchlight is rotating (dθ/dt), we can use the relationship between the distances and angles in a right triangle.

Using the definition of sine, we can write:
sinθ = opposite/hypotenuse
sinθ = 15/20
sinθ = 0.75

To find dx/dt, the rate at which the vehicle is moving, we are given a speed of 4 m/s. Since the vehicle is moving along the path, dx/dt represents the rate at which the vehicle is changing its position in the x-direction.

Using the relationship between sine and cosine:
cosθ = sqrt(1 - sin²θ)
cosθ = sqrt(1 - 0.75²)
cosθ = 0.6614

Now, to find dx/dθ, we can use the reciprocal of the derivative of x with respect to θ:
dx/dθ = 1/(sin²θ/15cosθ)
dx/dθ = 15cosθ/sin²θ

Next, we need to find dθ/dt. We can use the chain rule of differentiation:
dθ/dt = (dx/dt)(dθ/dx)

Plugging in the values we know:
dθ/dt = (4sin²θ/15cosθ)(15cosθ/sin²θ)
dθ/dt = 4/15 rad/s

Therefore, the searchlight is rotating at a rate of approximately 0.135 rad/s when the vehicle is 15 m from the point on the path closest to the searchlight.