Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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

Answer 4

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?

Answer 5

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