Calculating Light Beam Speed with Inverse Trig Functions

[imull]

Homework Statement

A patrol car is 50 ft from a long warehouse. The revolving light on top of the car turns at a rate of 30 rotations per minute. How fast is the beam of light moving along the warehouse wall when the beam makes a 45° angle with the line perpendicular from the light to the wall?

Homework Equations

The Attempt at a Solution

First I started out by setting θ as a function of x : tanθ=(x/50)→θ=arctan(x/50). So when θ=45°, x=50ft. dθ/dt is 30(2∏)=60∏. So taking the derivative: dθ/dt=[(50)/(50²+x²)]dx/dt. After substituting values, 60∏=(0.01)dx/dt→dx/dt=6000∏ ft/min. Is this right?

 

[Dick]

Homework Statement

A patrol car is 50 ft from a long warehouse. The revolving light on top of the car turns at a rate of 30 rotations per minute. How fast is the beam of light moving along the warehouse wall when the beam makes a 45° angle with the line perpendicular from the light to the wall?

Homework Equations

The Attempt at a Solution

First I started out by setting θ as a function of x : tanθ=(x/50)→θ=arctan(x/50). So when θ=45°, x=50ft. dθ/dt is 30(2∏)=60∏. So taking the derivative: dθ/dt=[(50)/(50²+x²)]dx/dt. After substituting values, 60∏=(0.01)dx/dt→dx/dt=6000∏ ft/min. Is this right?

Yes, it is right. But you didn't really need to go through the arctan stuff. Just differentiate both sides of 50tanθ=x.

 

[imull]

Okay, thank you. The reason that I did the arctan stuff was because we're doing arctan derivatives, so I figured my teacher would want me to do it that way.