Rate of Change Using Inverse Trig Functions

[Amrator]

Homework Statement

A spectator is standing 50 ft from the freight elevator shaft of a building which is under construction. The elevator is ascending at a constant rate of 20 ft/sec. How fast is the angle of elevation of the spectator's line of sight to the elevator increasing when the elevator is 50 ft above the ground?

Homework Equations

Derivative of arctan
g'(x) = 1/[(x^2)+1]

h=height

The Attempt at a Solution

I don't know how to use latex.

I used implicit differentiation. Since the elevator is moving at a velocity of 20 ft/sec, I plugged that in for dh/dt. Because they wanted me to use the derivative of arctan formula, I'm assuming the input is h/50 (y/x). After I plugged in h/50 into the formula, I plugged in the height they gave me, 50, into h. I multiplied that by 20 (dh/dt) and got 0.2 rad/sec.

Did they want me to find the time when the elevator was 50 ft above the ground first?
Also I apologize for the small writing. It's arctan(h/50).

 

[Dr. Courtney]

I don't think the time when the elevator is 50 ft high is relevant.

 

[Amrator]

I don't think the time when the elevator is 50 ft high is relevant.

Am I allowed to ask if the result I got is correct? If not, that's fine.

 

[paradoxymoron]

I don't see anything wrong with your work.

 

[SammyS]

Am I allowed to ask if the result I got is correct? If not, that's fine.

The result looks to be correct.

Working it a somewhat different way, I get the same result.

 

[Amrator]

Alright. Thanks for the help, guys. I really appreciate it.