Question

A man is walking towards a wall at 3 ft/sec while looking at a picture on the wall, at a height of 5 feet from the floor. How is the angle of observation changing when the man is 8 feet away from the screen.

Answer 1

You have to have the height the man's eyes are frm the floor to do this.

angle of observation= arctan(eye-5)/d
where distance d is the horizontal distance. eye = height of eyes off floor.

you are given d'=3ft/sec

theta= arctan(eye-5/d)

4ye-5=d*tanTheta
take the deriviative of each side.
0=d' tanTheta+ d (sec^2 theta) dTheta/dt

so solve for dTheta/dt

dTheta/dt= -3tanTheta /(d sec^2 theta)

so at 8 feet, d=8, tanTheta=(eye-5)/8,
secTheta=8/sqrt(64+(eye-5)^2). Note if the eye height is very near 5 ft, then secTheta=appx 1.0

Answer 2

Bob is wrong



dx/dt = 3ft/s, dƟ/dt = ?
5ft/x = tan Ɵ
Differentiating each side with respect to the time: (-5ft)/x^2 dx/dt=〖sec〗^2 ϑ dϑ/dt
So dϑ/dt=(-5ft)/x^2 dx/dt.〖cos〗^2 θ
We want to evaluate the change at the particular instant when x = 8ft (and so the hypotenuse is √89 and the cosƟ = 8/ √89 )
dϑ/dt=(-5ft)/(64〖ft〗^2 ) 3 ft/sec.64/89=15/89sec
The angle is changing at 15/89 radians per second