The bottom of a large theator screen is 3 ft above your eye level and the top of the screen is 10 ft above your eye level. Assume you walk away from the screen (perpendicular to the screen) at a rate of 3 ft/s while looking at the screen. What is the rate of change of the viewing angle $\theta$ when you are 30 ft from the wall on which the screen hangs, if the floor is horizontal?
This is my attempt, but it's wrong… Why?
Let the angle $\alpha$ be between the floor and your eye level.
$$\tan(\alpha)=\frac{3}{x} \implies \frac{d \alpha}{dx}=\frac{-3}{9+x^2}$$
$$\tan(\alpha + \theta) = \frac{13}{x} \implies \frac{d \theta}{dx} + \frac{d \alpha}{dx} = \frac{-13}{169+x^2}$$
$$\frac{d \theta}{dx} = \frac{-13}{169+x^2} + \frac{3}{9+x^2}$$
But when I plug in $x=30$, I get $\frac{d \theta}{dx} = -0.00886$ when the answer is $\frac{d \theta}{dx}=-0.0201$
Where did I go wrong? I can't figure it out. I've even checked my derivatives using online calculators and everything is right… so something must be wrong with my logic?
Best Answer
I think there are two issues with your work.
(1) The $10$ in your diagram should be a $7$ (so that the total height is $10$).
(2) Second, their answer is using a time derivative (with respect to $t$), not an $x$ derivative. So you'll have to take into account that $\frac{dx}{dt}=3$ ft/sec.