Calculus rate problem using trig A person is approaching a 500-foot tower on a trolley car at the rate of ten miles per hour, looking at the top of the tower. At what rate must he be raising his head (or line of sight) when the car is 500 feet from the tower on level ground?
I am supposed to find the rate of the angle but no matter how I approach it my answer always differs from the book answer. Help will be greatly appreciated
 A: Before I start solving the actual problem, let's specify some basic parameters: Our units of distance, time and angle are going to be feet, minute and radians, respectively. For notational convenience, let $h$ be the height of the tower, $x$ be the distance of the car from the foot of the tower(i.e, its distance on level ground) and $\theta$ be the angle of elevation from the car to the top of the tower.
The car is approaching the tower(effectively reducing the distance) at a speed of $10$ miles/hour, which is equivalent to $880$ feet/min. By definition, this is the velocity of the car. $$\frac{dx}{dt}=-880\, \mathrm{feet/min}$$
(The negative sign indicates the reducing distance between tower and the car)
Now in our right triangle, $$\tan\theta=\frac{h}{x}\implies x\tan\theta=h$$
Differentiating both sides w.r.t time and applying chain rule, we get: $$x\cdot\sec^2\theta\cdot\frac{d\theta}{dt}+\tan\theta\cdot\frac{dx}{dt}=0$$
Substituting $\sec\theta=\frac{\sqrt{h^2+x^2}}{x}$ and $\tan\theta=\frac{h}{x}$, we get: $$\frac{d\theta}{dt}=-\frac{h}{h^2+x^2}\cdot\frac{dx}{dt}$$
Substituting the value of all parameters for $h=x=500\,\mathrm{feet}$, we obtain the answer: $$\frac{d\theta}{dt}=\frac{1320}{1500}\,\mathrm{rad/min}=0.88\,\mathrm{rad/min}$$ You can convert this result to degrees using suitable conversion factor.
