How did I mess up this Implicit differentiation of tangent? So, I was doing work in this calculus book that I've had for a couple years now, and have stumbled on this question, and can't seem to see where I messed up.
The textbook question: A rocket travels vertically at a speed of 1200 km/h. The rocket is tracked through a telescope by an observer located 16 km from the
launching pad. Find the rate at which the angle between the telescope
and the ground is increasing 3 min after lift-off.
My work: V = 1200 = dh/dt; b = 16; t = 1/20
tan(th) = h / b
dt(tan) = dt( h/b )
t*sec^2(t) = (1/b)*dh/dt = (1/16)*1200 = 75
sec(t) = sqrt(75/t)
sec(1/20) = sqrt(75*20)
1.001 = 9.746
As you can see, 1.001 is not actually equivalent to 9.746...Unless there's some fancy new maths that I've not seen before.
Jokes aside, I...I am not actually certain where I messed up. I am definitely thinking that it was on either the where I took the derivative of tan, as (having just checked this right now) (1/20*sec(1/20))^2 is even further from 75.
But, honestly, I don't know what went wrong.
Nor, honestly, am I entirely sure for what I am trying to solve for. The change in the angle means to take the derivative of it, right? Is a dsin/dt supposed to pop out, or something?
 A: As you point out, $\tan{\theta}=\frac{h}{16}$ and $\frac{dh}{dt}=1200$
Taking the derivative gives us $\sec^2{\theta}\frac{d\theta}{dt}=\frac{1}{16}\frac{dh}{dt}=75\implies\frac{d\theta}{dt}=75\cos^2(\theta)$ 
where $\theta=\tan^{-1}{\frac{60}{16}}$
Note that this assumes that at $t=0$ the rocket was at $h=0$. If the rocket was at some initial height other than $0$ then you must take that into account when calculating $\theta$ at $t=\frac{1}{20}$.
A: You correctly started with 
$$\tan(\theta(t)) = \frac{h(t)}b.$$
I've explicitely added which variables depend on $t$ to make that clear, as it may be part of your confusion.
Then you diffirentiated both sides with regard to $t$:
$$\frac{d(\tan(\theta(t)))}{dt} = \frac{d\frac{h(t)}b}{dt}.$$
The right hand side you got correct
$$\frac{d\frac{h(t)}b}{dt}= \frac1b\frac{dh(t)}{dt}=\frac{V}b$$.
But on the left hand side, you seem to have mixed up $t$ and $\theta$, as your result of differentiation does not contain $\theta$ any more! 
The correct result of the differentitation, using the chain rule, is
$$\frac{d(\tan(\theta(t)))}{dt} = \frac{d\theta(t)}{dt}\frac1{sec^2(\theta(t))}$$
Finally we get
$$\frac{d\theta(t)}{dt}\frac1{sec^2(\theta(t))} = \frac{V}b$$
or, equivalently 
$$\frac{d\theta(t)}{dt} = \frac{V}b{sec^2(\theta(t))}$$
We know almost anything on the right hand side, the exception is $\theta(t)$. But we can get this value from the first equation, so the final steps to solve the problem are
1) Get $\theta(t)$ from 
$$\theta(t) = \arctan\left(\frac{Vt}b\right).$$
2) Calculate the asked for rate of changes as
$$\frac{d\theta(t)}{dt} = \frac{V}b{sec^2(\theta(t))}$$
where know every thing on the right hand side is known.
