Finding rate of change of angle of elevation

Question

Question-

From a building of height $48$ m, a man is walking away at a speed of $2$ m/s. Find the rate of change of angle of elevation, when he is at a distance of $36$ m from the base of the tower.

What I did-

$$\theta=\arctan(\frac{48}{l})$$

$$\frac{d\theta}{dt}=\frac{48\ln l}{1+\frac{48^2}{l^2}}\times\frac{dl}{dt}$$

Now substituting the values given, I get, $123.84$ which is not the correct answer. Can someone please tell me the mistake that I am making?

The answer in my book is $-2/75$ radians/sec. Also, why is the answer coming up in radians?

Answer 1

If you went directly from $$\theta=\arctan\left(\frac{48}{l}\right)$$ to $$\frac{d\theta}{dt}= \frac{48\ln l}{1+\frac{48^2}{l^2}}\,\frac{dl}{dt}$$ all in one step the way you showed your work in the question, I would advise not to try that sort of thing any more. I’m sure some people can do it with justified confidence in their result, but it’s fine if you never become such a person. (Admittedly my opinion on that last point may be a bit biased since I am not such a person myself.)

If you show your intermediate steps someone might be able to pinpoint your error, but my guess is that you integrated $1/l$ in a place where you should have differentiated it.

As for why radians rather than degrees, working in degrees forces you to keep inserting annoying conversion factors whenever you want to differentiate or integrate anything trigonometric. It’s much simpler to do calculus with radians. Most books will just assume radians unless there is a reason to convert a result to a different scale.

Answer 2

Your angle $\theta$ is a function of time. The distance between the man and the building is the function $f(t)=2t$. He is going to be $36\ \text{m}$ away from the building at the time $t=18\ \text{sec}$ because it takes $18$ seconds to cover a distance of $36$ meters if you're moving with a speed of $2\ \text{m/sec}$. Likewise, when he is $36\ \text{m}$ away from the building, the cosine of the angle of elevation $\theta$ at that time is going to be equal to $\frac{36}{\sqrt{48^2+36^2}}$. Here's the rest of the calculations:

$$ \tan{\theta}=\frac{48}{2t}\implies\\ \frac{d}{dt}\left(\tan{\theta}\right)=\frac{d}{dt}\left(\frac{24}{t}\right)\implies\\ \sec^2{\theta}\frac{d\theta}{dt}=-\frac{24}{t^2}\implies\\ \frac{d\theta}{dt}=-\cos^2{\theta}\frac{24}{t^2}=-\left(\frac{36}{\sqrt{48^2+36^2}}\right)^2\frac{24}{18^2}=-\frac{2}{75}\ \text{rad/sec}. $$

In calculus, we measure angles in radians. That's why the answer is in radians.

Answer 3

$$\tan \theta = \frac {48}{L}$$

$$\frac{dL}{dt} = 2$$

$$\frac{d}{dt}\tan \theta = \frac{d}{dt}\frac{48}{L}$$

$$\sec^2\theta\frac{d\theta}{dt} = \frac{-48}{L^2}\frac{dL}{dt}$$

$$\frac{d\theta}{dt} = \frac{1}{\sec^2\theta}\frac{-48}{L^2}\frac{dL}{dt}$$

$$\frac{d\theta}{dt} = \cos^2\theta\ \frac{-48}{L^2}\frac{dL}{dt}$$

$$\frac{d\theta}{dt} = \cos^2\theta\ \frac{-96}{1296}$$

$$\frac{d\theta}{dt} = \cos^2(\arctan\frac{48}{36}) \frac{-96}{1296}$$

$$\frac{d\theta}{dt} = -.02667\ \text{rad/sec}$$