I can do related rates problems a little bit, but I've been given one that requires me to use a rate of $\frac{-\pi}{6}$ radians per second to figure out how fast a plane is going. Since I assume that plane is moving in a straight line, I'm not sure how to proceed.

I think the answer might be to find out how fast the given rate is moving in terms of horizontal speed, and then apply implicit differentiation on that to find the answer, but I'm not sure.

I could use the angular velocity formula, $\omega = \frac{\theta}{t}$, but I'm worried that's not the right way since it gives an average.

How can I convert a rate with radians to a rate involving purely horizontal/straight movement?

Problem Text:

A plane flies horizontally at an altitude of 5 km and passes directly over a tracking telescope on the ground. When the angle of elevation is π/3, this angle is decreasing at the rate of π/6 radians per minute. How fast is the plane traveling at that time?

I have worked through it, and I've discovered that the plane, the telescope, and the distance of the plane from the ground can be formed into a right triangle. Since, $\frac{\pi}{3}$ radians is $60^\circ$ and a right triangle has a $90^\circ$ corner too, that means the the last corner must also be $60^\circ$.

Using the Law of Sines, I have found the hypotenuse is $\frac{10\sqrt{3}}{3}$ kilometers long and the other two edges are both $5$ kilometers.

I haven't been able to get further than that.

  • 1
    $\begingroup$ Note you have the hypotenuse correct, but "the other two edges are both $5$ kilometers" is not correct. The height is a constant $5$, but the horizontal distance at that point is $\frac{5}{\sqrt{3}}$ kilometers. $\endgroup$ Mar 12, 2019 at 2:34

2 Answers 2


Draw a picture to see more clearly what's going on:

enter image description here

Both the angle of elevation and the distance $x$ are functions of time. Express $x$ is terms of $\theta$:

$$\cot{\theta}=\frac{x}{5}\implies x=5\cot{\theta}$$

Implicitly differentiate both sides with respect to time:


Our goal is to find how fast the plane is moving ($\frac{dx}{dt}$) when the angle of elevation is $\pi/3$. We also know that the rate with witch the angle of elevation is changing is constant. It's $-\pi/6$ rad/min. There is a minus sign in front of it because they're saying that it's deceasing. Well, just plug all those quantities into the expression that we got:

$$ \frac{dx}{dt}=-5\cdot\csc^2{\left(\frac{\pi}{3}\right)}\cdot\left(-\frac{\pi}{6}\right)=5\cdot 2^2\cdot\frac{\pi}{6}=\frac{10\pi}{3}\ km/min $$

So, the plane is traveling at that moment with a velocity of $\frac{10\pi}{3}$ km/min.


Use a Cartesian coordinate system where the tracking telescope is at $(0,0)$. Thus, using polar coordinates, we have

$$x = r\cos\theta \tag{1}\label{eq1}$$ $$y = r\sin\theta = 5 \tag{2}\label{eq2}$$

Since the height is not changing, the speed of the airplane is determined strictly by $\frac{dx}{dt}$. We want to determine this value when $\theta = \frac{\pi}{3}$. Since $r$ and $\theta$ are implicitly functions of time $t$, differentiate \eqref{eq1} and \eqref{eq2} wrt $t$ using the product & chain rules to get

$$\frac{dx}{dt} = \frac{dr}{dt}\cos\theta - r\sin\theta \frac{d\theta}{dt} \tag{3}\label{eq3}$$ $$\frac{dy}{dt} = \frac{dr}{dt}\sin\theta + r\cos\theta \frac{d\theta}{dt} = 0 \tag{4}\label{eq4}$$

At $\theta = \frac{\pi}{3}$, we have a $30^o-60^o-90^o$ triangle, so since the height is a fixed $5$, the radius at that point is $\frac{10}{\sqrt{3}}$, as the OP stated. Also, it's given that $\frac{d\theta}{dt} = -\frac{\pi}{6}$. In addition, note that $\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$ and $\sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$. Substituting these values into \eqref{eq3} gives

$$\frac{dx}{dt} = \frac{dr}{dt}\left(\frac{1}{2}\right) - \left(\frac{10}{\sqrt{3}}\right)\left(\frac{\sqrt{3}}{2}\right)\left(-\frac{\pi}{6}\right) = \frac{dr}{dt}\left(\frac{1}{2}\right) + \frac{5\pi}{6} \tag{5}\label{eq5}$$

Substituting the values into \eqref{eq4} gives

$$\frac{dr}{dt}\left(\frac{\sqrt{3}}{2}\right) + \left(\frac{10}{\sqrt{3}}\right)\left(\frac{1}{2}\right)\left(-\frac{\pi}{6}\right) = 0 \tag{6}\label{eq6}$$

You can now move the second term to the right, divide both sides by $\frac{\sqrt{3}}{2}$ to get $\frac{dr}{dt}$ and then substitute its value into \eqref{eq5}. I trust you can finish the rest yourself.


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