Here is a problem my father gave me while he was studying Calculus.

A lighthouse is $4$ miles from point $P$ along a straight shoreline and the light from this lighthouse makes $4$ revolutions per minute. How fast, in miles per minute is the beam of light moving along the shoreline when it is $3$ miles from point $P$.

Here is the solution given in the book.

There is a diagram with the lighthouse at the top and the shoreline at the bottom.

L (the lighthouse)

: angle $\theta$






$P$------------------------------ $x$ -------------------------------------------$Q$

$Q$ is the point where the light hits the shore

y is 4

The hypotenuse $LQ$ is $5$

$x = 3$


In the diagram, label the important variables as $x$, $y$, and $\theta$. We are given that $y=4$ (constant), $x=3$, and $\frac{d\theta}{dt} = 4 \frac{\text{rev}}{\text{min}}$. You want $\frac{dx}{dt}$. Since we are given information about angles, you want a trigonometric function using $x$ and $y$. Our equation would be $\tan \theta = \frac{x}{y}$.

We can immediately plug in the value of y because it is a constant so we get

$\tan \theta = \frac{x}{4}$ [equation 0]

Taking the derivative, we get $\sec^2\theta \cdot\frac{d\theta}{dt} = \frac{1}{4} \frac{dx}{dt}$. It is important to understand that $\frac{d\theta}{dt}$ must be measured using radians, so we get $\frac{d\theta}{dt} = 4 \frac{\text{rev}}{\text{min}} (2\frac{\pi}{\text{rev}}) = 8 \frac{\pi}{\text{min}}$.

And although we do not know the value of theta, we know that the hypotenuse of the triangle is $5$ so $\sec \theta = \frac{5}{4}$. Putting it together:

$sec^2\theta\cdot\frac{d\theta}{dt} = \frac{1}{4} \frac{dx}{dt}$ [equation 1]

$(\frac{5}{4})^2\cdot(8 \pi) = \frac{1}{4} \frac{dx}{dt}$

$\frac{dx}{dt} = 50 \pi \frac{\text{miles}}{\text{min}}$

Question: If $\frac{d\theta}{dt} is in radians per minute, then why is \frac{dx}{dt} in miles per min?

Update: We now realize that the units don't have to match. You could start with a relationship like $T = s$ where $T$ is degrees Celsius and $t$ is seconds after time $0$. Clearly temperature and time are different units but we can still have an equation to relate the quantities. Taking the derivative just creates a new equation that relates the rates but the rates can have different units.

  • $\begingroup$ According to your transcribed solution, $\frac{dx}{dt}$ has units of miles per minute, not miles per hour. $\endgroup$ – Matthew Leingang Jan 27 '19 at 21:08
  • $\begingroup$ Thanks, I fixed it. $\endgroup$ – user637421 Jan 27 '19 at 21:11
  • $\begingroup$ Other than that, the reason for the different units on $\frac{dx}{dt}$ and $\frac{d\theta}{dt}$ is the different units on $x$ and $\theta$: one is a length, the other an angle. $\endgroup$ – Matthew Leingang Jan 27 '19 at 21:12
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    $\begingroup$ In short, because radians per minute is a valid unit of angular velocity and miles per minute is (in only three countries, though) a valid unit of (linear) velocity $\endgroup$ – Hagen von Eitzen Jan 27 '19 at 21:12

Radians are not actually a unit. You can compute angles in radians by the distance along the edge of a circle divided by the radius of the circle. So the units of the angle $\theta$ are miles per mile and hence radians are unitless.

Your equation is $\ \tan\theta = x/y\ $ where $y$ is 4 miles. $\ $ So, $$\sec^2(\theta) \frac{d\theta}{dt} = \frac{1}{y} \frac{dx}{dt}.$$ Solving for $\frac{dx}{dt},$ we get $$\begin{align} \frac{dx}{dt} &= y \sec^2(\theta) \frac{d\theta}{dt} \\ &= 4\; \mathrm{miles}\cdot (5/4)^2 \cdot 8 \pi /\mathrm{minute}\\ &= 50\pi\; \mathrm{miles/minute}. \end{align} $$

With regard to your other comment, the units need to match. If $T(t)=\alpha t^2$ measures the temperature at time $t$ in Celsius degrees and $t$ is measured in seconds, then $\alpha$ needs to have units of degrees Celsius per second squared and the rate of change of temperature $T'(t)= 2\alpha t$ will have units of Celsius degrees per second.

  • $\begingroup$ If radians are not a unit, what about degrees? Aren't degrees just a constant multiple of radians? $\endgroup$ – user637421 Jan 27 '19 at 23:26
  • $\begingroup$ What are the units in tan theta = x / y? Suppose the units of theta were feet, what would the units of tan theta be? $\endgroup$ – user637421 Jan 27 '19 at 23:37
  • $\begingroup$ I think it might be more accurate to say radians are a dimensionless unit—they're a ratio of lengths, but always specify an angle (which is a dimensionless quantity). $\endgroup$ – timtfj Jan 28 '19 at 1:35
  • $\begingroup$ The $\tan$ function takes in a unitless input and gives a unitless output. Alternatively, it could take an input in degrees and give a unitless output. Recall that the output of $tan$ is the opposite length divided by the adjacent length, so it would have units of feet divided by feet or centimeters/centimeters which are both unitless. $\endgroup$ – irchans Jan 28 '19 at 2:43
  • $\begingroup$ I think it comes down to the definition of dimension. You can say the squaring function y = x ^ 2 takes a unitless input and gives a unitless output. But if we use 3 feet as the input, we end up saying the output is 9 feet squared and the units somehow change. Some people say time is a fourth dimension. I can imagine a point in space at a given time and if we assign numbers to colors I could think of the color of the point as a fifth dimension. I like to think of dimension as a property. I can see angles having a dimension in that they measure an amount of rotation around a point. $\endgroup$ – user637421 Jan 28 '19 at 6:29

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