# Why are the units miles per min in this related rates question?

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 $$\theta$$

:

:

$$y$$

:

:

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

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

y is 4

The hypotenuse $$LQ$$ is $$5$$

$$x = 3$$

Solution:

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. • According to your transcribed solution,$\frac{dx}{dt}$has units of miles per minute, not miles per hour. – Matthew Leingang Jan 27 at 21:08 • Thanks, I fixed it. – user637421 Jan 27 at 21:11 • 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. – Matthew Leingang Jan 27 at 21:12 • 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 – Hagen von Eitzen Jan 27 at 21:12 ## 1 Answer 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. • If radians are not a unit, what about degrees? Aren't degrees just a constant multiple of radians? – user637421 Jan 27 at 23:26 • What are the units in tan theta = x / y? Suppose the units of theta were feet, what would the units of tan theta be? – user637421 Jan 27 at 23:37 • 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). – timtfj Jan 28 at 1:35 • 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. – irchans Jan 28 at 2:43
• 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. – user637421 Jan 28 at 6:29