The question is

A triangle has sides of length 4cm and 9cm. The angle between them is increasing at a rate of 1$^\circ$ per minute. Find the rate in cm$^2$ per minute at which the area of the triangle is increasing when the angle is 45$^\circ$.

My solution was (and it is supposedly correct):

$ A = \frac{1}{2}\cdot 4 \cdot 9\cdot \sin(\theta) = 18 \sin(\theta) $

$ \frac{\Delta A}{\Delta \theta} = 18\cos(\theta) $

$ \frac{\Delta \theta}{\Delta t} = 1^\circ = \frac{\pi}{180}^c $

$ \frac{\Delta A}{\Delta t} = \frac{\Delta A}{\Delta \theta} \cdot \frac{\Delta \theta}{\Delta t} = \frac{\pi}{180} \cdot 18\cos(\theta) = \frac{\pi}{10} \cos(\theta) $

Let $ \theta = \frac{\pi}{4} $

$ \frac{\Delta A}{\Delta t} = \frac{\pi \sqrt{20}}{2}cm^2/min $

But I was wondering, why did we have to convert the angle from degrees to radians? I only converted to radians simply because I happened to prefer working with radians over degrees.

Obviously, by chain rule, if we stayed in degrees, we would have a different expression for $ \frac{\Delta A}{\Delta t} $ and therefore a different answer for the rate of change. However, the rate of change shouldn't change simply because we used a different unit of measurement right? After all, $ 1^\circ = \frac{\pi}{180}^c $ and always will be.

Is there a fundamental reason why the angle must be measured in radians? My hypothesis is that the units have to match when working with related quantities but I can't seem to make the connection between units of angle and units of area.

  • $\begingroup$ May be you know this, but I say it anyway. If you measure $\theta$ in degrees, then $\dfrac{\Delta A}{\Delta\theta}$ will be different. The derivative of $\sin\theta$ is $\cos\theta$ only when you measure $\theta$ in radians. If you use degrees, then another application of chain rule will cancel the change you noticed: $$\frac{\Delta\theta(\text{radians})}{\Delta\theta(\text{degrees})}=\frac{\pi}{180}.$$ $\endgroup$ – Jyrki Lahtonen Sep 28 '13 at 5:58

How do you know $\frac{d}{d\theta}\sin(\theta) = \cos(\theta)$? Is it still true if you measure $\theta$ in degrees instead of radians?

It's actually a remarkably subtle point, and I've seen many good mathematicians get confused by it.

I'll leave out the details of proving the derivative formula for now, but the crux of the matter is that you need to know that $\sin(\theta)$ is very close to $\theta$ when $\theta$ is small, and this is only true if you measure $\theta$ in radians. The best explanation I've seen for that is actually here on math exchange: How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?

An important implicit point there is that the area of a wedge of a circle $\frac{\theta}{2} r^2$, which only holds if $\theta$ is measured in radians.

  • $\begingroup$ When $\theta$ is measured in radians, then, for small $\theta$, $\sin \theta \approx \theta$. But when $\theta$ is measured in degrees, then, for small $\theta$, $\sin \theta \approx \dfrac{\pi}{180}\theta$. $\endgroup$ – steven gregory Nov 23 '19 at 0:03

Since one foot equals $12$ inches, then one foot per second equals $12$ inches per second.

Since $\pi$ radians equals $180^\circ$ then $\pi$ radians per second equals $180^\circ$ per second.

and so on...


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