How can I differentiate correctly in this problem so that the units work out correctly?

Question

I have this homework problem:

Two sides of a triangle are 4 m and 5 m in length and the angle between them is increasing at a rate of 0.06 rad/s. Find the rate at which the area of the triangle is increasing when the angle between the sides of fixed length is π/3.

Calculus Early Transcendentals 7E, by James Stewart, p. 249

I think I went about solving it the right way, but my units didn't work out the way I expected them to:

$$height_{triangle} = 4m \times \sin \theta$$ $$A_{triangle} = \frac{1}{2} \times 5 m \times 4 m \times \sin \theta = 10m^2 sin \theta$$ $$\frac{dA}{dt} = \frac{d}{dt}(10 m^2 \times \sin \theta)$$ $$\frac{dA}{dt} = 10 m^2 \times \cos \theta \times \frac{d\theta}{dt}$$ $$\frac{dA}{dt} = 10 m^2 \times \cos \frac{\pi}{3} \times 0.06 \frac{rad}{s} = 0.3 \frac{m^2 rad}{s}$$

The answer in the book, $0.3 \frac{m^2}{s}$, makes a lot more sense. Can you please point out what I am missing here? Can I ignore the $rad$ for some reason?

Edit: Looks like Berkeley solved it the same way I did, but they ignored the units entirely! Any ideas?

Answer 1

Are you using "rad" to mean radians? It is typical to suppress the unit "radian". So, you have done it correctly. For example we never say $\cos(\pi/3)$ radians.

Answer 2

Hope this helps.

the area of a triangle is A = $0.5 b h$. Then $b = 5m$ and sin$\theta$ = $h/4$.

Thus, $A = 1/2(5)(4sin$$\theta$) = 10sin$\theta$

Then given $dA/dt = 0.06rad/s$, so $dA/dt = dA/d\theta$* $ d\theta/dt$ = (10cos$\theta$)(0.06) = 0.6cos$\theta$.

Then when $\theta$ = $\pi/3$ , we have $dA/dt = 0.06$(cos$\pi/3$) = 0.6(1/2) = $0.3m^2/s$