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?
Best Answer
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.