I have this exercise:
Point B moves from point A to point C at 2 cm/sec in the accompanying diagram. At what rate is Theta changing when x = 4 cm?
I solved it this way: the theta angle equals Pi minus the sum of the angles alpha and beta, so its rate of change is minus the sum of the rates of changes of alpha and beta. I found the rate of alpha by differentiating:
$$ \cot{\alpha} = \frac{CB}{6} \\ -\csc^2{\alpha} \alpha' = \frac{1}{6}CB' = \frac{1}{6}(-2) = \frac{-1}{3} \\ \alpha' = \frac{\sin^2{\alpha}}{3} = \frac{\sin^2{\pi/2}}{3} = \frac{1}{3} $$
The same way I found the rate of change of beta, but AB' = 2 cm/sec in that case:
$$ \tan{\beta} = \frac{3}{AB} \\ \sec^2{\beta} \beta' = -1*\frac{3}{AB^2}AB' \\ \sec^2{\beta} \beta' = -1*\frac{3}{16}(2) = \frac{-3}{8} \\ \beta' = \frac{-3\cos^2{\beta}}{8} = \frac{-3\cos^2{\arctan{3/4}}}{8} \\ \beta' = \frac{-3*64}{8*100} = \frac{-3*8}{100} = \frac{-6}{25} $$
So, the answer I found was $\theta' = \frac{6}{25} - \frac{1}{3} = \frac{18 - 25}{75} = \frac{-7}{75}$. This seems correct even from the intuitive point: the alpha angle is increasing, as x moves up, and beta is decreasing. So the rate should be, as I found it.
But the correct answer from the textbook is 43/75, which looks as they just added both rates with positive signs. I'm suspecting that this is incorrect, but could someone, please, cross-check my solution?