From Calculus by Varberg and Purcell, 6th Edition, 1992. I'm Painstakingly trying to teach beginning calculus to myself, but on this question I'm stumped:
"A weight connected to a spring moves along the x-axis so that its coordinate at time t is: $ \ x \ = \ \sin (2t) + \sqrt3 \cos (2t) \ \ . $ What is the farthest the weight gets from the origin?"
I found $ \ \frac{dx}{dt} \ = \ 2 \cos (2t) - 2·\sqrt3 \sin (2t) \ \ . $
Setting this equal to zero to find a critical point, $ \cos (2t) \ =\ \sqrt3 \sin (2t) \ \ . $
Even if I've done the derivation correctly (a big if), how do I translate this simple equation into a distance at time t?
My limited trigonometry skills are not helping. (I tried dividing through by $ \ \sin (2t) \ $ to end up with $ \ \tan (2t) \ = \ \frac{1}{\sqrt3} \ \ , $ which even if correct, still leaves me lost just as far in the thicket.)
As always, any help is greatly appreciated, as I'm on my own out here. Thank you to all.
You have\begin{align}x(t)&=\sin(2t)+\sqrt3\cos(2t)\\&=2\left(\frac12\sin(2t)+\frac{\sqrt3}2\cos(2t)\right)\\&=2\left(\cos\left(\frac\pi3\right)\sin(2t)+\sin\left(\frac\pi3\right)\cos(2t)\right)\\&=2\cos\left(\frac\pi3+2t\right),\end{align}and therefore the maximum value of $x$ is $2$.