Will P. answered • 06/06/19
UChicago Grad For Math, Spanish, and Test Prep
First, let's write down the quantity that we want to maximize. In this case, we have a point moving a round with 2 coordinates for its position, so it has 2 coordiantes for its acceleration, a(t) = (a_x(t),a_y(t)), which depends on the time. Now, we can't really "maximize" a vector, so we are going to maximize the magnitude of the acceleration, or the length of the acceleration vector, which is |a(t)| = sqrt{(a_x)^2 + (a_y)^2}. Now let's try and write this down; taking two derivatives of the position with respect to time gives us the acceleration, so
a_{x}(t) = d^2/dt^2 [ cos(2*pi* t)] = -4picos(2pi t) and a_{y} = d^2/dt^2 [ 2sin(2*pi *t) ] = -8pi*sin(2*pi*t), and then |a(t)| = \sqrt{(-4pi)^2cos(2pi t)^2 + (-8pi)^2sin(2pi t)^2 } = \sqrt{16\pi^2 + 48\sin(2\pit)^2} where we used the basic sin^2+cos^2 = 1 identity to simplify our result. Then, we can find the max here without even using calculus, since we know that the square root is a strictly increasing function, and sin(2\pi t) is maximized when 2\pit = \pi/2 (on the unit circle, this is when the sin of the angle is 1, or in plain english, sin(\pi/2) = 1). So, the acceleration is maximized when 2\pi t = \pi/2 --> t = 1/4. Plugging in the time t to the acceleration and position equations above gives exactly the results that you wrote above under Ans.
