Question

A particle moves around the circle x^2 + y^2 = 25 at constant speed, making one revolution in 2 seconds. Find the acceleration vector when it is at (3, 4).

Answer 1

First you need to get position vectors for x and y using polar coordinates:
$x(t) = 5 cos (\pi t) \\ \\ y(t) = 5 sin (\pi t)$

Acceleration is the 2nd derivative of position, Find 2nd derivatives of both x and y:
$x''(t) = -5\pi^2 cos (\pi t) \\ \\ y''(t) = -5\pi^2 sin (\pi t)$

Finally, evaluate acceleration at (3,4)
x = 3 when cos = 3/5
y = 4 when sin = 4/5
Substitute into acceleration functions:
$x''(t) = -5\pi^2(\frac{3}{5}) = -3\pi^2 \\ \\ y''(t) = -5\pi^2 (\frac{4}{5}) = -4\pi^2$

Final Answer:
$( -3\pi^2 , -4\pi^2)$