The time rate of change of the displacement (velocity) of a robot arm is ds/dt = 8t/(t^2 + 4)^2. Find the expression for the displacement as a function of time if s = -1 m when t = 0 s.
Find the moment of inertia of a plate covering the first-quadrant region bounded by y^2 = x, x = 9 and the x-axis with respect to x-axis.
This is a book example, so the solution is here. I am hung up on one step though.
ds = (int)/(8tdt)((t^2 + 4)^2) = 4 (int) (t^2 + 4)^-2(2t dt)
This next step throws me off. What happened to (2t dt)??
it just disappears. Can someone explain? Thanks.
s = 4(1)(-1))(t^2 + 4)^-1 + C
Ix =k (int) y^2(x2 - x1)dy
9ky^2 dx = 9kx = 9k1/2x^2
4.5x^2 = 4.5 * 3^2
4.5*9 = 40.5
However, the answer is 162/5k = 32.4
so where'd I go wrong? thanks.