s(t) = t^4 - 2t^2 + 1
t is greater than or equal to 0
Find the interval(s) in which the robotic arm is slowing down in the negative direction.
To determine when the robotic arm is slowing down in the negative direction, we need to find the intervals where the velocity function, v(t), is decreasing and negative.
The velocity function is the derivative of the position function, s(t). So, let's start by finding the velocity function by taking the derivative of s(t):
s(t) = t^4 - 2t^2 + 1
To find v(t), we differentiate s(t) with respect to t:
v(t) = d/dt (t^4 - 2t^2 + 1)
= 4t^3 - 4t
Now, we need to determine when v(t) is decreasing and negative.
1. First, find the critical points of v(t) by setting its derivative equal to zero and solving for t:
4t^3 - 4t = 0
We can factor out 4t:
4t(t^2 - 1) = 0
From this equation, we find two critical points:
t = 0
t^2 - 1 = 0 => t = 1 and t = -1
2. Next, we can create a number line with the critical points and test intervals to determine the sign of v(t) in each interval:
Interval 1: (-∞, -1)
Pick a test point t < -1, e.g., t = -2
Substitute t = -2 into v(t):
v(-2) = 4(-2)^3 - 4(-2) = -32 + 8 = -24
Since v(-2) is negative, v(t) is negative in this interval.
Interval 2: (-1, 0)
Pick a test point -1 < t < 0, e.g., t = -0.5
Substitute t = -0.5 into v(t):
v(-0.5) = 4(-0.5)^3 - 4(-0.5) = -0.5 + 2 = 1.5
Since v(-0.5) is positive, v(t) is positive in this interval.
Interval 3: (0, ∞)
Pick a test point t > 0, e.g., t = 1
Substitute t = 1 into v(t):
v(1) = 4(1)^3 - 4(1) = 0
Since v(1) is zero, we cannot determine the sign of v(t) in this interval.
Based on the number line, we can conclude that the robotic arm is slowing down in the negative direction in the interval (-∞, -1).