Consider a particle moving along the x-axis where
x(t)
is the position of the particle at time t,
x' (t)
is its velocity, and
x'' (t)
is its acceleration.
x(t) = t3 − 12t2 + 21t − 9, 0 ≤ t ≤ 10
Find the open t-intervals on which the particle is moving to the right. (Enter your answer using interval notation.)
To find the open t-intervals on which the particle is moving to the right, we need to determine when the velocity of the particle, x'(t), is positive. When the velocity is positive, it means that the particle is moving in the positive direction along the x-axis (i.e., to the right).
Let's start by finding the derivative of the given position function, x(t), to obtain the velocity function:
x'(t) = d/dt (t^3 - 12t^2 + 21t - 9)
Taking the derivative of each term:
x'(t) = 3t^2 - 24t + 21
Now we need to find the open t-intervals when x'(t) > 0. To do this, we set x'(t) greater than zero and solve for t:
3t^2 - 24t + 21 > 0
We can factor this quadratic inequality as follows:
(t - 1)(3t - 21) > 0
Now we have two factors: (t - 1) and (3t - 21). We need to determine the sign of each factor to find the intervals where the inequality is satisfied.
1. (t - 1) > 0:
Solving this inequality gives t > 1. So, the factor (t - 1) is positive for t > 1.
2. (3t - 21) > 0:
Solving this inequality gives t > 7. So, the factor (3t - 21) is positive for t > 7.
To find the open t-intervals where both factors are positive, we take the intersection of the two intervals: t > 1 and t > 7.
Since t > 7 already includes t > 1, the intersection is simply t > 7.
Therefore, the particle is moving to the right on the open t-interval (7, ∞) (using interval notation).
moving to the right means x is increasing.
So, where do you have x'(t) > 0?