A particle's position along the x-axis is given by x(t) = t^4/24 - t^3/2 + 2t^2 - 1. What is the maximum acceleration on the interval 0 <= t <= 4?
I found the acceleration to be (t^2 - 6t + 8)/2 and the derivative of the acceleration to be t - 3. I thought the maximum should occur at t = 3, but the second derivative of the acceleration is positive, so that would be a minimum. I'm not sure what to do here.
I agree with you.
I also found the 2nd derivative to be (1/2)t^2 - 3t + 4
and the third derivative to be t - 3
since the 2nd derivative is a quadratic opening upwards, you would have a minimum at t = 3
Could be a misprint
To find the maximum acceleration on the interval 0 <= t <= 4, we need to find the critical points of the acceleration function and determine whether they correspond to a maximum or a minimum.
First, let's find the critical points by setting the derivative of the acceleration function equal to zero:
t - 3 = 0
Solving this equation gives us t = 3, which is a potential critical point.
Next, let's evaluate the second derivative of the acceleration function to determine the nature of the critical point at t = 3. The second derivative is given by:
d^2/dt^2 (t^2 - 6t + 8)/2
Taking the derivative of the acceleration function with respect to t, we get:
d^2/dt^2 (t - 3) = 1
Since the second derivative is positive (equal to 1), this means the critical point at t = 3 is a local minimum, not a maximum.
To determine the maximum acceleration, we now need to evaluate the acceleration function at the endpoints of the interval, t = 0 and t = 4.
At t = 0:
Acceleration = (0^2 - 6(0) + 8)/2 = 8/2 = 4
At t = 4:
Acceleration = (4^2 - 6(4) + 8)/2 = (16 - 24 + 8)/2 = 0
Comparing these values, we see that the maximum acceleration on the interval 0 <= t <= 4 is 4, which occurs at t = 0.
Therefore, the maximum acceleration is 4.
To find the maximum acceleration on the interval 0 <= t <= 4, you need to find the critical points of the acceleration function and then determine whether each critical point is a maximum or minimum using the second derivative test. Let's go through the process step by step.
Step 1: Calculate the acceleration function.
Based on the given position function, x(t) = t^4/24 - t^3/2 + 2t^2 - 1, the acceleration can be obtained by taking the second derivative of x(t) with respect to time (t).
Taking the derivative of x(t) gives us the velocity function:
v(t) = (d/dt) x(t) = (d/dt) (t^4/24) - (d/dt) (t^3/2) + (d/dt) (2t^2) - (d/dt) 1
v(t) = t^3/6 - (3/2)t^2 + 4t
Now, taking the derivative of v(t) gives us the acceleration function:
a(t) = (d/dt) v(t) = (d/dt) (t^3/6) - (d/dt) (3/2)t^2 + (d/dt) 4t
a(t) = (1/6)t^2 - 3t + 4
Step 2: Find the critical points.
To find the critical points, we need to solve for t when a(t) = 0.
(1/6)t^2 - 3t + 4 = 0
This is a quadratic equation, and you can either factor it or use the quadratic formula to find its roots.
Step 3: Determine the nature of the critical points using the second derivative test.
Once you have the critical points, you need to determine whether each critical point is a maximum or minimum using the second derivative test.
Taking the derivative of a(t) gives us the second derivative:
a'(t) = (d/dt) (1/6)t^2 - 3t + 4
a'(t) = (1/3)t - 3
To determine the nature of the critical points, substitute the values of the critical points for t into a'(t). If a'(t) is positive, the critical point is a minimum; if a'(t) is negative, the critical point is a maximum.
In this case, when t = 3, a'(3) = (1/3)(3) - 3 = 0.
Since a'(3) = 0 and the second derivative is positive, the critical point at t = 3 is a minimum, not a maximum.
Therefore, there is no maximum acceleration on the interval 0 <= t <= 4.