An object moves in a straight line. Its velocity, in m/s, at time t is v(t)=4t^2/(1+t^2),t≥0.
Determine the maximum and minimum velocities on the interval 1≤x≤4.
Find the derivative using the quotient rule.
Find the critical numbers.
Evaluate v(t) at the critical numbers and at 1 and 4
Lowest value is the minimum, highest the maximum
When I did this I got v'=8t/(1+t^2)^2 so t=0 is a critical number
v(0)=0
V(1)=2
v(4)=64/17
therefore v(4)=64/17 is the maximum
v(0)=0 is the minimum
Look at the graph on your calculator to verify your answers
but on the interval 1≤t≤4
v(1) is min, v(4) is max.
To determine the maximum and minimum velocities on the interval [1, 4], we need to find the critical points of the velocity function v(t) on this interval.
Step 1: Find the derivative of v(t) with respect to t.
v'(t) = d/dt (4t^2/(1+t^2))
To find the derivative of this function, we can use the quotient rule:
v'(t) = [(1+t^2)*(d/dt)(4t^2) - (4t^2)(d/dt)(1+t^2)] / (1+t^2)^2
Simplifying this expression, we get:
v'(t) = [(1+t^2)*(8t) - (4t^2)(2t)] / (1+t^2)^2
= (8t + 8t^3 - 8t^3) / (1+t^2)^2
= 8t / (1+t^2)^2
Step 2: Find the critical points by setting v'(t) = 0 and solving for t.
Set v'(t) = 0:
8t / (1+t^2)^2 = 0
Since the numerator is 0, we have:
8t = 0
This implies t = 0.
Step 3: Check for endpoints of the interval [1, 4] and any other points of interest.
We need to check the velocities at the interval endpoints and at the critical points between the interval endpoints to determine the maximum and minimum velocities.
Let's substitute the values t = 1 and t = 4 into our velocity function v(t):
v(1) = 4(1)^2 / (1 + (1)^2)
= 4/2
= 2 m/s
v(4) = 4(4)^2 / (1 + (4)^2)
= 64/17 ≈ 3.765 m/s
Since we only have one critical point, t = 0, and it lies outside the interval [1, 4], we don't need to consider it in this case.
Step 4: Identify the maximum and minimum velocities.
From our calculations, we found that v(1) = 2 m/s and v(4) ≈ 3.765 m/s. Therefore, the maximum velocity on the interval [1, 4] is 3.765 m/s, and the minimum velocity is 2 m/s.