An object moves in a straight line. Its velocity, in m/s, at time t is v(t)=4t^2/(4+t^3), t>(greater or equal)0. Determine the maximum and minimum velocities over time interval 1<t<4
I got minimum as 1= 4/5. However for maximum at t=4, I got 16/17 but the book says 4/3?
To find the maximum and minimum velocities over the time interval 1 < t < 4, we need to first find the critical points of the function v(t) = 4t^2 / (4 + t^3).
To find the critical points, we need to calculate the derivative of v(t) and set it equal to zero:
v'(t) = (d/dt) (4t^2 / (4 + t^3)) = 0.
To simplify the calculation, we can use the quotient rule:
v'(t) = [(4 + t^3) * (d/dt) (4t^2) - 4t^2 * (d/dt) (4 + t^3)] / (4 + t^3)^2.
Differentiating the respective terms:
v'(t) = [(4 + t^3) * (8t) - 4t^2 * (3t^2)] / (4 + t^3)^2,
v'(t) = (32t + 8t^4 - 12t^4) / (4 + t^3)^2,
v'(t) = (8t - 4t^4) / (4 + t^3)^2.
Now, we set v'(t) equal to zero and solve for t:
(8t - 4t^4) / (4 + t^3)^2 = 0.
Setting the numerator equal to zero:
8t - 4t^4 = 0.
Factor out a common term t:
t(8 - 4t^3) = 0.
Setting each factor equal to zero:
t = 0 (extraneous solution, as t > 0 by the given condition),
8 - 4t^3 = 0.
Solving for t:
4t^3 = 8,
t^3 = 2,
t = ∛(2),
t ≈ 1.26.
Now, to classify these critical points as maximum or minimum, we can consider the second derivative test. We need to calculate the second derivative v''(t):
v''(t) = (d^2/dt^2) [(8t - 4t^4) / (4 + t^3)^2].
Differentiating the respective terms:
v''(t) = (8 - 16t^3) / (4 + t^3)^2.
Now, we substitute our critical points t = 0 and t ≈ 1.26 into v''(t) and determine the sign of v''(t) to classify the points:
At t = 0, v''(0) = (8 - 0) / (4 + 0)^2 = 8 / 16 = 0.5 > 0, so it is a minimum.
At t ≈ 1.26, v''(∛2) = (8 - 16(∛2)^3) / (4 + (∛2)^3)^2 = (8 - 16∛2) / (4 + (∛2)^3)^2 < 0, so it is a maximum.
Now, we can substitute these critical points into the original function v(t):
For the minimum velocity:
v(0) = 4(0)^2 / (4 + (0)^3) = 0 / 4 = 0 m/s.
For the maximum velocity:
v(∛2) = 4(∛2)^2 / (4 + (∛2)^3) = 4(2^(2/3)) / (4 + 2) = (4 * 2^(2/3)) / 6.
To compare this result with the book's answer, we can simplify it further:
(4 * 2^(2/3)) / 6 ≈ 1.08.
Therefore, according to the calculations, the maximum velocity over the time interval 1 < t < 4 is approximately 1.08 m/s, not 4/3 as stated in the book.
To find the maximum and minimum velocities over the time interval 1 < t < 4, we need to first determine the critical points of the velocity function. We can do this by finding where the derivative of the velocity function is equal to zero or is undefined.
Given that the velocity function is v(t) = (4t^2) / (4 + t^3), let's find its derivative:
v'(t) = [(8t)(4 + t^3) - (4t^2)(3t^2)] / (4 + t^3)^2
We can set v'(t) = 0 and solve for t to find the critical points:
[(8t)(4 + t^3) - (4t^2)(3t^2)] / (4 + t^3)^2 = 0
Simplifying the equation:
8t(4 + t^3) - 12t^2 (t^2) = 0
8t(4 + t^3) - 12t^4 = 0
8t + 8t^4 - 12t^4 = 0
t(8 - 4t^3) = 0
From this equation, we have two critical points: t = 0 and t = 2. In the given time interval, only t = 2 lies within the range 1 < t < 4. Now, we need to check the endpoints of the interval as well, which are t = 1 and t = 4.
To find the maximum and minimum velocities, we substitute these values into the velocity function:
v(1) = (4(1)^2) / (4 + (1)^3) = 4 / 5
v(2) = (4(2)^2) / (4 + (2)^3) = 16 / 17
v(4) = (4(4)^2) / (4 + (4)^3) = 4 / 3
Therefore, the minimum velocity over the interval 1 < t < 4 is 4 / 5, and the maximum velocity is 4 / 3.
It seems there was an error in the book, and you were correct in finding the maximum velocity as 16 / 17 at t = 2.