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?
Perhaps your derivative is wrong
Using the quotient rule, I got a derivative of
-4t(t^3 - 8)/(4+t^3)^2
setting that to zero gave me
t - 2 or t = 0
for the given domain, t=2
v(2) = 16/12 = 4/3
To determine the maximum and minimum velocities over the given time interval, we first need to find the critical points of the function. We can do this by finding when the derivative of the function is equal to zero or undefined.
First, let's find the derivative of the velocity function, v(t):
v'(t) = d/dt (4t^2 / (4 + t^3))
To simplify this, 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
v'(t) = ( (4 + t^3)(8t) - (4t^2)(3t^2) ) / (4 + t^3)^2
Now, let's set v'(t) equal to zero and solve for t:
( (4 + t^3)(8t) - (4t^2)(3t^2) ) / (4 + t^3)^2 = 0
Simplifying the equation, we get:
(4 + t^3)(8t) - (4t^2)(3t^2) = 0
Expanding and rearranging the equation:
8t^4 + 8t^4 + 32t - 12t^4 = 0
4t^4 + 32t = 0
t(4t^3 + 8) = 0
From this equation, we can see that t = 0 is a critical point. However, since t > 0 in this case, we can ignore this solution.
Next, let's find when the derivative is undefined. We need to check when the denominator of the derivative equation, (4 + t^3)^2, equals zero:
4 + t^3 = 0
t^3 = -4
Taking the cube root of both sides:
t = -∛4
Since t must be greater than or equal to zero, we can ignore this solution as well.
Now, we need to evaluate the function v(t) at the critical points and the endpoints of the given time interval (1 < t < 4).
v(1) = 4(1)^2 / (4 + (1)^3) = 4/5
v(4) = 4(4)^2 / (4 + (4)^3) = 16/17
Therefore, the minimum velocity over the interval 1 < t < 4 is 4/5, which matches your result.
However, the book states that the maximum velocity at t = 4 is 4/3, which is different from your result of 16/17. Let me explain how to find the correct value.
To determine the maximum velocity, we need to consider not just the critical points, but also the endpoints of the given time interval.
So, let's evaluate the function v(t) at the endpoints:
v(1) = 4/5
v(4) = 16/17
Comparing these values, we can see that the maximum velocity is indeed 16/17, as you calculated correctly. It seems there might be an error in the book's answer where it states the maximum velocity at t = 4 as 4/3.
In conclusion, according to our calculations, the maximum velocity over the time interval 1 < t < 4 is 16/17, and the minimum velocity is 4/5.