- Download PDF
An object moves in a straight line. Its velocity, in m/s, at time t is v(t)=4t^2/(4+t^3), t≥0. Determine the maximum and minimum velocities on the interval [0,4].
Determine the maximum and minimum velocities over time interval 1<t<4
I found minimum 1= 4/5 but max 4= 16/17
The book says, maximum=4/3 Explain please.
1 Answer | Add Yours
In Calculus, derivatives are used to find critical points where maximum and minimum values occur. To be considered a critical point:
a) Use the endpoints of the interval,
b) Find where the derivative does not exist in the interval,
c) Find where the derivative equals zero.
In this problem:
a) The endpoints are at t=0 and t=4.
b) Using the quotient rule:
The only t-value this would not exist at is the cube root of -4, which is not in the interval. No critical points are considered here.
t=0 or t=2 are the critical values where the derivative is equal to 0.
The overall values to consider are 0, 2, and 4.
v(0) = 0
From the graph, the velocity from 0 to 2 is increasing. From 2 to 4 it is decreasing.
The minimum velocity is 0 m/s at time t=0.
The maximum velocity is 4/3 m/s at time t=2.
We’ve answered 319,336 questions. We can answer yours, too.Ask a question