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An object moves in a straight line. Its velocity, in m/s, at time t is...

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bobby455 | Student, Undergraduate | eNoter

Posted February 27, 2012 at 4:30 AM via web

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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.

Tagged with calculus, math

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taylormath | High School Teacher | (Level 2) Adjunct Educator

Posted February 27, 2012 at 5:48 AM (Answer #1)

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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:

`v'(t)=((4+t^3)(8t)-(4t^2)(3t^2))/((4+t^3)^2)`

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.

c)  `((4+t^3)(8t)-(4t^2)(3t^2))/((4+t^3)^2)=0`

`32t+8t^4-12t^4=0`

`-4t^4+32t=-4t(t^3-8)=0`

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

`v(2)=16/(4+8)=16/12=4/3`

`v(4)=64/(4+64)=64/68=16/17`

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.

Sources:

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