Related RA

1. Jul 13, 2011

tagrico

edit: meant to put rates of change as the title, not related rates.
1. The problem statement, all variables and given/known data
A particle moves on a vertical line so that its coordinates at time t is s(t) = 4t^3 - 12t^2 +7, where t≥0
a) when is the particle moving upward and when is it moving downward
b) find the total distance that the particle travels in the time interval 0≤t≤ 3

3. The attempt at a solution
so i'm pretty sure I would take the derivative to find the velocity function, but how do I find out when it's moving upward and when it is moving downward? positive = upward and negative = downward?

2. Jul 13, 2011

Dick

Sure, they are sort of implying since the motion is along a vertical line that s(t) is distance along that line upward from the origin. So positive derivative is upwards.

3. Jul 13, 2011

tagrico

When I take the derivative I get 12t^2 -24t. Are you saying this can only be a positive function?

4. Jul 13, 2011

Dick

No! I'm saying for values of t where the derivative is positive motion is upwards, for values of t where the derivative is negative, it's downwards. The question is when (for what values of t) is it upwards and for which is it downwards.

5. Jul 13, 2011

tagrico

Gotcha! I'm stuck on which values of t is would be upward and downward. Other than plugging in random values of t into the velocity function, how would I go about doing this?

6. Jul 14, 2011

HallsofIvy

Staff Emeritus
Factor it. $12t^2- 24t= 12t(t- 2)$. That will be positive when the two variable factors, t and t-2, have the same sign, negative when they have different signs.

Also, "a- b" is positive if and only if a> b, negative if and only if a< b so the signs of t- 2 and t= t- 0 depende upon whether t> 2 and t> 0. If t< 0, then t< 2 also so t and t- 2 are both negative. If 0< t< 2, t is positive and t- 2 is negative. If t> 2 then t> 0 also so t and t- 2 are both positive.