# Thread: How to tell where this function is increasing?

1. ## How to tell where this function is increasing?

I am given a position equation
$s(t)=2t^3-24t^2+90t+7$ for t is greater than or equal to 0. I am asked to find the values of t where the speed is increasing.

If I'm correct, speed is the absolute value of velocity, which is the derivative of position. I got:
$y=$| $6t^2-48t+90$|
(not sure how to input absolute value symbol)

How do I know where the speed is increasing without plugging it into the calculator? To find where velocity increases, I found the zeros and then used a sign line, but I don't think that works for speed...or does it?

2. Originally Posted by pocketasian
I am given a position equation
$s(t)=2t^3-24t^2+90t+7$ for t is greater than or equal to 0. I am asked to find the values of t where the speed is increasing.

If I'm correct, speed is the absolute value of velocity, which is the derivative of position. I got:
$y=$| $6t^2-48t+90$|
(not sure how to input absolute value symbol)

How do I know where the speed is increasing without plugging it into the calculator? To find where velocity increases, I found the zeros and then used a sign line, but I don't think that works for speed...or does it?
how do you find where a function is increasing? you find its derivative and see where the derivative is positive, right? Do the same thing here. To tell where the speed is increasing, we must find its derivative (that is, the acceleration function) and see where it's positive.

3. Originally Posted by pocketasian
How do I know where the speed is increasing without plugging it into the calculator?
speed is increasing whenever velocity and acceleration have the same sign.

4. Originally Posted by skeeter
speed is increasing whenever velocity and acceleration have the same sign.
That's what I would think, but how do I confirm analytically the interval that speed is increasing on?

5. Find the zeros of the equation:

$s'(t)=6t^2-48t+90 \Longrightarrow s'(t)=6(t^2-8t+15)$

Test around the critical values obtained above, and you will see in what intervals your equation is positive, and which are negative.

6. Originally Posted by ANDS!
Find the zeros of the equation:

$s'(t)=6t^2-48t+90 \Longrightarrow s'(t)=6(t^2-8t+15)$

Test around the critical values obtained above, and you will see in what intervals your equation is positive, and which are negative.
If I did that, wouldn't I actually be finding where the velocity is increasing and decreasing? That's what I did at first to see that it increases from 0 to 3, and from 5 to infinity, but my answer did not match up with the book's. Since the problem is asking for the intervals on which speed is increasing, wouldn't that be different seeing how speed is the absolute value of velocity and not the derivative of position?

7. $s(t)=2t^3-24t^2+90t+7$ t \ge 0

$v(t) = 6t^2 - 48t + 90 = 6(t-3)(t-5)
$

$v(t) > 0$ ... $0 \le t < 3$ and $t > 5$

$v(t) < 0$ ... $3< t < 5$

$a(t) = 12t - 48 = 12(t-4)$

$a(t) < 0$ ... $0 \le t < 4$

$a(t) > 0$ ... $t > 4$

v(t) and a(t) both positive for $t > 5$

v(t) and a(t) both negative for $3 < t < 4$

those are the intervals where the particle is speeding up.

8. My bad. If you are looking for the intervals where speed is increasing then you need to find the derivative of speed which is the a(t). All finding the zeros of the velocity function will do is to tell you where velocity is positive and negative. If you take the absolute value of the acceleration function (its a line so its easy to visualize), you'll see acceleration decreasing from like 0 to 3 (or somewhere around there), and increasing from then on (assuming the thing doesn't stop.