Question

An object moves along the x-axis, its position at each time t> 0 given by X(t)= 1/4t^4- t^3 + t^2 Determine the time interval(s), if any, during which the object moves right.

Answer 1

When \(x'(t) > 0\) it is moving right.

Answer 2

Do you know how to find derivatives @kizi239 ?

Answer 3

but i need to find the time intervals

Answer 4

Do you know how to find derivatives, yes or no?

Don't worry, I'll walk you through it, but I need to know what you can do.

Answer 5

yes i can

Answer 6

To give you a bit more confidence, remember that the solution to an inequality is an interval.

Answer 7

ok....

Answer 8

For example, the inequalty \(x < 2\) is the interval \((-\infty, 2)\)

Answer 9

ohhhh

Answer 10

So first of all, what is \(x'(t)\)?

Answer 11

is it 0?

Answer 12

Well, we are looking for when \(x'(t) = 0\), what I mean is, what is the derivative of \(x(t)\). Remember that \(x(t) =\frac{1}{4}t^4- t^3 + t^2\)

Answer 13

@kizi239 So can you take the derivative?

Answer 14

[t ^{3}-3t ^{2} + 2t\]

Answer 15

Great. Now when we want to factor this function.
If we can get it into the form: \[
(t-r_1)(t-r_2)(t-r_3) < 0
\]It'll be easier to find the intervals.

Answer 16

\[
x'(t) = t ^{3}-3t ^{2} + 2t < 0 \\
(t^{2}-3t + 2)(t) < 0
\]We want to use the quadratic formula to find the roots of the polynomial \(t^{2}-3t + 2\)

Answer 17

ok

Answer 18

@kizi239 Will you do it?

Answer 19

-2 and 8

Answer 20

Wait, does that make sense? \(-2 + 8 = 6 \quad -2 \times 8 = -16\).
Ummmm

Answer 21

\[
\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(2)}}{2(1)}

\]

Answer 22

thats wat i did

Answer 23

\[
\begin{split}
\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(2)}}{2(1)}
&=\frac{3\pm\sqrt{9-8}}{2} \\
&=\frac{3\pm1}{2} \\
&=\frac{4}{2},\quad \frac{2}{2} \\
&= 2, \quad 1
\end{split}
\]

Answer 24

Maybe you messed up somehow.

Answer 25

ohh i made a mistake srry

Answer 26

So anyway, we end up with: \[
(t-2)(t-1)(t) <0
\]

Answer 27

txs so much

Answer 28

Forgive me, I made a mistake... it should be \[
(t-2)(t-2)(t) > 0
\]

Answer 29

ok txs

Answer 30

that 2 should be a 1, sorry somehow I'm messing up.

we're still not done yet though!

Answer 31

So we are multiplying three factors: \((t-1)\), \((t-2)\), and \(t\).
We want them to be greater than \(0\).
So we have to remember our rules about multiplying negatives positives...

We want either all of the factors to be positive.
Or we want one factor to be positive and two to be negative. The two negatives will cancel out.

Answer 32

They told us at the beginning that \(t>0\), so we know that the \(t\) factor will ALWAYS be positive.

Answer 33

So we want the remaining factors to both be positive or both be negative.
Basically: \[
t-1 > 0, \quad t-2 > 0
\]O: \[
t-1 < 0, \quad t-2 < 0
\]Does that make sense?

Answer 34

This gives us \[
t > 1, \quad t > 2
\]Which is just \(t > 2\).
Or \[
t < 1, \quad t < 2
\]Which is just \(t < 1\).

Answer 35

That gives us 4 inequalities all together: \[
t > 0 \\
t < 1 \\
t > 2 \\
t < \infty
\]Which is just \[
0 < t < 1 \\
2 < t < \infty
\]In interval notation, this is just \[
\forall t \in [0,1]\cap [2, \infty) \quad v'(t) > 0
\]

Answer 36

@kizi239

Think you got it?