# A particle travels at 2 units per second along the curve y=x^3-3x^2+2. How long does it take to travel from the local maximum to the local minimum?

Dec 21, 2017

#### Explanation:

the answer is the curve length divided by the speed

to find curve length, first find the x-values of the local min and max (there's either one of each or neither as y(x) is a cubic).

to find local min/max, first find $y ' \left(x\right)$:
$= 3 {x}^{2} - 6 x$

$y ' \left(x\right)$ equals 0 at the local min/max:
$3 {x}^{2} - 6 x = 0$
$3 x \left(x - 2\right) = 0$
$x = 0 , 2$

so the curve length is from $x = 0$ to $x = 2$

curve/arc length using formula = ${\int}_{0}^{2} \sqrt{1 + {\left(f ' \left(x\right)\right)}^{2}} \mathrm{dx}$
$= {\int}_{0}^{2} \sqrt{1 + {\left(3 {x}^{2} - 6 x\right)}^{2}} \mathrm{dx}$
$= {\int}_{0}^{2} \sqrt{1 + 9 {x}^{4} - 36 {x}^{3} + 36 {x}^{2}} \mathrm{dx}$

i'm not sure how to integrate that, so if you use a calculator,
${\int}_{0}^{2} \sqrt{1 + 9 {x}^{4} - 36 {x}^{3} + 36 {x}^{2}} \mathrm{dx}$ is about 4.592 units.

time elapsed: 4.592/2 seconds = about 2.296 seconds

Dec 21, 2017

Approximately $2.3$ seconds.

#### Explanation:

This is an interesting problem. Let's start by finding the local max/min.

$y ' = 3 {x}^{2} - 6 x$

This will have critical values when $y ' = 0$.

$0 = 3 {x}^{2} - 6 x$

$0 = 3 x \left(x - 2\right)$

$x = 0 \mathmr{and} 2$

We are guaranteed that one of these will be the local max and the other local min, because those are the only two options considering the nature of the function.

Now we want to find the arc length of the curve from $0$ to $2$, because this will be the distance travelled.

The formula for arc length is

$A = {\int}_{a}^{b} \sqrt{1 + {\left(y '\right)}^{2}} \mathrm{dx}$

In our case

$A = {\int}_{0}^{2} \sqrt{1 + {\left(3 {x}^{2} - 6 x\right)}^{2}} \mathrm{dx}$

$A = {\int}_{0}^{2} \sqrt{1 + 9 {x}^{4} - 36 {x}^{3} + 36 {x}^{2}} \mathrm{dx}$

Which according to the integral calculator has no elementary solution. But using an approximation calculator, or a computer algebra system, you should get the arc length to be approximately $4.592$ units. Since $s = \frac{d}{t}$, where $s$ is speed , $d$ is distance and $t$ is time, $t = \frac{d}{s}$, thus $t = \frac{4.592 u}{2 \frac{u}{s}} \approx 2.3 s$

Hopefully this helps!