# Distance formula

• Feb 22nd 2008, 10:55 AM
XIII13Thirteen
Distance formula
Distance: $s=-16t^2+800t$

find t, for s increasing, for s'=0, and for s decreasing

Am I figuring maximums and minimums here?

$y'=-32t+800$ = $y'=-t+25$

so I have a critical point at x=25

$y''=-1$

So our singular critical point is a maximum. So t>25 we start to decrease?

I have a multiple choice solution set

a.) (0,25]
increasing
{0,50}
s=0

b.)[25,50]
decreasing
{0,50}
s=0

c.) {0,50}
s=0

d.) None of these

Keep in mind. He likes to trick us and I've noticed a lot of "none of these" resulting from things like positive and negative signs in the equation. Part of the thing confusing me is how he wrote up the solution set
• Feb 22nd 2008, 11:19 AM
galactus
As you probably know, it is a concave down parabola with vertex at
V(25,10000).

It follows that $f'(x) < 0 \;\ if \;\ 25 < x < {\infty}$

and $f'(x) > 0 \;\ if \;\ {-\infty}

Since f(x) is continuous at x=25:

Therefore it is decreasing over the interval $[25,\infty)$ and increasing over the interval $(-\infty, 25]$

$f''(x) < 0---->f''(x) = -32$, so it is concave down. We can tell that from the negative coefficient of x^2.