text: Distance formula
  
  • Feb 22nd 2008, 10:55 AM
  XIII13Thirteen
  Distance formula
  Distance: $\displaystyle s=-16t^2+800t$
  
  find t, for s increasing, for s'=0, and for s decreasing
  
  Am I figuring maximums and minimums here?
  
  $\displaystyle y'=-32t+800$ =$\displaystyle y'=-t+25$
  
  so I have a critical point at x=25
  
  $\displaystyle y''=-1$
  
  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 $\displaystyle f'(x) < 0 \;\ if \;\ 25 < x < {\infty}$
  
  and $\displaystyle f'(x) > 0 \;\ if \;\ {-\infty}<x<25$
  
  Since f(x) is continuous at x=25:
  
  Therefore it is decreasing over the interval $\displaystyle [25,\infty)$ and increasing over the interval $\displaystyle (-\infty, 25]$
  
  $\displaystyle f''(x) < 0---->f''(x) = -32$, so it is concave down. We can tell that from the negative coefficient of x^2.