Problem Eleven: 1999-2000
Calculus I (AB): Stop ... Hubble Time (Absolute Extrema)

       The Hubble Space Telescope was deployed on April 24, 1990, by the space shuttle Discovery. A model for the velocity of the shuttle during this mission, from liftoff at t = 0 seconds until the solid rocket boosters were jettisoned at t = 126 seconds, is given by

(in feet per second). Using this model, estimate the absolute maximum and mimimum values of the acceleration of the shuttle between liftoff and the jettisoning of the boosters.

Difficulty Rating: 
Photo courtesy of NASA
Solution:
First of all, we want to find the absolute extrema of acceleration, the derivative of velocity. Remember that absolute extrema can occur at critical numbers or at endpoints of a closed interval, so we need to find the critical numbers of acceleration, which is done by finding the derivative of acceleration and determining where the derivative is zero and where is is nonexistent.

Now, we know the extrema will occur at either the endpoints (x = 0, 126) or at the newly-found critical number (x = 23.11571941). We plug each of these into the acceleration to see which given acceleration its absolute maximum and minimum values. We get that a(0) = 23.61, a(23.1157) = 21.52288, and a(126) = 62.86858. Thus, the absolute minimum on the interval is 21.52288, and the absolute maximum us 62.86858.

Calculus II (BC): Fundamental Theorem Madness!
      Find a function f and a number a such that .
Difficulty Rating: 
Solution:
In order to do this problem, we first take the derivative of both sides to find f(x):
Now we can plug f(x) into the equation, simplify, integrate, and find a:


This week's problems are taken from Calculus: Concepts and Contexts, by James Stewart, who, ironically, shares the name of the running back I should have played this week in fantasy football since he was running against the Pittsburgh Steelers, but did I think of that? Noooo.
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