Tim Cahill:
"I am in Ellison's Cave, about to rappel down Incredible Pit, the second-deepest save pit in the continental United States. The drop is 440 feet, about what you'd experience from the top of a 40-story building. If you took the shaft in a free fall, you'd accelerate to more than 100 miles an hour and then - about five seconds into the experience-you'd decelerate to zero. And die."
Solution: Start with

take antiderivatives of both sides to get

Since the velocity is 0 at the beginning, we get

so that C = 0. Again taking antiderivatives, we get

Since the height at the beginning is 0, we get

so that D = 0. We now have our three equations:

How long does it take to fall 440 ft? That's the same question as asking, 'When will s(t) = -16t² = -440?' We get t = ²(27.5), which is 5.244, to three decimal places. That's about 5 seconds, so that part of the story is accurate. To find out how fast Tim is falling at that time, we need to plug t = ²(27.5) into our velocity equation, and get

which is about -167.81 ft/sec. To find out how fast this is in miles per hour, we need to change units several times:

so that is roughly about right, too.

The top of the figure is the curve y = 4sin(x), the bottom of the figure is the curve y = - x²/4, and the right side is a vertical line segment. However, the executive committee believes it is horrible, and forces him to agree to making it smaller. He agrees only to moving the vertical line segment left to the point where the line segment is as short as possible. To what point will have have to move it? (If at some point you get stuck, you might consider using Newton's method to approximate the solution to some equation.)
Solution: Making the line segment as short as possible means the vertical distance between the graph of y = 4sin(x) and y = -x²/4 as short as possible. In other words, minimize V = 4sin(x) - (-x²/4). By looking at the graph of 4sin(x), we see that the furthest left we would want to move the segment is x = Pi, and the furthest right we would want to move it would be x = 2*Pi. We start by taking the derivative, and getting

There are no places where V'(x) is undefined; setting it equal to 0, however, is a problem. We can't solve it! Algebraically, that is. Let's use Newton's method: f(x) = 4cos(x) + x/2, f'(x) = -4sin(x) + 1/2, and then x_n+1 = x_n - f(x_n)/f'(x_n). We have to start with a guess, and since we're only considering numbers between Pi and 2*Pi, I'll start with a guess of x₀ = 5. We summarize our work in the table below:

Since the numbers aren't changing anymore, I'll say that the critical point (when V'(x) = 0) is x = 4.16483091406. To finish the problem, we make a little table with all of the critical points, plus the endpoints, and evaluate them using V, and see which is smallest:

So, the shortest the segment ever gets is 0.9213, and it happens when x = 4.165...

where a > 0 is a constant.

1. Find P'(x), and construct a sign chart for the first derivative.
   Solution: Using the chain rule, we get P'(x) = -(1+a*e^-x)^-2*(-a*e^-x) = a*e^-x*(1+a*e^-x)^-2 = a*e^-x/(1+a*e^-x)²

   Since a fraction is 0 only when the top is 0, and our top is a*e^-x, which is never 0, this fraction is never 0. Since this fraction is undefined only when the bottom is 0, and our bottom is always at least 1, this fraction is never undefined. This means on our number line, there are no numbers to put, and we only have to do one test value, and I'll use x = 0:

   x	P'(x)
   0	+

   So, our derivative is always positive, so our function P(x) is always increasing.

2. Find P''(x), and construct a sign chart for the second derivative.
   Solution:Taking the derivative of our derivative a*e^-x*(1+a*e^-x)^-2 using the chain rule, we get P''(x) = -a*e^-x*(1+a*e^-x)^-2 -2a*e^-x*(1+a*e^-x)^-3*(-a*e^-x)
   = a*e^-x*(1+a*e^-x)^-3* (-1-a*e^-x+2a*e^-x)
   = a*e^-x*(1+a*e^-x)^-3*(-1+a*e^-x)
   = a*e^-x*(-1+a*e^-x)/(1+a*e^-x)³

   Now, a fraction is 0 only when the top is 0. Since the top is already factored, we only need to set each part = 0, and see what we get:
   When we set a*e^-x = 0, we get no solutions.
   When we set (-1+a*e^-x) = 0, we get x = ln(a).
   Our fraction is only not defined when the bottom = 0, but our bottom is always at least 1, so is never 0.
   Conclusion: The only critical value we get is x = ln(a). So, when we put ln(a) on our number line, we get two parts. Below, we summarize our work:

   Interval	Test value x	V''(x)	+/-
   x < ln(a)	ln(a/2) (which is less than ln(a))	a*(2/a)*(-1+a*2/a)/(1+a*(2/a))3 = 2/27	+
   x < ln(a)	ln(2a) (which is bigger than ln(a))	a*2a*(1+2a2)/(1+2a2)3	-

   Conclusion: Our function is always increasing, is concave up before x = ln(a), and is concave down afterwards.

3. Find

   lim	P(x)
   x ->

   and

   lim	P(x)
   x -> -

   Solution: As x -> , e^-x -> 0, so the whole fraction gets closer and closer to 1.
   As x -> -, e^-x gets larger and larger; the whole fraction thus gets smaller and smaller.

4. We need to draw a graph that way to the left gets closer and closer to 0, way to the right gets closer and closer to 1, is increasing the whole time, is concave up before ln(a), and is concave down afterwards. One sample graph:
   
5. This is a good graph for populations since they increase very fast at the beginning, but has a limiting value. (Usually because of lack of room, or food, or because the larger the population, the more likely some of the population will be subject to predators.)

Solution: We are being asked to minimize the surface area of this container, which is, in total, the area of the bottom (Pi*r²) + area of the side (2*Pi*r*h) + area of the hemisphere (2*Pi*r²), so

We'd like to take the derivative, but there are two variables, and we need to get rid of one. There is one fact we haven't used yet: The volume of the whole container is 1000 cm³. But we also know the volume of the whole container is the volume of the cylinder part (Pi*r²*h) + volume of the hemisphere part (2/3*Pi*r³), so

Solving this for h, we get

Plugging this value of h into our formula for T, the total surface area, we get

So this is the function we need to take the derivative of, and do a sign chart for, etc.

This derivative is undefined when r = 0 (which is okay; our container has no volume then, either!). To solve the equation T'(r) = 0, we need to factor out what the two terms have in common. I choose to factor out (10/3)*r^-2, and get

Setting this equal to 0 is now the same as setting each part equal to 0; we get no solution at all when we set 10/(3*r²) = 0, since a fraction is equal to 0 only when the top is equal to 0, and 10 is not 0!. Setting the second part equal to 0 gives us

which is approximately equal to 5.759. To do a sign chart, we don't worry about values of r < 0, since r must be positive for us to have a container at all: (Remember, if you can, let your calculator do the plugging-and-chugging!)

This shows us that before r = 5.759, the total surface area is decreasing, and after r = 5.759, the total surface area is increasing. So, the minimum surface area occurs when r = 5.759, and when h = 5.759.