Scribe Post 12/7

     We began a marvelous Friday with a fairly difficult quiz on optimization, related rates, linearization, and the like. After about 50 minutes, the quiz ended, the groans began, and class moved on.

     After settling back into our seats, we took a look at question 4 from Mr. O'Brien's AP Calculus exam from 1990!  Mr. O'B got all excited, recollecting his heroic score of a 6 on the AP exam, the only person to ever score higher than a perfect score. Who could have guessed that the question was all about related rates! An excited murmur traveled across the room, as we began to realize that it's not even halfway through the year, yet we have all the knowledge to fully answer a question on the terrifying AP exam. We began to do the three parts of question 4.


     Given that the radius "r" of a sphere is increasing at a constant rate of 0.04 centimeters per second, part A asked us to find the rate of increase of the sphere's volume when its radius measured 10 centimeters. The ever-important and oft-forgoten equation for volume of a sphere,

was also given in the problem. We went to work, first differentiating this equation for the volume of the sphere.

It is crucial to realize that we have plugged NO values in yet, we simply differentiated the entire equation. You can think of it as implicit differentiation if you'd like to, for that is what it technically is. It is also important to note that the constant, (4/3)π, remains unchanged when you take the derivative. From here, we simplified the derivative a little more, and then plugged the given values in.

But wait, how did we get the units? This is a key part of related rates: since volume is measured in centimeters cubed and time is measured in seconds, the answer must be measured in centimeters cubed per second. It makes a lot of sense - just match up the units and you're good!


     Part A is done, easy! On to part B. This asked us the rate of increase of the area of a cross section through the center of a sphere when the volume of the sphere is 36π. Let's start with a picture! That often makes things easier.

We're looking for the rate of increase of the area (of a section that is a circle), so we first must remember the equation for the area of a circle. Easy!

What do you know, it's differentiable! Let's take the derivative, in the same manner as we did in part A.

Hmm, that looks good but we're kinda stuck. We're trying to find the rate of increase of the area and we know the rate of increase of the radius, but we don't know the radius. But! The fact that the volume of the sphere was 36π was given at the beginning of the problem, so we can solve for the radius!

 Easy enough! Since we've already differentiated the equation for the rate of change of the area, let's plug in values.

And we're done! Onto part C.

     This part asked for the radius when the volume and radius of the sphere were increasing at the same rate. Let's set both of those things equal to each other.

Next, let's plug in the values we've calculated above.

 Simplify a little further.

 Finally, square root both sides of the equation to find r.

Remember, since r is in centimeters the answer will be as well. We're done! Not too bad, certainly not as scary as many of us thought an AP exam would be. To check these answers, or to see another person's working of this same problem, check out this site and type in "1990 AB4 Solution"in the "Find" menu.

     Also, please remember that the finely crafted opportunity day is on THURSDAY, so keep up with your IW and come to class ready to REVIEW next week! For additional help on related rates problems, check out this awesome applet about a melting snowball. If worse comes to worst, you can always count on our buddy PatrickJMT!

P.S. Ever wondered what he really looks like?

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