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?

Next scribe will be ?

Scribe Post 12/5

At the start of class, O'Brien told us that unit three was finally drawing to a close, and that our final topic, related rates, would occupy our final two classes before the test, and that we would be supercorrecting the test going into our Christmas -- Hanukka -- Quanza -- Saturnalia... HOLIDAY break. Surely that's the best gift of all: forgiveness for our failures on opportunity day.

But before class could begin, a scribe had to be chosen, for Eddie, awake until 1:30 the night before slaving away on his scribe post, had forgotten to name his successor. The class quickly divided into two factions pleading not to be chosen: the Phyzards and the Bio-Nerds, complaining about "supercorrections" and "bio outlines" respectively. Even Gabe and Duncan of no-supercorrections did not step forward. Eddie was overwhelmed, but just as O'Brien was about to choose randomly from the mob, Crockett rose above the masses, a beacon of hope (see fig. 1) and volunteered his evening for the benefit of Phyzards and Bio-Nerds alike. Crockett's hair flowed back in a sudden classroom breeze and women swooned at his courage and selflessness.

fig. 1: Crockett emerges just as the Phyzards and Bio-Nerds are at each others throats.

Related Rates

O'Brien then gave us a problem:
Given a circular cone of height 10cm and radius r, increasing at a rate of 1 cm/s, what is the rate of change of the volume when the radius is 24?
O'Brien gave us a sort of loose procedure for problems like these:
1. Write what you know.
in this case we know that:
r = 24 cm
dr/dt = 1 cm/s
h = 10 cm
2. Draw a picture.
Drawing pictures makes it easier to find relationships. A helpful tactic is to superimpose your drawing over a set of axes, then maybe write some equations for lines that you see.

3. Find relationships.
To do this problem, we have to know that the volume of a cone is equal to 1/3Bh, or 1/3πr^2h
O'Brien told us that if we have equations for our variables, it's not always best to immediately substitute those in. However, if the value is a constant, we should substitute in immediately in order to avoid messy product rules.
Because h is constant, we substituted that in, leaving us with:
V = 10π/3 * r^2

4. Do calculus.
This is the easy part. If you've done the rest of the problem in a 
We're trying to find the time rate of change of V, so we have to differentiate the function we just found with respect to t. "But our equation for V is in terms of r," you might claim, but implicit differentiation allows us to take the derivative anyway:
V' = 10π/3 * 2r * dr/dt
What's that? We have a dr/dt term! But that's okay, because dr/dt and r are both given values:
V' = 10π/3 * 2(24) * 1


This video is a pretty good explanation of related rates problems:

There is a series of videos by PatrickJMT on related rates problems that you can find here:

ALEX CRANS WILL BE THE NEXT SCRIBE