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