We started off class with the question, “The side of a square is increasing at 2 cm/sec. How fast is the area increasing when the side is 4 cm long?” This brilliant question by O’Brien given with a wondrous animation on GeoGebra stumped our class. We kept asking questions like “at what time does the area start increasing? 0 seconds?” and “can’t you just tell us how to do it?” After a while of working on it our class decided on our answer: 8 cm^2/sec, which was wrong.
When O’Brien started actually helping us with this confusingly simple problem he threw our question right back at us: “Does it really matter at what number of seconds you start? 0 seconds? 1 seconds? 16 seconds?” He used a table (shown below) to prove that no, it doesn’t matter, no matter time you start at the volume is increasing at a steady rate.
We further proved this by looking at the rate of change at a moment in time instead of over 2 seconds. We did this by again, using a table (shown below) and looking at the times 1 second and 1.0001 seconds.
We noticed that for the change between the two times and between the two areas were both reeally close to zero, so we thought zero/zero = zero right? Or undefined? Then we thought, but wait, these two numbers aren’t equal to zero, they’re both just really close which means we’ll actually get a value here:
Once we did this we realized we just found the derivative by using the limit in indeterminate form. The derivative for this particular problem at this moment in time is about 16.0004.
After all this work O’Brien got his evil smile on to let us know there’s actually a much easier way to do all this. Thanks for letting us know before hand O’Brien. To use this easier way, we must first find the equation to this problem. By looking at our tables and the question asked in this problem it’s easy to see that our equation must be:
To solve for the rate of the area we must differentiate this equation with respect to t (time) (basically implicit differentiation):
Once again we proved that the rate or derivative at a moment in time is 16 cm^2/sec (above when we found the rate at a moment in time is was about 16). Important things to notice in the math above is that once we found the equation and differentiated it we plugged in the values we knew for s and ds/dt which were given in the original problem. It said “the side of a square is increasing at 2 cm/sec” which is ds/dt and that we are to solve for “when the side is 4 cm long” which is s.
Simple enough right?
The basic idea behind related rates that I’ve found O’Brien’s teachings and most of the sites and videos I’ve looked at to agree with is that each variable is now a function of time, or t, just like we did above. That means when we take the derivatives, we won’t be taking them in respect to each other like we used to, but instead with respect to t. This is important to know because some of the variables will not be changing with respect to t. These variables are like constants, and the derivative of them with respect to t is 0. When I say the derivative “with respect to t” I just mean that when you take the derivative of an equation you must multiple the variables in both sides by d/dt, for an example you can see we did this in the problem above.
And seeing as the name of the game is “related rates” it would be important to know if you didn’t already that something per something else, for instance cm/sec, would be a rate. So in an equation, whenever you see anything like that, it’s a rate and most likely important.
On the back of the exploration the Steps for every related rates problem: were given and I’ve found that almost every site or video on related rates uses basically the same steps so take note! They are important and helpful!
1. Label given information (Sketch where appropriate)
2. Note which quantities are variable, and which are constant
3. Identify rates d( )/dt which are given, and which are needed
4. Differentiate
5. Plug in date - ALWAYS AFTER YOU DIFFERENTIATE
The “beautiful man” (as Scotty would put it) Patrick JMT put it another way which I found helpful, he used an acronym: DREDS
Diagram
Rates
Equation
Derivative
Substitute specific info
He did that in this video, the rest was also helpful
Also for future reference Patrick JMT actually has a website and he has videos on there for every math concept imaginable so if you’re ever stuck look here! and *GASP* you can actually kind of see what he looks like!
And then while looking for other sites and videos on Related Rates which I thought would be helpful I’ve found this is actually quite the enjoyable experience because the people who make youtube videos on math problems are all so...interesting especially since most, unlike Patrick JMT who you only shows he’s hand in the videos, actually have their whole body in it. So there’s the expected nerdy, wonderfully awkward people trying to explain a math concept in a youtube video and then there’s this guy, Dr. Bob, who looks like he could be a body builder or a wrestling coach or something and is instead, on youtube being a math nerd, who, and I quote, said that he has a method for related rates problems that will “let us tear them down from start to finish.” Enjoy.
I hope I did a decent job explaining Related Rates and you all enjoyed my scribe post! The next scribe will be Sarah Mayberry, good luck!
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