I had a great discussion with my Calculus class about how the exponential sequence 2^x is
1, 2, 4, 8, 16, ... but that there is nothing special about the base 2. There has to be a number "a", such that (3^a)^x = 2^x, because there has to be an exponent "a" such that 3^a = 2. Together they found that number, and we said that g(x) = 3^(0.631x) is therefore an approximation of f(x) = 2^x. Similarly, there must be an exponent k such that (e^k)^x = 2^x. So, the kids took a minute or so to find that k = ln(2), so h(x) = e^(0.693x) must also be an approximation of f(x) = 2^x. We discussed how h is easily differentiable now that we are Chain Rule pros.
I then set the kids loose on an activity that asks them to think flexibly about the exponential form. The reason why I like this activity is because the kids distinguish between an exact form that is "natural" for the situation versus the usual e^(kx) approximation. (The kids know that they can substitute k with an exact expression without losing precision, however.) For example, for #1 in the worksheet, where the kids are looking at bacteria that double every 6 minutes, both the kids and I think that it's natural to think of the sequence as {10, 20, 40, 80, ...} and to observe the general form y = 10(2)^x. To fix the problem that we want it to double only ONCE by x = 6 minutes, not doubling 6 times, we divide our exponent by 6 to get y = 10(2)^(x/6). For me, this is the natural way of writing the equation and testing it initially to make sure that it fits the bill. If they then want to differentiate it, they then turn it into base e, where e^k = 2^(1/6), so that y = 10(e)^(kx) replaces y = 10(2)^(x/6). The kids figure out that k = (1/6)*ln(2) or around k = 0.116. So, y = 10(e)^(0.116x) is an approximation of our exact function, and it has the benefit of being easily differentiable.
In thinking about exponential form as being fluid, the kids can consider equivalent compound-interest scenarios. I gave them a couple of scenarios to play with and to explain, in order to get at that idea. I am pretty happy with the level of understanding they have with this concept, seeing that it's the second time this year we've seen exponential compounding. After that, they didn't seem to have much trouble working through our practice quiz on the exponential topic. Overall, I am pretty happy with the way our differentiation technique unit has gone. We're moving a little bit slower than I had hoped, but their understanding of the connections between concepts has been really great!!! Both they and I are still excited to walk into this class everyday, and that's a good feeling. I anticipate that by January, we'll be wrapped up with all the differentiation techniques (including related rates problems, which I've been sprinkling into the mix periodically), and the kids will be ready to start thinking backwards and/or to do a differentiation project.
Stay tuned!
Saturday, November 23, 2013
Friday, November 15, 2013
Friday Fun Day
For my calculus class, Fridays we meet in a different classroom. I like the change of scenery and it always makes me think about doing some extra fun, irregular exercise things on Fridays with the kids.
They've been working a lot on Chain Rule, but ironically only with polynomials embedded in polynomials, polynomials embedded in sine and cosine, and vice versa. They've also done an itty bitty related rates. We haven't seen the product or the quotient rule, and just the Chain Rule has kept us really busy.
Anyhow, on this Friday Fun Day, I thought of introducing the idea that reciprocal trig functions can be differentiated using Chain Rule. I told them that in reality, they won't often use those formulas, but after my hint at the Chain Rule, they were able to successfully differentiate f(x) = csc(x) and g(x) = sec(x).
Then, with only 10 minutes left in the class, I asked them to pair up and to assign one person as partner A, and the other as partner B. I asked each person to pull out their graphing calculator and to graph y=e^x. Then, they together pick an x value. Partner A needs to find f(x) and Partner B needs to find f'(x) using the calculator, and to compare answers to see what they notice.
In about 60 seconds, one kid said, "This is a trick!!"
haha. But, I wasn't prepared for their question that followed, as to WHY e^x is its own derivative. After class I did a bit of research and thought I'd post it here in case you Calculus teachers out there are wondering the same. The standard proof involves knowing how to differentiate ln(x), but as we haven't gotten there yet, I think this is a better explanation:
e is a value that comes from continuous compounding formula, namely the part (1 + 1/n)^n, limit taken as n approaches infinity.
e^x is therefore the limit as n approaches infinity of (1 + 1/n)^(nx)
By binomial expansion (which unfortunately my kids have never seen), this looks like:
1^(nx) + (nx choose 1) 1^(nx - 1)(1/n) + (nx choose 2) 1^(nx - 2)(1/n)^2 + (nx choose 3) 1^(nx - 3)(1/n)^3 + (nx choose 4) 1^(nx - 4)(1/n)^4 + ....
= 1 + (nx)(1/n) + [(nx)(nx - 1)/2] (1/n)^2 + [(nx)(nx - 1)(nx - 2)/(2*3)] (1/n)^3 + [(nx)(nx - 1)(nx - 2)(nx - 3)]/(2*3*4)(1/n)^4 + ....
As n approaches infinity, this becomes
e^x = 1 + x + x^2/2 + x^3/(2*3) + x^4/(2*3*4) + ...
So if you differentiate each term with respect to x, you get:
d(e^x)/dx = 0 + 1 + x + x^2/2 + x^3/(2*3) + .... which is the same as the original sequence.
BAM. It's a trick. ;)
PS. I am mad that I tried to tell my husband on Friday about my e^x discovery, and he said off-handedly (before I gave him any mathematical details), "Oh yeah, I remember expanding e^x and then each term in the expansion has a derivative term that maps back to the original expansion." DAMN. He's 32 and doesn't remember most things most times; why does he still remember this?!?! That's just not right.
PPS. Yes, we have this sort of conversations. It's very geeky.
They've been working a lot on Chain Rule, but ironically only with polynomials embedded in polynomials, polynomials embedded in sine and cosine, and vice versa. They've also done an itty bitty related rates. We haven't seen the product or the quotient rule, and just the Chain Rule has kept us really busy.
Anyhow, on this Friday Fun Day, I thought of introducing the idea that reciprocal trig functions can be differentiated using Chain Rule. I told them that in reality, they won't often use those formulas, but after my hint at the Chain Rule, they were able to successfully differentiate f(x) = csc(x) and g(x) = sec(x).
Then, with only 10 minutes left in the class, I asked them to pair up and to assign one person as partner A, and the other as partner B. I asked each person to pull out their graphing calculator and to graph y=e^x. Then, they together pick an x value. Partner A needs to find f(x) and Partner B needs to find f'(x) using the calculator, and to compare answers to see what they notice.
In about 60 seconds, one kid said, "This is a trick!!"
haha. But, I wasn't prepared for their question that followed, as to WHY e^x is its own derivative. After class I did a bit of research and thought I'd post it here in case you Calculus teachers out there are wondering the same. The standard proof involves knowing how to differentiate ln(x), but as we haven't gotten there yet, I think this is a better explanation:
e is a value that comes from continuous compounding formula, namely the part (1 + 1/n)^n, limit taken as n approaches infinity.
e^x is therefore the limit as n approaches infinity of (1 + 1/n)^(nx)
By binomial expansion (which unfortunately my kids have never seen), this looks like:
1^(nx) + (nx choose 1) 1^(nx - 1)(1/n) + (nx choose 2) 1^(nx - 2)(1/n)^2 + (nx choose 3) 1^(nx - 3)(1/n)^3 + (nx choose 4) 1^(nx - 4)(1/n)^4 + ....
= 1 + (nx)(1/n) + [(nx)(nx - 1)/2] (1/n)^2 + [(nx)(nx - 1)(nx - 2)/(2*3)] (1/n)^3 + [(nx)(nx - 1)(nx - 2)(nx - 3)]/(2*3*4)(1/n)^4 + ....
As n approaches infinity, this becomes
e^x = 1 + x + x^2/2 + x^3/(2*3) + x^4/(2*3*4) + ...
So if you differentiate each term with respect to x, you get:
d(e^x)/dx = 0 + 1 + x + x^2/2 + x^3/(2*3) + .... which is the same as the original sequence.
BAM. It's a trick. ;)
PS. I am mad that I tried to tell my husband on Friday about my e^x discovery, and he said off-handedly (before I gave him any mathematical details), "Oh yeah, I remember expanding e^x and then each term in the expansion has a derivative term that maps back to the original expansion." DAMN. He's 32 and doesn't remember most things most times; why does he still remember this?!?! That's just not right.
PPS. Yes, we have this sort of conversations. It's very geeky.
Thursday, November 14, 2013
Organizing Information for Related Rates
I started doing an itty-bitty bit of related rates with my kids today, and it was totally fun! I only did very basic problems with them so far, because they've been asking about where we would encounter chain rule. My goal this month/early next month is to finish off all the basic differentiation rules and skills, then to come back to related rates later to do a thorough study, so I gave them only the first few problems (Level 1) from Bowman today, just enough to wet their appetite.
And I gave them this grid to help them organize their thinking. I tried doing all the problems (up through Sam's problem on the two flies on the inflatable earth, which I consider Level 4), and they all fit into this grid pretty well. I hope that helps!
I LOVE RELATED RATES!!!!!
By the way, some of my kids solved for rate of change in radius like this:
dr/dt = dr/dA * dA/dt
others did:
dA/dt = dA/dr * dr/dt and divided both sides by dA/dr.
I like this! It's a good opportunity for us to come together as a class tomorrow to discuss equivalence of equations involving derivatives, so that they can learn to think more flexibly about the Chain Rule already.
I am also instituting very structured group work this term, which has been very helpful in approaching new tasks, at least in this class (and also in my Alg 2 classes). The kids haven't really commented on it, but I find that their discussions are more productive now that we are into our second grouping and they are more used to the idea of working with people and having assigned leadership roles. Excited about the promise of the rest of the year!!!
And I gave them this grid to help them organize their thinking. I tried doing all the problems (up through Sam's problem on the two flies on the inflatable earth, which I consider Level 4), and they all fit into this grid pretty well. I hope that helps!
I LOVE RELATED RATES!!!!!
By the way, some of my kids solved for rate of change in radius like this:
dr/dt = dr/dA * dA/dt
others did:
dA/dt = dA/dr * dr/dt and divided both sides by dA/dr.
I like this! It's a good opportunity for us to come together as a class tomorrow to discuss equivalence of equations involving derivatives, so that they can learn to think more flexibly about the Chain Rule already.
I am also instituting very structured group work this term, which has been very helpful in approaching new tasks, at least in this class (and also in my Alg 2 classes). The kids haven't really commented on it, but I find that their discussions are more productive now that we are into our second grouping and they are more used to the idea of working with people and having assigned leadership roles. Excited about the promise of the rest of the year!!!
Graphically Analyzing Inequalities and Equations Flexibly
I've been doing inequalities and equations in the coordinate plane with my Algebra 2 kids, and I just love it. I see extensions from this topic to lots of others.
We started by looking at the meaning of <, >, and = inside the coordinate plane. We established quickly that when two things are equal in the coordinate plane, they must overlap. For example, (6, 2)= (6, 2) because they overlap. Also, y = x + 3 is equal to y = x + 3 because they overlap.
Then, we discussed that < means "lower than" and > means "higher than."
For any given equation or inequality, we would graph it as Y1 and Y2.
For example, 2x + 3 = 3x - 8, means that Y1=2x + 3 has to overlap with Y2=3x - 8. The kids sketch the graph, label the intersection, and answer in terms of x: x = 11. Easy breezy.
For inequalities, the kids learn to first analyze them using graphs before reviewing how to solve them by hand. For example, 2x + 3 < 3x - 8, means that Y1=2x + 3 has to be lower than Y2=3x - 8. The kids sketch the graph, label the intersection, and draw a vertical dashed line at x = 11 to separate the plane into two sections. They put their hands to cover one of the halves of the plane and ask, "Is Y1 actually lower than Y2 here?" If so, they shade it.
It looks something like this:
and from that, the kids can conclude that the solutions for x are all to the right of x = 11, so they write down x > 11.
On the quiz for inequalities, my kids had 100% accuracy on all the equations and inequality questions because they were asked to show their work two ways, one by hand and one by the graph, for each equation or inequality. They needed to get the same answer using both methods, in order to gain full reflection points (as well as accuracy and process points). Most of the kids were also able to explain why two parallel lines would cause an equation to have no solution, and inequalities to have either infinite solutions or no solutions, depending on the direction of the inequality symbol.
What I love about this graphical emphasis is how extendable this is to different topics. Today in one class I asked the kids to solve -20 < -4x + 5 < 11, and half the class decided to solve it by hand and the other half decided to solve it by just looking at the graph. They determined that only the middle part of the graph satisfies the inequality on both sides, and from that they determined -1.5 < x < 6.25.
And the graphing is extendable to absolute-value equations and inequalities, which loop in the idea of graphical symmetry (BEFORE the kids ever see quadratic graphs!).*
If, for example, my kids are asked to solve 2|x - 8| = 10, they can at this point sketch by hand the absolute-value graph centered at x=8, know that its slope is 2, and sketch another horizontal line at y=10. They can first assume the quantity (x - 8) is positive, so they can solve for x = 13 as one of the intersection points. Then, via symmetry around x = 8, they can then quickly figure out that x = 3 must be the other intersection point.
Although I plan to teach absolute-values a different way still, in order to reinforce the meaning of the various symbols, it has been delightful to preview the idea of functional symmetry and turning point before we ever see quadratics. The kids are doing so well on the abstract concepts so far! Totally kicking butt! YEAH.
*Also, the shading also is a good way of visualizing how to split the coordinate into different sections, delimited by x values, so their familiarity with this will really help out when we need to graph or analyze piecewise functions later this year.
We started by looking at the meaning of <, >, and = inside the coordinate plane. We established quickly that when two things are equal in the coordinate plane, they must overlap. For example, (6, 2)= (6, 2) because they overlap. Also, y = x + 3 is equal to y = x + 3 because they overlap.
Then, we discussed that < means "lower than" and > means "higher than."
For any given equation or inequality, we would graph it as Y1 and Y2.
For example, 2x + 3 = 3x - 8, means that Y1=2x + 3 has to overlap with Y2=3x - 8. The kids sketch the graph, label the intersection, and answer in terms of x: x = 11. Easy breezy.
For inequalities, the kids learn to first analyze them using graphs before reviewing how to solve them by hand. For example, 2x + 3 < 3x - 8, means that Y1=2x + 3 has to be lower than Y2=3x - 8. The kids sketch the graph, label the intersection, and draw a vertical dashed line at x = 11 to separate the plane into two sections. They put their hands to cover one of the halves of the plane and ask, "Is Y1 actually lower than Y2 here?" If so, they shade it.
It looks something like this:
On the quiz for inequalities, my kids had 100% accuracy on all the equations and inequality questions because they were asked to show their work two ways, one by hand and one by the graph, for each equation or inequality. They needed to get the same answer using both methods, in order to gain full reflection points (as well as accuracy and process points). Most of the kids were also able to explain why two parallel lines would cause an equation to have no solution, and inequalities to have either infinite solutions or no solutions, depending on the direction of the inequality symbol.
What I love about this graphical emphasis is how extendable this is to different topics. Today in one class I asked the kids to solve -20 < -4x + 5 < 11, and half the class decided to solve it by hand and the other half decided to solve it by just looking at the graph. They determined that only the middle part of the graph satisfies the inequality on both sides, and from that they determined -1.5 < x < 6.25.
If, for example, my kids are asked to solve 2|x - 8| = 10, they can at this point sketch by hand the absolute-value graph centered at x=8, know that its slope is 2, and sketch another horizontal line at y=10. They can first assume the quantity (x - 8) is positive, so they can solve for x = 13 as one of the intersection points. Then, via symmetry around x = 8, they can then quickly figure out that x = 3 must be the other intersection point.
And they can then extend their graphical analysis to solving 2|x - 8| > 10 to get x > 13 or x < 3 without memorizing procedures.
Although I plan to teach absolute-values a different way still, in order to reinforce the meaning of the various symbols, it has been delightful to preview the idea of functional symmetry and turning point before we ever see quadratics. The kids are doing so well on the abstract concepts so far! Totally kicking butt! YEAH.
*Also, the shading also is a good way of visualizing how to split the coordinate into different sections, delimited by x values, so their familiarity with this will really help out when we need to graph or analyze piecewise functions later this year.
Tuesday, November 12, 2013
Product Rule via Geometry
I was researching some justification for the product rule, and I am amazed that I never knew this before. It's so obvious (and much easier) if you approach it from a geometric perspective, thinking about how the length, width, and area of the rectangle change. Here are my visuals in thinking through this process, and also how I am planning to scaffold it for my kids. I named the product formula A = ... because I think that is most intuitive to think of that product as the Area that changes as a result of each length or width function changing.
Do you introduce product rule this way? Do you have recommendations for me?
Do you introduce product rule this way? Do you have recommendations for me?
Tuesday, November 5, 2013
Blowing Kids' Minds, One Combined Function at a Time
In my Calculus class, recently a discussion came up about when a derivative trig function would have a midline that is not 0. We quickly discussed the algebraic result of this question, but I thought afterwards that this was just too juicy of a discussion to let slide by.
So, on our new practice quiz, I decided to throw in one such bonus problem. (Go to the end.) They were AMAZED by the resulting graph when they worked on this in class today. One algebra-whiz kid was like, "This is NOT allowed! You cannot mix trig function with other functions!!" So hilarious!! I LOVE MY KIDS. I was pleased with their sense of wonder and surprise, but even more pleased that they were still able to look at and compare features between the derivative and original function graphs, even though they thought what they were looking at was super weird and not intuitive.
At the end of class, I off-handedly asked a couple of the fastest-working kids today, "So what's the average value of the derivative function?" They said, "...-1?" And then I asked, "How does that show up inside the original graph?" Those kids' eyes got so big and they said, "It has an average derivative of -1! OMG, you can see it!!!"
Cool graphical connections!!! We'll have to revisit this bonus problem as a class tomorrow, to make sure that everyone can appreciate the juiciness of this connection before we move on to other algebra goodness.
So, on our new practice quiz, I decided to throw in one such bonus problem. (Go to the end.) They were AMAZED by the resulting graph when they worked on this in class today. One algebra-whiz kid was like, "This is NOT allowed! You cannot mix trig function with other functions!!" So hilarious!! I LOVE MY KIDS. I was pleased with their sense of wonder and surprise, but even more pleased that they were still able to look at and compare features between the derivative and original function graphs, even though they thought what they were looking at was super weird and not intuitive.
At the end of class, I off-handedly asked a couple of the fastest-working kids today, "So what's the average value of the derivative function?" They said, "...-1?" And then I asked, "How does that show up inside the original graph?" Those kids' eyes got so big and they said, "It has an average derivative of -1! OMG, you can see it!!!"
Cool graphical connections!!! We'll have to revisit this bonus problem as a class tomorrow, to make sure that everyone can appreciate the juiciness of this connection before we move on to other algebra goodness.
Exploring Sine and Cosine Derivatives
My onwards saga with differential calculus presented via explorations! (See previous post here.)
My kids recently did an exploration of sine and cosine derivatives. It was a little long, but I think well worth it. The idea is this:
* First, we spent a class period reviewing how to find a sine or cosine equation given a graph. This way I made sure that they'd have the pre-requisite exploration algebra skills.
* On the worksheet, they were given f(x) = sin(x) and asked to sketch its derivative using their knowledge of how to sketch derivative graphs. They're quite confident with this sketch, and then they decided that f'(x) = cos(x) by looking at its shape. I circulated and made sure that each kid used the dy/dx feature of their calculator to verify that the amplitude of f' is still 1, by making sure that the steepest part of f has a derivative value of 1.
* They then are given g(x) = cos(x), and again they started by sketching its derivative graph in order to determine that g'(x) = -sin(x) by inspection.
* They then made predictions for when the amplitude of the original function is not 1. In each case, they sketched the curve and then verified the max derivative value, in order to verify the amplitude of their resulting derivative graph.
* They repeated this for simple cases when the period changes. Based on their sketch of the derivative graph, they determined that the resulting derivative function would share the same period as the original functions, so they think k(x) = sin(2x) would have k'(x) = cos(2x). But, when they then checked the max derivative value along k(x), they realized that they were not correct. --WHY?? One of the kids immediately figured it out. I overheard him saying that because "sin(2x) is squished horizontally, that middle part of maximum steepness now becomes twice as steep as before, so obviously the amplitude will change." They checked this again using L(x) = cos(2x), whose max derivative (or steepness) is again 2, not 1, due to the horizontal compression. Brilliant!!
* They then tested their hypothesis out with M(x) = 3sin(2x), whose max steepness is now even steeper than 2.
* As a class, when we came together to discuss these exploratory problems together, I pulled up GeoGebra to guide our discussion. I entered something like M(x) = 3sin(2x) and had kids hypothesize just how steep that graph gets at x=0, since not everyone had quite reached the same conclusions before. If they said that the max derivative is 1, I entered y = 1x to show them that's unfortunately less steep than our curve is at x = 0. If they said 2, I entered y = 2x to show it's still less steep than our curve. Someone then guessed that the slope of the tangent at that point is 6, and we graphed it to see that y = 6x is indeed as steep as the curve gets. We then typed in M'(x) into the GeoGebra input bar, which automatically generates the formula AND the graph for the derivative. We discussed why the amplitude changes to 6, but why the period stays the same as that of M(x).
BAM! My kids are awesome!! They're able to now differentiate basic wave equations without knowing the first thing about chain rule. (Not yet, that's next, after we reinforce some other parts of our abstract analysis.) And they can explain the outcome of these sine and cosine derivative graphs! Yeah!!
My kids recently did an exploration of sine and cosine derivatives. It was a little long, but I think well worth it. The idea is this:
* First, we spent a class period reviewing how to find a sine or cosine equation given a graph. This way I made sure that they'd have the pre-requisite exploration algebra skills.
* On the worksheet, they were given f(x) = sin(x) and asked to sketch its derivative using their knowledge of how to sketch derivative graphs. They're quite confident with this sketch, and then they decided that f'(x) = cos(x) by looking at its shape. I circulated and made sure that each kid used the dy/dx feature of their calculator to verify that the amplitude of f' is still 1, by making sure that the steepest part of f has a derivative value of 1.
* They then are given g(x) = cos(x), and again they started by sketching its derivative graph in order to determine that g'(x) = -sin(x) by inspection.
* They then made predictions for when the amplitude of the original function is not 1. In each case, they sketched the curve and then verified the max derivative value, in order to verify the amplitude of their resulting derivative graph.
* They repeated this for simple cases when the period changes. Based on their sketch of the derivative graph, they determined that the resulting derivative function would share the same period as the original functions, so they think k(x) = sin(2x) would have k'(x) = cos(2x). But, when they then checked the max derivative value along k(x), they realized that they were not correct. --WHY?? One of the kids immediately figured it out. I overheard him saying that because "sin(2x) is squished horizontally, that middle part of maximum steepness now becomes twice as steep as before, so obviously the amplitude will change." They checked this again using L(x) = cos(2x), whose max derivative (or steepness) is again 2, not 1, due to the horizontal compression. Brilliant!!
* They then tested their hypothesis out with M(x) = 3sin(2x), whose max steepness is now even steeper than 2.
* As a class, when we came together to discuss these exploratory problems together, I pulled up GeoGebra to guide our discussion. I entered something like M(x) = 3sin(2x) and had kids hypothesize just how steep that graph gets at x=0, since not everyone had quite reached the same conclusions before. If they said that the max derivative is 1, I entered y = 1x to show them that's unfortunately less steep than our curve is at x = 0. If they said 2, I entered y = 2x to show it's still less steep than our curve. Someone then guessed that the slope of the tangent at that point is 6, and we graphed it to see that y = 6x is indeed as steep as the curve gets. We then typed in M'(x) into the GeoGebra input bar, which automatically generates the formula AND the graph for the derivative. We discussed why the amplitude changes to 6, but why the period stays the same as that of M(x).
BAM! My kids are awesome!! They're able to now differentiate basic wave equations without knowing the first thing about chain rule. (Not yet, that's next, after we reinforce some other parts of our abstract analysis.) And they can explain the outcome of these sine and cosine derivative graphs! Yeah!!
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