Invariable Calculus Project II

As Rocky hinted in his comment to my last post, \displaystyle f(x)=\frac{k}{x}  also has the constant area property.  Following a lead from Cohen, et al’s Student Research Projects in Calculus, I discovered long ago that with nothing more than knowledge of the product rule, the quotient and/or chain rule, and a healthy dose of some patient algebra manipulations, students could actually determine all functions which have the property that right triangles formed by the x– and y-axes and tangent lines to those functions have constant area.  This morning, I discovered a nice CAS approach that makes the problem accessible to far more students.

This makes another great project for calculus students who’ve just learned algebraic rules for differentiation:

Determine an equation for any twice-differentiable function–y=g(x)–with the property that all tangent lines to g in Quadrant I, along with the x- and y-axes, form triangles of constant area.

(Very minor) HINT:  At some point in solving this problem, you’ll need to make and use some assumptions about the values of a, g(a), g'(a), and g''(a).

SOLUTION ALERT!  Don’t read further if you want to solve this problem for yourself.

Assumptions:  Let (a,g(a)) be any arbitrary point on y=g(x) in Quadrant I.  This makes a>0 and g(a)>0.  I also know g'(a)<0 because otherwise both of the x– and y-intercepts of the tangent line would not be positive, making the triangle’s area negative.  Finally, if g''(a)=0, then g would be a linear function, and there would be only one triangle.  To keep the problem interesting, I’m going to assume g''(a)\ne 0.

Setting up:  We no longer have a specific function, so everything must be in generalities.  A generalized equation for a tangent line to any function y=g(x) at x=a is

y-g(a)=g'(a)\cdot (x-a).

From here, the generalized x-intercept is \displaystyle a-\frac{g(a)}{g'(a)}, and the y-intercept is g(a)-a\cdot g'(a).  [Side note, the x-intercept is also the same form used in Newton’s Method for root approximations, a connection I’ll make later in the term when I’m teaching AP Calculus.]  Combining the generalized intercepts, I can write a generic area formula.

\displaystyle Area = \frac{1}{2} \cdot \left( a - \frac{g(a)}{g'(a)} \right) \cdot \left( g(a) - a \cdot g'(a) \right)

Differentiating and Cleaning Up:  Whenever I’ve used this problem in the past, my students and I have always used paper & pencil techniques.  While I’m quite comfortable with my algebraic manipulation skills, the significant majority of my students struggle with this part of the assignment.  As I was writing this post today, I finally had a technology insight that I should have years ago, given my long interest in CAS.  So, I’ll show the next portion in two different approaches, first traditional, and then via CAS.

APPROACH 1 – TRADITIONAL PAPER & PENCIL:  Trying to keep some brevity in this expanding post, I skip a few algebra steps below while providing some guiding explanations.  Finding a common denominator in the Area equation and recognizing a common factor leads to

\displaystyle Area(a) = - \frac{1}{2} \cdot \frac{\left( a \cdot g'(a) - g(a) \right) ^2}{g'(a)} .

Applying the quotient rule with respect to a gives

\frac{d(Area(a))}{da} = -\frac{1}{2} \cdot \frac{g'(a)\cdot 2(a\cdot g'(a)-g(a))(1\cdot g'(a)+a\cdot g''(a) - g'(a))- (a\cdot g'(a)-g(a))^2\cdot g''(a)}{(g'(a))^2} .

Remember that I seek functions whose tangent lines create constant area triangles, so \displaystyle \frac{d(Area(a))}{da} = 0.  Using this on the left and canceling some terms on the right gives

\displaystyle 0 = -\frac{1}{2} \cdot \frac{2a\cdot g'(a)\cdot g''(a)(a\cdot g'(a)-g(a))-(a\cdot g'(a)-g(a))^2 \cdot g''(a)}{(g'(a))^2} .

Pulling out common factors and cleaning up a little more turns this into a completely factored form.

\displaystyle 0 = - \frac{1}{2} \cdot \frac{(a\cdot g'(a)-g(a))\cdot g''(a)\cdot (a\cdot g'(a)+g(a))}{(g'(a))^2}

APPROACH 2 – CAS:  Because the algebra was going to be complicated enough to make viewing on a handheld calculator very difficult, I used my TI-nSpire CAS Computer software to tackle the problem.  In the image below, I defined the area function in line 1 and differentiated with respect to a in line 2.  Equating the final equation to 0 gives the last algebraic line above from APPROACH 1.

Invariable1

Notice that I did not need to define a differentiation technique or to manipulate the factoring.  The results on both lines automatically accomplish the factoring I worked so carefully to establish earlier in APPROACH 1.  This is a beautiful example of what I see as a central benefit of CAS:  Keeping users focused on the mathematics of the problem situation. 

Some students might actually be curious about how the challenge of differentiating line 1 could end up as relatively “clean” as the result in line 2.  GOOD!  CAS also inspires creative thinking.

Other than the differentiation step, everything else in APPROACH 1 was simple algebra.  Complicated, perhaps, but simple.  In fact, I don’t think it’s mathematics at all; it’s algebraic arithmetic.  I’m not disparaging the work or the approach, but I see mathematics as pattern recognition and big thinking.  I think CAS is completely justified in this problem.

Applying the Zero Product Property:  Our initial assumptions clear the denominator because g'(a)<0.  Because g''(a)\ne 0, I can eliminate that term, too.  With a and g(a) both positive and g'(a)<0, the (a\cdot g'(a)-g(a)) term must be negative and therefore can be eliminated.  That drops the initially complicated differential equation to

\displaystyle 0 = a\cdot g'(a)+g(a).

Finally–the Solution:  Depending on how much your students know, this last equation can be solved three different ways:  A) recognizing differentiation rules, B) solving a separable differentiable equation, or C) using a CAS solver.  I typically assign this problem so early in a calculus course that they have no idea what a differential equation is, making  the first approach the only available technique.  But this is also a great problem to introduce after learning about separable DEs.

APPROACH A:  If you look carefully, you can recognize the right side as the result of the product rule applied to a\cdot g(a).  (In my experience, most students need some time, encouragement, and occasionally some hints to “see” this.)  Because the product rule result equals zero, the original expression must have equalled a constant.  That means a\cdot g(a) = C for any constant, C.  Solving gives \displaystyle g(a)=\frac{C}{a}.  That means Rocky’s suggested family of functions at the top of this post, \displaystyle f(x)=\frac{k}{x} not only produces triangles of constant area, it’s the only family of functions that does!  Very cool!

APPROACH B:  Rewriting the result of the Zero Product Property simplification using xs and ys gives \displaystyle 0=x\cdot \frac{dy}{dx} +y.  The variables can be rearranged to give \displaystyle -\frac{dx}{x}=\frac{dy}{y}.  Integration gives -ln(x)+ln(C)=ln(y) for any random constant, ln(C).  Logarithm properties lead to y=\displaystyle \frac{C}{x}, as before.

APPROACH C:  While I like the pattern recognition insights from the previous two approaches, the solution can also be found using a CAS.

Invariable3On the TI-nSpire, c1 represents any random constant, so the DE solver again gives the same results.

Conclusion:  No matter what approaches you take, this problem shows that the only functions that have the property of their tangent lines producing constant area triangles.

Shiny.

Advertisements

One response to “Invariable Calculus Project II

  1. Pingback: Non-Calculus approach to Invariable Calculus Project | CAS Musings

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s