# Tag Archives: differential equations

## 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.

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.

On 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.