As Rocky hinted in his comment to my last post, 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––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 , and .

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

**Assumptions: **Let be any arbitrary point on in Quadrant I. This makes and . I also know 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 , then *g* would be a linear function, and there would be only one triangle. To keep the problem interesting, I’m going to assume .

**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 at is

.

From here, the generalized *x*-intercept is , and the *y*-intercept is . [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.

**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

.

Applying the quotient rule with respect to *a* gives

.

Remember that I seek functions whose tangent lines create constant area triangles, so . Using this on the left and canceling some terms on the right gives

.

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

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 . Because , I can eliminate that term, too. With *a* and both positive and , the term must be negative and therefore can be eliminated. That drops the initially complicated differential equation to

.

**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 . (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 for any constant, *C*. Solving gives . That means Rocky’s suggested family of functions at the top of this post, 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 *x*s and *y*s gives . The variables can be rearranged to give . Integration gives for any random constant, . Logarithm properties lead to , 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.