# Tag Archives: tangent lines

## Non-Calculus approach to Invariable Calculus Project

I shared my posts (here and here) on the Invariable Calculus Project in the AP Calculus Community.  Gary Litvin posted a response within the Community offering there a great non-calculus alternative solution to the original problem of the area of the triangle formed by the x- and y-axes and any Quadrant I tangent line to $\displaystyle \frac{1}{x}$.  Here’s a paraphrase of Gary’s approach.

Let $\displaystyle \frac{x}{a} + \frac{y}{b}=1$ be any Quadrant I tangent line to $\displaystyle y=\frac{1}{x}$.  (In case you don’t recognize it, this tangent equation uses the intercept form of a line–a is the x-intercept and b is the y-intercept.)  Because the line intersects the parabola in a single point, we can find that point by solving the system of equations defined by the two equations.  Substituting for y gives

$\displaystyle \frac{x}{a} + \frac{\frac{1}{x}}{b}=1$.

This is equivalent to $x^2 - a \cdot x+\frac{a}{b}=0$, a quadratic.  We could determine the value of x using the quadratic formula.  Because there is only one solution to this equation (there is only one point of intersection, the point of tangency), the discriminant must be zero.  That means

$\displaystyle (-a)^2 - 4\cdot \left( \frac{a}{b} \right)=0$

which can be rearranged to give $ab=4$ ($a=0$ is extraneous).  Therefore, the area of the triangle formed by the tangent line to $\displaystyle y=\frac{1}{x}$ and the coordinate axes is $\displaystyle Area=\frac{1}{2} ab=2$ no matter what the point of tangency.

Shiny.

## 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.

## Invariable Calculus Project

Here’s one of my favorite calculus projects.  I initially discovered it over 20 years ago in Cohen, et al’s superb Student Research Projects in Calculus.

For $x>0$, what is true about every triangle formed by the x- and y-axes and any tangent line to $\displaystyle y=\frac{1}{x}$ ?  Prove thy claim.

I’d love to say nothing more than that, but I usually don’t.  The problem sounds vague in its statement, but is pretty simple to solve.  The hidden property is a delightful surprise.  I encourage you to try it out for yourself before reading further.

I just assigned the problem to one of my classes of seniors.  The class is a one-semester introduction to calculus for primarily students who’ve never been in honors and largely aren’t enamored by mathematics.  Most take the class to get an introduction to statistics (fall) and calculus (spring) before likely taking a course in one of these two in college and–for most–never taking another math course.  With that background in mind, I’ve probably scaffolded this iteration of the problem more than I should.  Here’s the assignment I gave them this week.

WARNING!  Partial Solution Alert!  Don’t read further if you want to solve the problem for yourself.

I typically use this project early in my introduction to derivatives and walk students through a little review and data gathering to help them discover the surprising hidden property.  While I don’t expect my students to do this, my default approach to geometric-type problems is to use a dynamic geometry package.  The animation below shows what happens when I varied the point of tangency while tracking the base, height, and area of the resulting triangle.

Well, I hope that animation screams something.  The x– and y-intercepts are the base and height, respectively, of a right triangle.  While those intercepts obviously vary as the point of tangency changes, the area of the triangle always seems to be 4.  It never changes!  If you’ve any geometry sense, something like that just shouldn’t happen.  So, is this a universal property, or is my animation misleading or limited in some way?  That’s a good question, and it requires proof.  Can you prove this apparent property about tangent lines to $\displaystyle y=\frac{1}{x}$?

FINAL SOLUTION ALERT!  Don’t read further if you want to prove this property for yourself.

For $\displaystyle f(x)=\frac{1}{x}$, $\displaystyle\frac{d}{dx}\left(f(x)\right)=\frac{-1}{x^2}$, so an equation for the tangent line to f at any point $x=a$ is

$\displaystyle \left(y-\frac{1}{a}\right)=\frac{-1}{a^2}\left(x-a\right)$.

The x-intercept of this generic line is $\left(2a,0\right)$, and its y-intercept is $\displaystyle \left(0,\frac{2}{a}\right)$.  Therefore, the area of the triangle formed by the x-and y-axes and the tangent line to f at any point $x=a$ is

$\displaystyle Area=\frac{1}{2}\cdot base\cdot height = \frac{1}{2}\cdot 2a\cdot\frac{2}{a}=2$.

Cool!  The triangle’s area is always 2, completely independent of the point of tangency!

EXTENSION:

Are there any other functions that have a similar property, or is $\displaystyle y=\frac{1}{x}$ alone in the mathematical universe for having constant area triangles?  Well, that’s a problem for another post.

## Tangent Perspectives

I assigned AP Calculus BC 1975 problem #7 to my class a couple weeks ago.  I got a 100% legitimate answer I didn’t expect from a student, so I thought I’d share.  It’s what can happen when you encourage students to follow their instincts.

Paraphrasing, the students first had to find an equation of a line through the origin tangent to the graph of $y=ln(x)$.  Most had no problems concluding that this was $\displaystyle y=\frac{x}{e}$.

The next part asked if the tangent was above or below $y=ln(x)$.  In class, we had discussed why the position of tangent lines was dependent on the underlying function’s concavity, so I fully expected successful solutions to end up at $\displaystyle y''=\frac{-1}{x^2}$ which is negative for all $x\neq 0$ therefore making the original curve concave down and the tangent line above.  Most successful solutions did this, but one was different.

Paraphrasing M’s work, he concluded that if the tangent line was entirely on one side, then $\displaystyle g(x)=\frac{x}{e}-ln(x)$ must have an extremum.  From there, $\displaystyle g'(x)=\frac{1}{e}-\frac{1}{x}=0$ confirms the tangency point at $x=e$ from earlier, but this time as a critical point on g.  From here, he concluded that $\displaystyle g''(x)=\frac{1}{x^2}>0$ for all $x\neq 0$ making his critical point a global minimum.  From the construction of g, the tangent line then had to be above $y=ln(x)$.

Admittedly, M’s algebra work took a bit longer, but what impressed me was his completely different visualization of the problem.  I’m betting he didn’t remember the down-concavity-means-tangent-above factoid from class, so he had to invent his own approach.  And he did this by turning a concavity problem into an optimization problem.  Nice.