The history of the calculus is a fascinating story, inspired by the search for solutions to interesting problems. We do our students a disservice when we fail to share with them some of this exciting history. During 2005-07, with support from the National Science Foundation, I developed modules for teaching calculus concepts in a way that integrates the historical evolution of these concepts. This article contains three examples, which include several interactive applets produced using the software GeoGebra, a free, open-source software package. In addition, many of the exercises provided require GeoGebra and/or a computer algebra system (CAS) such as Maple, Mathematica, or a Texas Instruments (TI) Voyage 200 calculator.

Module 1: Curve Drawing Then and Now

Module 2: Tangent Lines Then and Now

Module 3: Optimization Problems Then and Now

- Heron's “Shortest Distance” Problem
- Snell's Law and the Principle of Least Time
- L'Hôpital's Pulley Problem and the Principle of Least Potential Energy
- Regiomontanus' Hanging Picture Problem
- Galileo and the Brachistochrone Problem

*Editors' note:* This article was originally published in *Convergence* in August of 2007. We are pleased to "reprint" it in June/July of 2014 with GeoGebra applets hosted by GeoGebraTube. In addition to viewing these applets in the webpages of this article, you may view and download all of the applets that appear in the article directly from GeoGebraTube at http://tube.geogebra.org/m/jdRxWdw6?doneurl=%2F.

It was Rene Descartes (1596-1650) who dramatically changed the mathematical landscape with his book *Discours de la méthode pour bien conduire sa raison et chercher la vérité dans les sciences* (*A Discourse on the Method of Rightly Conducting the Reason and Seeking Truth in the Sciences*). In *La Géométrie,* one of three appendices of this book, Descartes showed how to describe geometric objects like curves by means of an algebraic equation. This enabled him to take advantage of the power of algebra to solve geometric problems. This idea of describing points in the plane with coordinates appears also in a work by Pierre de Fermat from 1637. In fact, while Descartes used this idea to derive algebraic equations of certain curves that he was interested in, it was Fermat who first used equations to define new curves. Later, Jacob Bernoulli (1654-1705) and Isaac Newton (1643-1727) generalized this concept to other kinds of coordinate systems such as polar.

The curves that Descartes was interested in were ones that could be described by some mechanical motion. The mechanism depicted in Figure 1 appears in *La Géométrie* as an example of such curves, traced out by points *D, F,* and *H* as the angle *XYZ* expands and contracts. Descartes showed how to derive the equations of these curves.

**Figure 1.** A curve-drawing device from Descartes' *Géométrie*

*La Géométrie* was a difficult work to read, as Descartes left much to the reader to work out for himself. He wrote

I hope that posterity will judge me kindly, not only as to the things which I have explained, but also to those which I have intentionally omitted so as to leave to others the pleasure of discovery.

The popularity of Descartes' *La Géométrie* was due largely to Frans van Schooten (1615-1660), a Dutch professor of mathematics who translated Descartes' work into Latin and wrote a commentary on it filling in many gaps. Van Schooten was particularly interested in conic section drawers and wrote a treatise on them. He devised several instruments for drawing conic curves, as shown in Figure 2. Like Descartes' instruments, these generally consisted of a collection of straight rods hinged together in some way.

**Figure 2.** Above: Frans van Schooten's conic section drawers for (left to right) parabola, ellipse, and hyperbola. Directly below: Click on "Animate" to see van Schooten's Parabola Drawer in action; his Ellipse Drawer is further illustrated in Figures 6 and 8, below.

Today's graphing calculators easily generate graphs of curves by using an algebraic equation to generate the coordinates of many points on the curve and then plotting them. Many, many points must be plotted in order to obtain an accurate graph, so this would not have been a very practical method before the age of computers. In contrast, the ancients drew curves by constructing specialized instruments or "curvedrawers." In the same way that ruler and compass can be used to draw straight lines and circles in a continuous movement, the curve drawers devised by van Schooten and others were used to draw other curves of interest, particularly the conic sections, namely parabolas, ellipses, and hyperbolas.

In the exercises in this first module, several ellipse drawers from the history of mathematics are explored. These include the familiar string construction, as well as an ellipse drawer due to Proclus (411-485) and another one attributed to Archimedes (287-212 BCE).

**Figure 3.** String Construction of an Ellipse (click image to animate)

In the string construction in Figure 3, above, two ends of a string are attached to two fixed pins \(F\) and \(F^{\prime}.\) Then a pen holds the string taut as it rotates around the pins. The first person known to have written about the string construction of the ellipse was Anthemius of Tralles in the fifth century.

**Figure 4.** Archimedes' Trammel Construction of an Ellipse (click image to animate)

The next ellipse drawer, attributed to Archimedes, is the trammel construction. In Figure 4, above, the point \(B\) on segment \(AP\) is attached to the horizontal axis in such a way that it can slide along the axis. Similarly, the point \(A\) is attached to the vertical axis and allowed to slide along it.

**Figure 5.** Proclus' Ellipse Drawer (click image to animate)

The next ellipse drawer, illustrated in Figure 5, above, would have been known to Proclus, who knew that an ellipse could be generated by tracing a point \(P\) inside a circle that rolls without slipping inside and tangent to another circle whose radius is twice as long. Such a construction brings to mind the popular spirograph game that children use to generate cycloidal curves.

**Figure 6.** Van Schooten's Ellipse Drawer (click image to animate)

Our last ellipse drawer is due to the Dutch mathematician Frans van Schooten. Van Schooten was very interested in conic section drawers and wrote a treatise on them. He devised several instruments for drawing conic curves (see Figure 2, above), including the ellipse drawer shown in Figure 6, above. These conic section drawers were illustrated in the fourth book of his work *Exercitationes mathematicae libri quinque *(*Five Books of Mathematical Exercises*), published in 1657. Van Schooten had published the fourth book, *Organica conicarum sectionum in plano descriptio,* by itself in 1646.

Consider the string construction of an ellipse. Assume a coordinate system with \(F\) and \( F^{\prime}\) on the \(x\)-axis and the origin half way between \(F\) and \( F^{\prime}.\) Then the coordinates of \(F\) and \( F^{\prime}\) are \(({\pm}{c},0)\) for some \(c.\) If \(P=(x,y)\) is a point on the ellipse, then we must have \(|PF|+|PF^{\prime}|=d,\) where \(d\) is a constant (the length of the string in the string construction).

**Figure 7.** String Construction of Ellipse (applet for Exercise 1)

Use the distance formula to translate \(|PF|+|PF^{\prime}|=d\) into an algebraic equation involving \(x,\) \(y,\) \(c,\) and \(d.\) Solve for \(y\) to obtain two equations describing the top and bottom halves of an ellipse. Find the values of \(c\) and \(d\) such that the ellipse has \(x\)-intercepts at \(({\pm}{5},0)\) and \(y\)-intercepts at \((0,\pm3).\) Substitute these values into your equations and simplify. Use the sliders in the applet in Figure 7 above to set \(c\) and \(d\) and verify that the resulting ellipse has the correct intercepts. Enter your equations into the input field to obtain graphs and verify that you get the ellipse shown.

**Figure 8.** Van Schooten's Ellipse Drawer (applet for Exercise 2)

The applet in Figure 8 above shows the ellipse drawer devised by the Dutch mathematician Frans van Schooten. Click the small triangular Play button at the lower left hand corner to generate the curve. What should be the lengths of \(a=CD\) and \(b=DP\) so that the ellipse generated will have \(x\)-intercepts at \((\pm5,0)\) and \(y\)-intercepts at \((0,\pm3)?\) Use the sliders to set \(a\) and \(b\) to these values and verify that the resulting curve has the correct intercepts. Enter the equations that you found in Exercise 1 into the input field to verify that van Schooten's ellipse drawer does generate an ellipse.

**Figure 9.** Proclus' Ellipse Drawer (applet for Exercise 3)

The applet in Figure 9 above shows the ellipse drawer devised by Proclus. Click the Play button at the lower left hand corner to generate the ellipse. What should be the lengths of \(a=OA\) and \(b=PC\) so that the curve generated will have \(x\)-intercepts at \((\pm5,0)\) and \(y\)-intercepts at \((0,\pm3)?\) Use the sliders to set \(a\) and \(b\) to these values and verify that the resulting curve has the correct intercepts. Enter the equations that you found in Exercise 1 into the input field to verify that Proclus's ellipse drawer does generate an ellipse.

**Figure 10.** Archimedes' Ellipse Drawer (applet for Exercise 4)

The applet in Figure 10 above shows the ellipse drawer devised by Archimedes. Click the Play button at the lower left hand corner to generate the curve. What should be the lengths of \(a=BP\) and \(b=AB\) so that the curve generated will have \(x\)-intercepts at \((\pm5,0)\) and \(y\)-intercepts at \((0,\pm3). \) Use the sliders to set \(a\) and \(b\) to these values and verify that the resulting curve has the correct intercepts. Enter the equations that you found in Exercise 1 into the input field to verify that Archimedes' ellipse drawer does generate an ellipse.

Carl B. Boyer, Historical Stages in the Definition of Curves, *National Mathematics Magazine,* Vol. 19, No. 6 (March 1945), pp. 294-310.

Jan Van Maanen, Seventeenth Century Instruments for Drawing Conic Sections, *The Mathematical Gazette,* Vol. 76, No. 476 (July 1992), pp. 222-230.

As long as mathematicians have been studying curves, they have been interested in the construction of their tangent lines. Apollonius, who lived over two thousand years ago and studied the conic curves extensively, found geometric methods for constructing tangent lines to parabolas, ellipses, and hyperbolas. In fact, Propositions 33 and 34 of Book I of Apollonius' great work, *The Conics,* give recipes for constructing tangents to these curves.

Apollonius' recipe for constructing tangent lines to parabolas is as follows:

**Proposition I-33.** *Let \(P\) be a point on the parabola with vertex \(E,\) with \(PD\) perpendicular to the axis of symmetry. If \(A\) is on the axis of symmetry so that \(AE=ED,\) then \(AP\) will be tangent to the parabola at \(P.\)*

**Figure 11.** Apollonius' recipe for constructing tangent lines to parabolas. Drag \(F,\) the focus, to change the shape of the parabola. Drag \(P,\) the point of tangency, to see the tangent line at any point on the parabola.

Apollonius' recipe for construction of tangent lines to ellipses and hyperbolas is as follows:

**Proposition I-34.** *Let \(P\) be a point on an ellipse or hyperbola and \(PB\) the perpendicular from the point to the main axis. Let \(G\) and \(H\) be the intersections of the axis with the curve and choose \(A\) on the axis so that \[\frac{|AH|}{|AG|}=\frac{|BH|}{|BG|}.\] Then \(AP\) will be tangent to the curve at \(P.\)*

**Figure 12.** Apollonius' recipe for construction of tangent lines to ellipses. Drag \(H\) horizontally and/or \(C\) vertically to change the shape of the ellipse. Drag \(P,\) the point of tangency, to see the tangent line at any point on the ellipse.

**Figure 13.** Apollonius' recipe for construction of tangent lines to hyperbolas. Drag \(H\) horizontally and/or \(C\) vertically to change the shape of the hyperbola. Drag \(P,\) the point of tangency, to see the tangent line at any point on (the right half of) the hyperbola.

Apollonius' recipes for constructing tangent lines to conic curves are easy to implement, but his methods were limited in that they could only be applied to a handful of curves. Of course, in the days of antiquity, there weren't that many curves that people were interested in studying, and so there wasn't that much of a need for a general method.

By the beginning of the seventeenth century, the collection of known curves had grown. Curves were often described as the path of a moving point; for instance, a circle can be viewed as the path traced out by a point on the outer edge of a spinning wheel. Indeed, the instruments that were used to draw curves, some of which were discussed in Module 1, underscored this notion of a curve as a path of a moving point. It was Gilles Personne de Roberval (1602-1675) who exploited this definition of a curve by viewing a tangent line as the direction in which the point is moving at a particular instant.

**Figure 14.** Tangent as direction of motion (click image to animate)

Consider, for example, a parabola, which, recalling Frans van Schooten’s Parabola Drawer from Module 1, can be defined as the path of point \(D\) as \(G\) moves along \(AC,\) as shown below in Figure 15.

**Figure 15.** Van Schooten's Parabola Drawer

As \(G\) moves along \(AC,\) the point \(D\) is being pulled simultaneously toward the point \(B\) and toward the horizontal line \(AC.\) Since \(|GD|=|BD|,\) the two forces are equal: if \(D\) moves a little toward \(B,\) it must move by the same amount toward the line \(AC\). By the parallelogram law, the direction of motion of \(D\) should be the angle bisector of \(\angle BDG.\) This angle bisector is actually the diagonal \(FH\) of the rhombus \(FGHB.\)

Both Apollonius and de Roberval used geometrical methods to construct tangent lines. One of the first algebraic methods is due to René Descartes, one of the inventors of analytic geometry. It was well known that a tangent line to a circle is always perpendicular to the radius of the circle. So Descartes' idea was to first find a circle tangent to the given curve. Then the tangent to that circle is the sought after tangent to the curve.

**Figure 16.** Descartes' Method: Tangent to curve is perpendicular to radius of tangent circle. (Click image to animate.)

It was Pierre de Fermat who came up with essentially the method used today to construct tangent lines, by viewing the tangent as the limit of secant lines through the given point \(A\) and a nearby point \(B\) on the curve, where \(B\) approaches and eventually becomes \(A.\)

**Figure 17.** Fermat's method of constructing tangent lines (click image to animate)

Fermat's method, although it worked in practice, was not without controversy. If \(A=(a,f(a))\) is a point on the curve \(y=f(x),\) and \(B=(a+e,f(a+e))\) is a nearby point, the slope of the line \(AB\) is perfectly well-defined as long as \(A\) and \(B\) are distinct, i.e. \(e\neq 0\), but how do we define the slope when \(e=0\) so that \(B=A?\) This seeming inconsistency invited much criticism, the most famous coming from the Irish philosopher and bishop, George Berkeley (1685-1753), in a pamphlet he published in 1734 called* The Analyst.* Here Bishop Berkeley ridiculed the increments \(e,\) calling them the “ghosts of departed quantities." In fact, Descartes developed his method discussed above as a way to get around these difficulties. You can see below how, as \(B\) approaches \(A,\) the circle with center \(O\) on the \(x\)-axis that goes through \(A\) and \(B\) is well-defined, even when \(A=B.\) However the line through \(A\) and \(B\) cannot be well-defined when \(A=B.\)

**Figure 18.** Descartes' method compared to Fermat's method (click image to animate)

Unfortunately, Descartes' method did not work very well except for some simple curves, because of the burdensome algebra involved in finding the unique circle with center on the \(x\)-axis that intersects the curve only once at \(A.\)

In spite of the problems with the logical foundations of the subject, nobody could dispute that the techniques worked. And so Fermat is credited with the invention of the differential calculus since he developed this method in 1629 and used it successfully to compute subtangents of a large collection of curves. It wasn't until the nineteenth century that the controversies and problems were satisfactorily addressed by the French mathematician Augustin-Louis Cauchy (1789-1857), who gave a precise definition of derivative in terms of a new concept called *limit.* This led to the definition that we use today.

**Figure 19.** This applet for Exercise 5 implements recipes of Apollonius to construct tangent lines to parabolas, ellipses, and hyperbolas.

Use the recipes prescribed by Apollonius and the applet in Figure 19, above, to construct the following tangent lines.

(a) The tangent line to the parabola \(y=\frac{x^2}{4}+1\) at the point \((4,5).\)

(b) The tangent line to the ellipse \(\frac{x^2}{25}+\frac{y^2}{9}=1\) at the point \((3,2.4).\)

(c) The tangent line to the hyperbola \(\frac{x^2}{16}-\frac{y^2}{9}=1\) at the point \((5,2.25).\)

In this exercise you will use de Roberval's method to construct tangent lines to several curves.

**Figure 20.** This applet for Exercise 6, part a, illustrates the standard construction of the parabola using focus \(F\) and directrix \(OB\).

(a) The applet in Figure 20, above, illustrates the construction of the parabola as the locus of points \(P\) that are equidistant from a fixed point \(F\) (the *focus*) and a fixed line \(OB\) (the *directrix*). You can see that as you drag the point \(B,\) the distances \(BP\) and \(PF\) are always equal.

There are two forces acting on \(P\): one toward \(F\) and one toward the horizontal line \(OB.\) Since \(|FP|=|BP|,\) the two forces are equal, meaning if \(P\) moves a little toward \(F,\) it must move by the same amount toward the line. Hence the direction of motion, by the parallelogram law, should be the angle bisector of \(\angle FPB.\) Construct the angle bisector to verify that this does indeed appear to be tangent to the curve at every point of the parabola. To do this:

Click on the angle bisector tool in the toolbar, then on the segment \(FP,\) and then on the segment \(PB.\) You'll get two lines, perpendicular to each other. Click the Move tool, then right-click on the line you don't want and select Show Object to hide it. You can drag \(B\) to verify that this construction works for every point on the parabola.

**Figure 21.** This applet for Exercise 6, part b, illustrates the standard construction of the ellipse using two foci \(F_1\) and \(F_2.\)

(b) The applet in Figure 21, above, illustrates the construction of the ellipse as the locus of points \(P\) such that the sum of \(P\)'s distances from two fixed points \(F_1\) and \(F_2\) is constant. Note that as you drag the point \(P,\) the sum \(|PF_1|+|PF_2|\) remains constant.

Since \(|PF_1|+|PF_2|\) is constant, each time \(P\) moves toward \(F_1\) it must move the same distance *away from* \(F_2.\) So there are two equal forces acting on \(P,\) one toward one of the foci, the second away from the other focus. So again you can construct the tangent line as an angle bisector. Construct the tangent to the ellipse at any point \(P.\) Make sure your construction works at every point of the ellipse.

**Figure 22.** This applet for Exercise 6, part c, illustrates the standard construction of the hyperbola using two foci \(F_1\) and \(F_2.\)

(c) The applet in Figure 22, above, is the construction of the hyperbola as the locus of points \(P\) such that the difference of \(P\)'s distances from two fixed points \(F_1\) and \(F_2\) is constant. Note that as you drag the point \(P,\) the difference \(|PF_1-PF_2|\) remains constant. Construct the tangent to the hyperbola at any point \(P.\) Make sure your construction works at every point of the hyperbola.

**Figure 23.** This applet for Exercise 6, part d, illustrates the cycloid as the path traced by a point on the circumference of a circle rolling along a horizontal line.

(d) The cycloid, illustrated in Figure 23, above, played a very important part in the history of the calculus, and its properties were studied extensively during the sixteenth and seventeenth centuries. It is the path traced out by a point on the circumference of a circle as the circle rolls on a horizontal line. Use de Roberval's method to construct the tangent line at any point \(P\) of the cycloid.

**Figure 24.** This applet for Exercise 6, part e, illustrates the epicycloid as the path traced by a point on the circumference of one circle rolling along the outside of another circle.

(e) The epicycloid, illustrated in Figure 24, above, is the path traced out by a point on the circumference of a circle as the circle rolls on the outside of another circle. Use de Roberval's method to construct the tangent line at any point \(P\) of the cycloid.

In this exercise, you will use Descartes' method to find tangent lines to the curve \(y=\sqrt{4x}.\) The method consists of first finding a circle tangent to the curve. Then the tangent line to the curve will be the same as the tangent line to the circle, which is easily constructed since it is perpendicular to the radius.

**Figure 25.** Applet for Exercise 7

(a) The applet shown above in Figure 25 contains: a graph of \(y=\sqrt{4x},\) the point \(A=(1,2)\) which is on this curve, a point \(O\) on the \(x\)-axis, and a circle with center \(O\) and radius \(OA.\) In general, depending on where point \(O\) is, this circle will cross the curve at two points \(A\) and \(B.\) Now drag \(O\) along the \(x\)-axis until the two points of intersection \(A\) and \(B\) merge into one. The resulting circle touches the curve at a single point and so is tangent to the curve. Now you can compute the slope of the tangent, since it is perpendicular to \(OA.\) Type the equation of the tangent line into the purple input box to graph it and verify that it is indeed tangent to \(y=\sqrt{4x}\) at \(A=(1,2).\)

(b) Repeat with \(A=(9,6)\) and enter the equation of the tangent line into the blue input box.

J. L. Coolidge, The Story of Tangents, *The American Mathematical Monthly,* Vol. 58, No. 7. (1951), pp. 449-462.

Paul R. Wolfson, The Crooked Made Straight: Roberval and Newton on Tangents, *The American Mathematical Monthly,* Vol. 108, No. 3 (2001), pp. 206-216.

Perhaps the most important application of the differential calculus is the solution of optimization problems, where one wants to find the value of a variable that maximizes or minimizes a certain quantity. In this module we look at the history of some of these problems and their solutions.

- Heron's “Shortest Distance” Problem
- Snell's Law and the Principle of Least Time
- L'Hôpital's Pulley Problem and the Principle of Least Potential Energy
- Regiomontanus' Hanging Picture Problem
- Galileo and the Brachistochrone Problem

Paul J. Nahin, *When Least is Best: How Mathematicians Discovered Many Clever Ways to Make Things as Small (or as Large) as Possible,* Princeton University Press, 2004.

V.M. Tikhomirov, *Stories about Maxima and Minima; translated from the Russian by Abe Shenitzer,* American Mathematical Society/Mathematical Association of America, 1990.

One of the first non-trivial optimization problems was solved by Heron of Alexandria, who lived about 10-75 C.E. Heron's “Shortest Distance” Problem is as follows:

Given two points \(A\) and \(B\) on one side of a straight line, to find the point \(C\) on the line such that \(|AC|+|CB|\) is as small as possible.

Thus in Figure 26, below, one possible point \(C\) is shown, as well as the length of the path \(A\) to \(C\) to \(B.\)

**Figure 26.** A possible path from \(A\) to \(B\)

**Figure 27.** Use this GeoGebra applet to explore the solution to Heron's problem. You can drag \(C\) to determine the approximate length of the shortest path attainable.

To solve this problem mathematically, Heron noticed that if \(B\) is reflected across the line, to say \(B^{\prime},\) then for any point \(C\) on the line, \(|CB|=|CB^{\prime}|,\) and hence minimizing \(|AC|+|CB|\) is equivalent to minimizing \(|AC|+|C B^{\prime}|.\) But clearly since the shortest path from \(A\) to \( B^{\prime}\) is a straight line, the point \(C\) that minimizes \(|AC|+|C B^{\prime}|\) should be the point of intersection of the line with the line segment \(AB^{\prime}.\) Any other path, such as \(A\) to \(C^{\prime}\) to \(B^{\prime},\) will clearly be longer.

**Figure 28.** Heron's solution of the “Shortest Distance” Problem

Note that \(m\angle 2=m\angle 3\) by construction, and \(m\angle1=m\angle 3\) since these are vertical angles. Therefore, \(m\angle 1=m\angle 2.\) This is the **equal angle law of reflection**. It was Euclid who, over three hundred years earlier, had noted the now well-known Reflection Law for light:

**Equal Angle Law of Reflection **(or **Euclid's Law of Reflection**)**.** If a beam of light is sent toward a mirror, then the angle of incidence equals the angle of reflection.

Heron appears to have been the first to observe that the Reflection Law implies that **light always takes the shortest path**.

As a matter of fact, the mirror in the Equal Angle Law of Reflection need not be flat. We may replace the line in Heron's problem by any concave curve (a curve is *concave* if it lies entirely on one side of any tangent line). In this case, the angles are measured with respect to the tangent line, and the same argument used by Heron shows that if \(C\) is such that the angle of incidence equals the angle of reflection, then \(|AC|+|BC|\) is minimized.

**Figure 29.** Law of Reflection for any concave curve

An interesting application of the Law of Reflection arises in the case of a light beam sent toward a parabolic mirror, where the light beam is parallel to the axis of the parabola. A parabolic mirror is one whose surface is generated by rotating a parabola about its axis. Suppose the parabola has focus \(F\) and directrix \({\mathcal L},\) and that the light beam \(\overrightarrow{GA}\) hits the parabola at \(A.\) Recall from Roberval's construction of tangent lines to parabolas that the tangent line at \(A\) bisects the angle \(\angle FAB.\) Hence \(m\angle 1=m\angle 2.\) Since \(\angle 2\) and \(\angle 3\) are vertical angles, \(m\angle 2=m\angle 3.\) Hence \(m\angle 1=m\angle 3.\) So by the Equal Angle Law of Reflection, the light beam will be reflected in the direction \(\overrightarrow{AF}.\) This will be true for any point \(A\) on the parabola.

**Figure 30.** Light rays directed toward a parabolic mirror parallel to the parabola's axis are all reflected toward the focus \(F\) of the parabola.

There is a famous story about Archimedes in which he is said to have used a parabolic mirror to defeat the Roman General Marcellus. Supposedly, he tilted the mirror toward the sun in such a way that all the sun's rays when reflected off the mirror went through the focal point. The heat generated at that point caused a fire to ignite and destroy the entire Roman fleet.

**Figure 31.** Wall painting from the Stanzino delle Matematiche in the Galleria degli Uffizi (Florence, Italy). Painted by Giulio Parigi (1571-1635) in the years 1599-1600.

**Figure 32.** Euclid's Law of Reflection (applet for Exercise 8)

Use calculus and a CAS to find the point \(C\) on the \(x\) axis for which \(|AC|+|CB|\) is the shortest, where \(A=(0,4)\) and \(B=(10,12).\) Use the applet in Figure 32, above, to enter your solution and verify Euclid's Law of Reflection.

Euclid's Law of Reflection, formulated around 300 BCE, tells us what happens when a beam of light is reflected off a mirror. The Dutch physicist Willebrord Snell (1580-1626) was interested in the phenomenon of **refraction,** which is the change in direction that occurs when a beam of light crosses a boundary from one medium (say air) into another (say water). He observed that when the light beam enters the denser medium at some point \(C\) on the line, its velocity decreases and its path bends toward the normal to the line at \(C.\)

Snell's experiments led him to the discovery of the following relationship, known as **Snell's Law**:

** Figure 33.** **Snell's Law:** \(\displaystyle{\frac{\sin \theta_1}{\sin \theta_2}={\rm constant}.}\)

**Figure 34.** Use this GeoGebra applet to investigate Snell's Law. You can drag \(C\) to determine the approximate location of the point that minimizes the time of travel along the path from \(A\) to \(C\) to \(B.\)

In 1637, Pierre de Fermat became interested in finding a theoretical derivation of Snell's Law. Recall that Heron had shown that if a light beam traveled from \(A\) to \(B\) by first reflecting off a line \({\mathcal L},\) the path taken was the one that would take the minimum time. Fermat reasoned that the same “least time” principle would govern that path from \(A\) to \(B\) across the boundary \({\mathcal L}.\)

Fermat used the differential calculus (techniques which he himself developed by reasoning that the slope of a tangent line at a local maximum or minimum must be zero) to determine the quickest path. Suppose \(A=(a,b)\) lies above the \(x\)-axis and \(B=(c,d)\) lies below the \(x\)-axis, as in the picture below:

**Figure 35.** In modern terms and notation, Fermat wished to minimize the time of travel of a light beam along the path from \(A\) to \(C\) to \(B.\)

We wish to find the point \(C=(x,0)\) on the \(x\) axis that minimizes the time of travel of a light beam along the path \(A\) to \(C\) to \(B,\) assuming its velocity above the \(x\) axis is \(v_1,\) and its velocity below the \(x\) axis is \(v_2.\) Since \[\mbox{time}=\frac{\mbox{distance}}{\mbox{velocity}},\] the time of travel is

\[T(x)=\frac{\sqrt{(x-a)^2+b^2}}{v_1}+\frac{\sqrt{(x-c)^2+d^2}}{v_2}\]

If we take the derivative and set it equal to \(0,\) we have

\[\frac{\frac12\left[(x-a)^2+b^2\right]^{-1/2}\cdot 2(x-a)}{v_1}+\frac{\frac12\left[(x-c)^2+d^2\right]^{-1/2}\cdot 2(x-c)}{v_2}=0,\]

which can be rewritten as

\[\frac{x-a}{v_1\sqrt{(x-a)^2+b^2}}+\frac{x-c}{v_2\sqrt{(x-c)^2+d^2}}=0.\]

But since \[\sin\theta_1=\frac{x-a}{\sqrt{(x-a)^2+b^2}}\quad\quad{\rm and}\quad\quad\sin\theta_2=\frac{c-x}{\sqrt{(x-c)^2+d^2}},\] the above equation can be rewritten as

\[\frac{\sin\theta_1}{v_1}-\frac{\sin\theta_2}{v_2}=0\]

or, equivalently,

\[\frac{\sin\theta_1}{\sin\theta_2}=\frac{v_1}{v_2},\]

which is of course Snell's Law. But whereas Snell had conjectured his law based on observation, Fermat succeeded in giving a mathematical proof.

Fermat was led to investigate Snell's Law after reading a paper by René Descartes on the subject. Fermat thought Descartes' work did not make sense, and he could not understand how Descartes was able to arrive at the correct result using faulty and illogical arguments. Indeed, Fermat's criticisms could be interpreted as an accusation that Descartes, who made no mention of Snell in his work, had somehow gotten hold of Snell's experimental data, conjectured the correct law from the data, and then, working backwards, invented an argument that made little sense but led to the correct conclusion. Needless to say, Descartes did not take kindly to Fermat's accusations!

**Figure 36.** Applet for Exercise 9, part a

(a) Use calculus (and a CAS if needed) to compute the value of the \(x\)-coordinate of the point \(C\) on the \(x\)-axis that minimizes the time of travel along \(A\) to \(C\) to \(B\) with \(A=(1,3),\) \(B=(4,-2),\) \(v_1=4\) units/sec, and \(v_2=5\) units/sec. Use the applet in Figure 36 to enter your solution to verify Snell's Law.

(b) Repeat part a with \(A=(-3,4),\) \(B=(2,-5),\) \(v_1=1\) unit/sec, and \(v_2=2\) units/sec.

(c) Use calculus and a CAS to find the coordinates of the point \(C\) on the curve \(y=-x^2,\) so that \(|AC|+|BC|\) is a minimum, where \(A=(0,4),\) and \(B=(5,-3).\) Illustrate your solution in GeoGebra, and measure the incidence and reflection angles to verify that they are equal. Your sketch should include the following:

- a plot of the curve \(f(x)=-x^2\)
- a plot of the two points \(A\) and \(B\)
- a plot of the point \(C\) that solves the problem, and
- a graph of the tangent line to the curve at \(C.\)

The first calculus textbook was written by the Marquis Guillaume-Francois-Antoine de l'Hôpital (1661-1704). Best known for the famous rule that bears his name that is used to compute limits of indeterminate forms, l'Hôpital was very interested in mathematics and particularly in the new techniques of the calculus that had recently been developed by Isaac Newton (1643-1727) and Gottfried Wilhelm Leibniz (1646-1716). Consequently, he hired one of Leibniz's best students, Johann Bernoulli (1667-1748), to teach the calculus to him. L'Hôpital then took the notes he compiled from Bernoulli's lectures and organized them into a textbook called *Analyse des Infiniment Petits.* The book contained the original results of Newton, Leibniz, and Johann Bernoulli, as well as Johann's brother Jakob Bernoulli (1654-1705). L'Hôpital has been justly criticized for giving little or no credit to these mathematicians, presenting the mathematics as if it were his own. Indeed, many have suggested that l'Hôpital 's Rule should really be called Bernoulli's Rule, since it was taught to him by Johann! Johann Bernoulli was forced to keep silent during l'Hôpital 's lifetime since l'Hôpital had essentially paid him for the right to use Bernoulli's results in any way he wanted. After l'Hôpital's death, however, Bernoulli began to take credit for practically the entire book.

Being a textbook, *Analyse des Infiniment Petits* contains a number of interesting problems that are used to demonstrate the techniques of the new calculus. One of these problems is the famous Pulley Problem, illustrated below.

**Figure 37.** L'Hôpital's Pulley Problem

A (weightless) cable of length \(r\) is attached to the ceiling at point \(A.\) The other end of the cable is attached to a pulley at point \(C\). The end of another cable of length \(l\) is attached to the ceiling at point \(B,\) where \(|AB|=d\) and \(d>r\). At the other end of the second cable (point \(D\)) is attached a weight. This second cable passes over the pulley, so \(|BC|+|CD|=l,\) and the whole system is allowed to adjust itself to reach a state of stationary equilibrium. The question is: How far below the ceiling is the final position of the weight? The solution relies on the **principle of minimum potential energy**. This means that the system will be stable when its potential energy is the least; that is, when the weight hangs as far below the ceiling as possible. Use the applet in Figure 38, below, to explore l'Hôpital's Pulley Problem and identify the approximate equilibrium position for the example shown.

**Figure 38. **Use this applet to explore l'Hôpital's Pulley Problem.

**Figure 39.** L'Hôpital's Pulley Problem for Exercise 10

In Figure 39, above, \(|AB|=d,\) \(|AC|=r,\) and \(|BC|+|CD|=l.\) Let \(x\) be the horizontal distance between \(G\) and \(B.\)

(a) Find an expression for \(|GA|.\)

(b) Find an expression for \(|GC|.\)

(c) Find an expression for \(|BC|.\)

(d) Find an expression for \(|CD|.\)

(e) Find an expression for \(|GD|.\)

(f) According to the principle of minimum potential energy, the system will reach equilibrium when \(|GD|\) is maximized. Use calculus and a CAS to find the value of \(x\) for which \(|GD|\) is a maximum.

(g) Compute \(x\) when \(d=12,\) \(r=7,\) and \(l=14.\)

**Figure 38 **(repeated from above)**.** Use this applet for Exercise 10, part h.

(h) The applet in Figure 38 illustrates the Pulley Problem. Drag \(C\) to locate the equilibrium position where \(|GD|\) is a maximum. Does \(x=|GB|\) agree with your solution above?

Johann Müller (1436-1476), commonly known today as Regiomontanus, was perhaps the most influential mathematician of the fifteenth century. Although his contributions were mainly in the area of trigonometry, we are interested here in the following problem that he posed in a letter he wrote in 1471:

At what point on the ground does a perpendicularly suspended rod appear largest?

In other words, which point \(P\) on line \({\mathcal L}\) in Figure 40 below maximizes the angle \(APB?\)

**Figure 40.** Regiomontanus' Problem: Which point \(P\) on \({\mathcal L}\) maximizes \(m\angle APB?\)

This problem was the first optimization problem that had appeared in the history of mathematics since the days of antiquity. It was probably inspired by questions of perspective that Renaissance artists of that time period were grappling with. It is often stated in modern calculus textbooks as follows:

A painting is hung flat against a wall. How far from the wall should one stand to maximize the viewing angle at his eye subtended by the painting?

**Figure 41.** Modern equivalent of Regiomontanus' Problem: How far from a wall should an observer stand to maximize his viewing angle \(APB\) (or \(\theta\)*)* of a picture of height \(|AB|\) on the wall? Use the applet to identify the point \(P\) that maximizes \(m\angle APB\) for the example shown. Drag \(P\) horizontally to adjust the distance of the man from the wall. Next, note that you may adjust as well the height of the man, the height of the picture above the floor, and the height of the picture itself, respectively, by dragging points \(P,\) \(A,\) and \(B\) vertically.

**Figure 42.** To solve the Hanging Picture Problem more generally, we introduce a coordinate system and assign variables as shown.

In the picture in Figure 42, above, the line \({\mathcal L}\) (from Figure 40) is drawn at the height of the viewer's eye. We can impose a coordinate system on the scenario, with line \({\mathcal L}\) as the \(x\)-axis, and \(A\) and \(B\) (from Figure 40) on the \(y\)-axis, say \(A=(0,a)\) and \(B=(0,b).\) Then the height of the painting is \(b-a\) and the bottom edge of the painting is \(a+h\) units above the floor, where \(h\) is the height of the viewer's eye above the floor.

The original solution to this problem, like Heron's solution to the shortest distance problem, was geometric and did not use calculus. Consider a circle that goes through points \(A\) and \(B\) and is tangent to the line \({\mathcal L}\) (the \(x\)-axis). Let \(P\) be the point of tangency. Then \(P\) is the point on \({\mathcal L}\) that maximizes angle \(APB.\)

**Figure 43.** Geometric solution to Regiomontanus' Problem: \(P\) should be such that line \({\mathcal L}\) is tangent to the circle through \(P,\) \(A,\) and \(B.\)

To see this, let \(P^{\prime}\) be another point on \({\mathcal L}.\) We claim that \(m\angle 5>m\angle 3.\) Let \(H\) be the point of intersection of the circle and \( P^{\prime}B.\) Then \(m\angle 2=m\angle 5\) since both angles subtend the same chord. Also, \(m\angle 1 +m\angle 2=m\angle 1+m\angle 3+m\angle 4,\) since both equal \(180\) degrees. Subtracting \(m\angle 1\) from both sides gives \(m\angle 2=m\angle 3+m\angle 4.\) Hence

\[m\angle 5=m\angle 2=m\angle 3+m\angle 4>m\angle 3,\]

as was to be shown.

Use calculus and a CAS to find the solution of Regiomontanus' Hanging Picture Problem. You need to find the value of \(x\) that maximizes \(\theta\) in the diagram below.

**Figure 44.** Regiomontanus' Hanging Picture Problem for Exercise 11

Here's one way to proceed: Since \(f(x)=\tan x\) is an increasing function, it is equivalent to maximize \[\tan\theta=\tan(\alpha-\beta)=\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}.\]

You should be able to rewrite this last expression all in terms of \(x,\) \(a,\) and \(b.\) Then use calculus to find the value of \(x\) that maximizes this quantity.

Suppose a six-foot tall adult comes to the museum with his three-foot tall child. They go into a room where a large 12-foot high painting is hung on the wall so that the bottom edge of the painting is 8 feet from the floor.

(a) How far from the painting should each of them stand in order to maximize their individual viewing angles?

(b) Compute each of their maximal viewing angles. Express your answer in degrees.

**Figure 45.** Applet for Exercise 12, part c. Drag \(I\) to adjust the height of the man. Drag \(A\) and \(B\) to adjust the height of the picture. Drag \(P\) to adjust the distance of the man from the wall.

(c) The applet in Figure 45 illustrates Regiomontanus' Hanging Picture Problem, with the line segment \(AB\) denoting the height of the picture, and the \(x\)-axis representing eye level. Check the box to show the circle through \(A\) and \(B\) tangent to the horizontal axis at eye level. Verify that the point of tangency is the solution to Regiomontanus' Problem by dragging the man's eye to that point and verifying that \(m\angle APB\) equals the maximal viewing angle you found above for the six-foot adult. Repeat for the three-foot child.

The last optimization problem that we discuss here is one of the most famous problems in the history of mathematics and was posed by the Swiss mathematician Johann Bernoulli in 1696 as a challenge “to the most acute mathematicians of the entire world.” The problem can be stated as follows:

Given two points on a plane at different heights, what is the shape of the wire down which a bead will slide (without friction) under the influence of gravity so as to pass from the upper point to the lower point in the shortest amount of time?

**Figure 46.** Brachistochrone Problem: Which path from \(A\) to \(B\) is traversed in the shortest time? (Click image to animate.)

This is the Brachistochrone (“Shortest Time”) Problem. This was a different kind of optimization problem, since instead of asking for the value of a variable, among all possible values, that would maximize or minimize something, it asked for the optimal function (or curve), among all possible curves. The differential calculus does not provide all the tools necessary to solve this problem. So for the moment, we would like to discuss Galileo Galilei's work relevant to this problem, which occurred in 1638, well before the Brachistochrone Problem had been explicitly stated.

Galileo studied motion under gravity, showing that a body falling in space traverses a distance proportional to the square of the time of descent. Using this law, he was able to compute the time of descent of an object falling along an inclined plane from point \(A\) to point \(B,\) assuming no friction.

**Figure 47.** Galileo's formula for time of descent along an inclined plane, as shown in the diagram:

The time to travel from \(A\) to \(B\) is \[\sqrt{\frac{2}{g}}\cdot\frac{d}{\sqrt{h}}\] where \(d\) is the distance between points \(A\) and \(B\) (in meters), \(h\) is the vertical distance between \(A\) and \(B\) (in meters), and \(g\) is the acceleration due to gravity, approximately \(9.80\) \(\mbox{m}/\mbox{sec}^2.\)

Clearly the shortest path from \(A\) to \(B\) is a straight line, but is this path the one that will take the shortest time? For example, is there some point \(C\) such that if the body were to fall following a straight line from \(A\) to \(C,\) and then a straight line from \(C\) to \(B,\) as shown in Figure 48, it would do so in less time than if it traversed a straight-line path from \(A\) to \(B?\)

**Figure 48.** Does the body fall from point \(A\) to point \(B\) in less time along the path \(A\rightarrow B\) or along the path \(A\rightarrow C\rightarrow B?\)

Galileo appeared to believe that the answer to the brachistochrone problem was an arc of a circle; that is, that the path \(A\rightarrow C\rightarrow B\) in Figure 48 should be replaced by a circle passing through points \(A\) and \(B.\) Was Galileo correct?

**Figure 49.** Use this GeoGebra applet to explore the Brachistochrone Problem by comparing the times of descent along different polygonal paths.

Use a CAS to define the following two functions:

\[T(a,b,c,d,u):=\frac{\sqrt{(a-c)^2 + (b-d)^2}\cdot\left(\sqrt{2g(b-d) + u^2}-u\right)}{g(b - d)}\]

\[V(a, b, c, d, u):=\sqrt{2g(b-d) + u^2}\]

where \(g\) has been predefined to be 9.8 (the acceleration due to gravity in meters per square seconds). The function \(T(a,b,c,d,u)\) computes the time of descent from \(A(a,b)\) to \(B(c,d)\) along a straight-line path, assuming no friction and initial velocity \(u.\) The function \(V(a,b,c,d,u)\) gives the terminal velocity. We assume distance is measured in meters and time in seconds.

(a) Compute \(T(0,0,5,-5,0)\) to find the time of descent from \(A(0,0)\) to \(B(5,-5)\) along an inclined plane, assuming no friction and an initial velocity of \(0.\)

(b) To compute the time of descent along the polygonal path \((0,0)\) to \((1,-2)\) to \((5,-5),\) you need to add the time it takes along \((0,0)\) to \((1,-2),\) which is \(T(0,0,1,-2,0),\) and the time it takes along \((1,-2)\) to \((5,-5),\) which is \(T(1,-2,5,-5,v_1)\) where \(v_1\) is the terminal velocity from the first part of the path, i.e. \(v_1=V(0,0,1,-2,0).\) Hence compute \[T(0,0,1,-2,0)+T(1,-2,5,-5,V(0,0,1,-2,0)).\]

(c) Suppose we want a polygonal path \((0,0)\) to \((x,y)\) to \((5,-5)\) where \((x,y)\) lies on the circle \((x-5)^2+y^2=25.\) Solving for \(y,\) where \(-5\leq y\leq 0,\) we obtain \(y=-\sqrt{25-(x-5)^2},\) so our point \((x,y)\) must be of the form \(\left(x,-\sqrt{25-(x-5)^2}\right)\) in order to lie on the quarter circle. The time of descent for this path then is

\[T\left(0,0,x,-\sqrt{25-(x-5)^2},0\right)+T\left(x,-\sqrt{25-(x-5)^2},5,-5,V\left(0,0,x,-\sqrt{25-(x-5)^2},0\right)\right)\]

Use a CAS and calculus to find the value of \(x\in(0,5)\) that minimizes the above quantity. Compare the time of descent along this polygonal path to the time along the straight line path.

(d) Let \(C\) be the point whose \(x\)-coordinate you found in part 3. Galileo knew that any polygonal path \((0,0)\) to \(C\) to \((5,-5),\) where \(C\) is on the circle, would take less time than the straight line path \((0,0)\) to \((5,-5).\) He then reasoned that if you added another point \(D\) on the circle to form the path \((0,0)\) to \(C\) to \(D\) to \((5,-5),\) the time of descent for this path would be less than the time for the path \((0,0)\) to \(C\) to \((5,-5).\) Find \(D=\left(x,-\sqrt{25-(x-5)^2}\right),\) so that the time of descent for the polygonal path \((0,0)\) to \(C\) to \(D\) to \((5,-5)\) is smallest. What is the total time for the path \((0,0)\) to \(C\) to \(D\) to \((5,-5)?\)

(a) Use calculus to find the value of \(x\) so that the time of descent along \((0,0)\) to \((x,-1)\) to \((5,-5)\) is the smallest. Let \(C=(x_1,-1)\) be the point that minimizes the time of descent. Also compute the terminal velocity \(v_1\) on the path \((0,0)\) to \(C.\)

(b) Use calculus to find the value of \(x\) so that the time of descent along \(C\) to \((x,-2)\) to \((5,-5)\) is the smallest, where the initial velocity at \(C\) is \(v_1.\) Let \(D=(x_2,-2)\) be the point that minimizes the time of descent. Also compute the terminal velocity \(v_2\) on the path \(C\) to \(D.\)

(c) Use calculus to find the value of \(x\) that minimizes the time of descent along \(D\) to \((x,-3)\) to \((5,-5)\), where the initial velocity at \(D\) is \(v_2.\) Let \(E=(x_3,-3)\) be the point that minimizes the time of descent. Also compute the terminal velocity \(v_3\) on the path \(D\) to \(E.\)

(d) Finally, use calculus to find the value of \(x\) that minimizes the time of descent along \(E\) to \((x,-4)\) to \((5,-5),\) where the initial velocity at \(E\) is \(v_3.\) Let \(F=(x_4,-4)\) be the point that minimizes the time of descent. Also compute the terminal velocity \(v_4\) on the path \(E\rightarrow F.\)

**Figure 50.** Diagram for Exercise 14, part e

(e) Compute the time of descent along the the path from \(A(0,0)\) to \(C\) to \(D\) to \(E\) to \(F\) to \(B(5,-5).\) How does this compare to the time of descent along the polygonal path \((0,0)\) to \(C\) to \(D\) to \((5,-5)\) that you computed in the previous exercise?

**Figure 51.** Applet for Exercise 14, part f

(f) The applet in Figure 51 contains the construction of an (upside-down) cycloid, a curve traced out by a point on the circumference of a circle as it rolls along a straight line. Drag \(C\) to rotate the circle. The curve goes through \(A(0,0),\) but it doesn't go through \(B(5,-5).\) You can find a cycloid that goes through \(B\) by adjusting the radius \(r\) of the rolling circle. What is the radius of the generating circle that produces a cycloid that goes through both \(A\) and \(B?\)

It turns out that this cycloid *is* the answer to the Brachistochrone Problem for any points \(A\) and \(B,\) not just \(A=(0,0)\) and \(B=(5,-5).\) Add the points on the polygonal path that you found above to the list of points in the applet. Your cycloid should be a pretty good fit.

The equation is \[\sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}=d.\] Solving for \(y\) we obtain \[y=\pm \frac{\sqrt{\left(-4 c^2+d^2\right) \left(d^2-4 x^2\right)}}{2 d}.\] The values of \(c\) and \(d\) for which the points \((0,\pm 5)\) and \((\pm 3,0)\) satisfy the equations(s) are \(c=4\) and \(d=10\). Substituting in these values and simplifying, the equations are
##### Exercise 2.

##### Exercise 3.

##### Exercise 4.

### Module 2: Tangent Lines Then and Now

##### Exercise 5.

##### Exercise 6.

##### Exercise 7.

### Module 3: Optimization Problems Then and Now

#### Heron's “Shortest Distance" Problem

##### Exercise 8.

#### Snell's Law and the Principle of Least Time

##### Exercise 9.

#### L'Hôpital's Pulley Problem and the Principle of Least Potential Energy

##### Exercise 10.

#### Regiomontanus' Hanging Picture Problem

##### Exercise 11.

##### Exercise 12.

#### Galileo and the Brachistochrone Problem

##### Exercise 13.

##### Exercise 14.

\[y=\pm\frac{3}{10}\sqrt{100-4x^2}.\]

We must have \(a+b=5\) and \(a-b=3\), so \(a=4\) and \(b=1\).

We must have \(\frac{a}{2}+b=5\) and \(\frac{a}{2}-b=3,\) so \(a=8\) and \(b=1\).

We must have \(a+b=5\) and \(a=3,\) so \(b=2\).

(a) Since the point \((4,5)\) is 4 units above the vertex, the \(y\)-intercept of the tangent line must be 4 units below the vertex, at \((0,-3)\).

(b) If the \(x\)-intercept \(A\) of the tangent line has coordinates \((x,0)\), then Apollonius' equation \[\frac{|AH|}{|AG|}=\frac{|BH|}{|BG|}\] becomes

\[\frac{x-5}{x+5}=\frac{2}{8}\]

Solving this for \(x\) we get \(x=\frac{25}{3}\) so the \(x\)-intercept is \((25/3,0)\).

(c) If the \(x\)-intercept \(A\) of the tangent line has coordinates \((x,0)\), then Apollonius' equation \[\frac{|AH|}{|AG|}=\frac{|BH|}{|BG|}\] becomes

\[\frac{4-x}{x+4}=\frac{1}{9}\]

Solving this for \(x\) we get \(x=3.2\) so the \(x\) intercept is \((3.2,0)\).

(a) For the parabola, the tangent line is the angle bisector of \(\measuredangle FPB\):

(b) For the ellipse, the tangent is the angle bisector of \(\measuredangle {F_1^{\prime}}PF_2\) where \({F_1^{\prime}}\) is obtained by rotating \(F_1\) about \(P\) 180 degrees.

(c) For the hyperbola, the tangent line is the angle bisector of \(\measuredangle F_1PF_2\).

(d) There are two ways to approach the tangent line to the cycloid. You can think of \(P\) as being pulled by two forces, one parallel to \(EG\) as the position of the circle moves forward, and another perpendicular to the radius \(PC\) as \(P\) rotates about \(C\). So we construct a line through \(P\) parallel to \(EG\), and then another line through \(P\) perpendicular to \(PC\). The angle bisector of the angle between these two lines will then be the tangent.

Alternatively \(P\) is being rotated about \(A\), so we can just construct the perpendicular of \(AP\).

(e) For the epicycloid, there are two forces pulling on \(P\), one perpendicular to \(CA\), and one perpendicular to \(AB\). The tangent line is the angle bisector of the angle between these two lines.

(a) The center of the circle at which \(A\) and \(B\) merge is \((3,0)\). The slope of the radius of \(-1\), so the slope of the tangent line, which is perpendicular, is 1. Hence the equation of the tangent line is \(y-2=1(x-1)\) or \(y=x+1\).

(b) The center of the circle at which \(A\) and \(B\) merge is \((11,0)\). The slope of the radius of \(-3\), so the slope of the tangent line, which is perpendicular, is \({\frac13}\). Hence the equation of the tangent line is \(y-6={\frac13}(x-9)\) or \(y=\frac13x+3\).

If \(C=(x,0)\), then we want to minimize

\[f(x)=|AC|+|CB|=\sqrt{(0-x)^2+(4-0)^2}+\sqrt{(10-x)^2+(12-0)^2}\] \[=\sqrt{x^2 - 20x + 244} + \sqrt{x^2 + 16}.\]

We use a CAS to compute the derivative:

\[f^{\prime}(x)=\frac{x-10}{\sqrt{144+(x-10)^2}}+\frac{x}{\sqrt{16+x^2}}\]

Then \(f^{\prime}(x)=0\Rightarrow x=\frac{5}{2}\). So the point \(C\) that minimizes \(|AC|+|CB|\) is \(C=\left(\frac52,0\right)\).

(a) If \(C=(x,0)\), then we want to minimize

\[f(x)=\frac{|AC|}{v_1}+\frac{|BC|}{v_2}=\frac{\sqrt{(x-1)^2+9}}{4}+\frac{\sqrt{(x-4)^2+4}}{5}\]

We use a CAS to compute the derivative:

\[f^{\prime}(x)=\frac{x-1}{4\sqrt{9+(x-1)^2}}+\frac{x-4}{5\sqrt{4+(x-4)^2}}\]

Then \(f^{\prime}(x)=0\Rightarrow x=2.57363\).

(b) We need to find the value of \(x\) that minimizes

\[f(x)=\frac{\sqrt{(x+3)^2+16}}{1}+\frac{\sqrt{(x-2)^2+25}}{2}\]

Taking a derivative, setting it equal to 0, and solving, we find the solution \(x=-1.743575120\).

(c) Let \(C=(x,-x^2)\). We need to find the value of \(x\) that minimizes \[f(x)=|AC|+|BC|=\sqrt{(0-x)^2+(4+x^2)^2}+\sqrt{(5-x)^2+(-3+x^2)^2}.\] Using a CAS to compute the derivative and solve \(f^{\prime}(x)=0\) numerically, we obtain the solution \(x=0.726582\).

(a) \(|GA|=d-x\).

(b) \(|GC|=\sqrt{r^2-(d-x)^2}\).

(c) \(|BC|=\sqrt{x^2+r^2-(d-x)^2}\).

(d) \(|CD|=l-\sqrt{x^2+r^2-(d-x)^2}\).

(e) \(|GD|=\sqrt{r^2-(d-x)^2}+l-\sqrt{x^2+r^2-(d-x)^2}\).

(f) Using a CAS to differentiate the expression with respect to \(x\) and setting it equal to 0, we obtain the solutions

\[x=\frac{4d^2 - r^2}{4d} \pm \frac{|r|\sqrt{8d^2 + r^2}}{4d}.\]

(g) Plugging in \(d=12\), \(r=7\), and \(l=14\) above we obtain \(x=5.925247326\).

(h) Yes, the maximum value of \(|GD|\) is \(10.60747,\) and it occurs when \(x=|GB|=5.925247,\) which agrees (to four decimal places) with the answer obtained above.

We need to find the value of \(x\) that maximizes \[\tan\theta=\frac{\frac{b}{x}-\frac{a}{x}}{1+\frac{b}{x}\cdot\frac{a}{x}}.\] Using a CAS to take a derivative and set it equal to zero, we obtain the solution \(x=\sqrt{ab}.\)

(a) The six-foot tall adult's eyes are two feet below the bottom of the painting. So \(a=2\) and \(b-a=12\). Hence \(b=14\) and so he should stand \(\sqrt{2\cdot 14}\approx5.291502622\) feet away from the painting.

The three-foot tall child's eyes are five feet below the bottom of the painting. So \(a=5\) and \(b-a=12\). Hence \(b=17\) and so he should stand \(\sqrt{5\cdot 17}\approx9.219544457\) feet away from the painting.

(b) Plugging \(a=2\), \(b=14\), and \(x=\sqrt{28}\) into the expression for \(\tan\theta\) we get \(\tan\theta=1.133893419,\) so \(\theta=0.8480620789\) radians. To convert to degrees we multiply by \(\frac{180\mbox{ deg}}{\pi\mbox{ radians}}\). Thus \[\theta=\frac{0.8480620789\cdot 180}{\pi}\approx 48.59037788\mbox{ degrees}.\]

Plugging \(a=5\), \(b=17\), and \(x=\sqrt{85}\) into the expression for \(\tan\theta\) we get \(\tan\theta=0.979067\), so \(\theta=0.774821\) radians. To convert to degrees we multiply by \(\frac{180\mbox{ deg}}{\pi\mbox{ radians}}\). Thus \[\theta=\frac{0.774821\cdot 180}{\pi}\approx 44.394\mbox{ degrees}.\]

c) For the 6-foot tall adult:

For the 3-foot tall child:

(a) \(T(0,0,5,-5,0)=1.428571428\) seconds.

(b) \(T(0,0,1,-2,0)+T(1,-2,5,-5,V(0,0,1,-2,0))=1.333079013\) seconds.

(c) Taking a derivative, setting it equal to \(0\), and solving we find that \(x = 0.4491013975\). The \(y\)-value is \(-\sqrt{25-(x-5)^2}=-2.071067818\). So \(C=(0.4491013975,-2.071067818)\).

The time of descent along this path is

\[T(0,0,0.4491013975,-2.071067818,0)\] \[+\,T(0.4491013975,-2.071067818,5,-5,V(0,0,0.4491013975,-2.071067818,0))\]

\[=1.330475856{\mbox{ seconds}}.\]

This is less than \(1.428571428\), the time along the path \((0,0)\to (5,-5)\).

(d) Let \[v_1=V(0,0,0.4491013975,-2.071067818,0)=6.371258058{\mbox{ m/sec}},\] the terminal velocity along \((0,0)\to(0.4491013975,-2.071067818)\). We need to find \(x\) that minimizes

\[T(0.4491013975,-2.0710678181,x,-\sqrt{25-(x-5)^2},v_1)\]

\[+\,T(x,-\sqrt{25-(x-5)^2},5,-5,V(0.4491013975,-2.0710678181,x,-\sqrt{25-(x-5)^2},v_1)).\]

Taking a derivative and setting it equal to 0, we find \(x=2.043116864\). The corresponding \(y\)-coordinate is \(-\sqrt{25-(x-5)^2}=-4.031977445\), so \(D=(2.043116864,-4.031977445)\).

The terminal velocity as the object reaches point \(D\) is \[v_2=V(0.4491013975,-2.071067818,2.043116864,-4.031977445,6.371258047)=8.889699541{\mbox{ m/sec}}.\]

The total time along the path \((0,0)\to C\to D\to (5,-5)\) is \[T(0,0,0.4491013975,-2.0710678181,0)\]

\[+\,T(0.4491013975,-2.0710678181,2.043116864,-4.031977445,6.371258058)\]

\[+\,T(2.043116864,-4.031977445,5,-5,8.889699541)\] \[=1.327598538{\mbox{ seconds}}.\]

(a) We need to find the value of \(x\) that minimizes \[T(0,0,x,-1,0)+T(x,-1,5,-5,V(0,0,x,-1,0)).\] Taking the derivative and setting it equal to zero, the solution is \(x=0.2434120319\). So \(C=(0.2434120319,-1)\).

The terminal velocity as the object reaches point \(C\) is \[V(0,0,0.2434120319,-1,0)=4.427188724.\]

(b) We need to find the value of \(x\) that minimizes \[T(0.2434120319,-1,x,-2,4.427188724)+T(x,-2,5,-5,V(0.2434120319,-1,x,-2,4.427188724).\] Taking the derivative and setting it equal to zero, the solution is \(x=0.876358533\). So \(D=(0.876358533,-2)\).

The terminal velocity as the object reaches point \(D\) is \[V(0.2434120319,-1,0.876358533,-2,4.427188724)=6.260990337.\]

(c) We need to find the value of \(x\) that minimizes \[T(0.876358533,-2,x,-3,6.260990337)+T(x,-3,5,-5,V(0.876358533,-2,x,-3,6.260990337)).\] Taking the derivative and setting it equal to zero, the solution is \(x=1.786631272\). So \(E=(1.786631272,-3)\).

The terminal velocity as the object reaches point \(E\) is \[V(0.876358533,-2,-3,6.260990337)=7.668115805.\]

(d) We need to find the value of \(x\) that minimizes \[T(1.786631272,-3,x,-4,7.668115805)+T(x,-4,5,-5,V(1.786631272,-3,x,-4,7.668115805)).\] Taking the derivative and setting it equal to zero, the solution is \(x=3.049531298\). So \(F=(3.049531298,-4)\).

The terminal velocity as the object reaches point \(F\) is \[V(1.786631272,-3,3.049531298,-4,7.668115805)=8.854377448.\]

(e) The time of descent is

\[T(0, 0, 0.2434120319,-1, 0) + T(0.2434120319,-1, 0.876358533,-2, 4.427188724)\]

\[+\,T(0.876358533,-2, 1.786631272,-3, 7.668115805) + T(1.786631272,-3, 3.049531298,-4, 8.854377448)\]

\[+\,T(3.049531298,-4, 5, -5, 8.854377448)\] \[=1.309306399\mbox{ seconds}.\]

This is less than \(1.327598538,\) the time along the polygonal path from the previous exercise.

(f) \(r\approx 2.86\)